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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Choose the most suitable answer from the given four alternatives

Question 1.
If \(\overline { a }\) and \(\overline { b }\) are parallel vectors, then [\(\overline { a }\), \(\overline { c }\), \(\overline { b }\)] is equal to
(a) 2
(b) -1
(c) 1
(d) 0
Solution:
(d) 0
Hint:
Since \(\overline { a }\) and \(\overline { b }\) are parallel ⇒ \(\overline { a }\) = λ\(\overline { b }\)
[ \(\overline { a }, \overline { c }, \overline { b }\)] = [λ\(\overline { b }, \overline { c }, \overline { b }\) ]
= λ[ \(\overline { b }, \overline { c }, \overline { b }\) ]
= λ(0) = 0

Question 2.
If a vector \(\overline { α }\) lies in the plane of \(\overline { ß }\) and \(\overline { γ }\), then
(a) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 1
(b) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = -1
(c) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0
(d) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 2
Solution:
(c) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0
Hint:
If \(\overline { α }\) lies in \(\overline { ß }\) & \(\overline { γ }\) plane
we have [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0

Question 3.
If \(\overline { a }\).\(\overline { b }\) = \(\overline { b }\).\(\overline { c }\) = \(\overline { c }\).\(\overline { a }\) = 0, then the value of [ \(\overline { a }, \overline { b }, \overline { c }\) ] is
(a) |\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
(b) \(\frac { 1 }{ 3 }\)|\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
(c) 1
(d) -1
Solution:
(a) |\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
Hint:

Question 4.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three unit vectors such that \(\overline { a }\) is perpendicular to \(\overline { b }\) and is parallel to \(\overline { c }\) then \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) is equal to
(a) \(\overline { a }\)
(b) \(\overline { b }\)
(c) \(\overline { c }\)
(d) \(\overline { 0 }\)
Solution:
(b) \(\overline { b }\)
Hint:

Question 5.
If [ \(\overline { a }, \overline { b }, \overline { c }\) ] = 1 then the value of

(a) 1
(b) -1
(c) 2
(d) 3
Solution:
(a) 1
Hint:

Question 6.
The volume of the parallelepiped with its edges represented by the vectors \(\hat { i }\) + \(\hat { j }\), \(\hat { i }\) + 2\(\hat { j }\), \(\hat { i }\) + \(\hat { j }\) + π\(\hat { k }\) is
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { π }{ 3 }\)
(c) π
(d) \(\frac { π }{ 4 }\)
Solution:
(c) π
Hint:
\(\left|\begin{array}{lll}
1 & 1 & 0 \\
1 & 2 & 0 \\
1 & 1 & \pi
\end{array}\right|\) = π\(\left|\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right|\)
= π (2 – 1) = π

Question 7.
If \(\overline { a }\) and \(\overline { b }\) are unit vectors such that [\(\overline { a }\), \(\overline { b }\), \(\overline { a }\) × \(\overline { b }\)] = \(\frac { 1 }{ 4 }\), then the angle between \(\overline { a }\) and \(\overline { b }\) is
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 4 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { π }{ 2 }\)
Solution:
(a) \(\frac { π }{ 6 }\)
Hint:

Question 8.
If \(\overline { a }\) = \(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + \(\hat { j }\), \(\overline { c }\) = \(\hat { i }\) and (\(\overline { a }\) × \(\overline { b }\))\(\overline { c }\) – λ\(\overline { a }\) + µ\(\overline { b }\) then the value of λ + µ is
(a) 0
(b) 1
(c) 6
(d) 3
Solution:
(a) 0
Hint:
\(\overline { a }\).\(\overline { c }\) = 1 and \(\overline { b }\).\(\overline { c }\) = 1
(\(\overline { a }\) × \(\overline { b }\))\(\overline { c }\) = (\(\overline { c }\) × \(\overline { a }\))\(\overline { b }\) – (\(\overline { c }\) × \(\overline { b }\))\(\overline { a }\) = λ\(\overline { a }\) + µ\(\overline { b }\)
⇒ µ = c; a = 1λ = -(\(\overline { c }\).\(\overline { b }\)) = -1
µ + λ = 1 – 1 = 0

Question 9.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are non-coplanar, non-zero vectors
such that [\(\overline { a }\), \(\overline { b }\), \(\overline { c }\)] = 3, then {[\(\overline { a }\) × \(\overline { b }\), \(\overline { b }\) × \(\overline { c }\), \(\overline { c }\) × \(\overline { a }\)]²} is equal to
is equal to
(a) 81
(b) 9
(c) 27
(d) 18
Solution:
(a) 81
Hint:

= 3⁴ = 81

Question 10.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three non-coplanar vectors such that \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) = \(\frac { \overline{b}+\overline{c} }{ √2 }\) then the angle between \(\overline { a }\) and \(\overline { b }\) is
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { 3π }{ 4 }\)
(c) \(\frac { π }{ 4 }\)
(d) π
Solution:
(b) \(\frac { 3π }{ 4 }\)
Hint:

Question 11.
If the volume of the parallelepiped with \(\overline { a }\) × \(\overline { b }\), \(\overline { b }\) × \(\overline { c }\), \(\overline { c }\) × \(\overline { a }\) as coterminous edges is 8 cubic units, then the volume of the parallelepiped with (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { b }\) × \(\overline { c }\)), (\(\overline { b }\) × \(\overline { c }\)) × (\(\overline { c }\) × \(\overline { a }\)) and (\(\overline { c }\) × \(\overline { a }\)) × (\(\overline { a }\) × \(\overline { b }\)) as coterminous edges is
(a) 8 cubic units
(b) 512 cubic units
(c) 64 cubic units
(d) 24 cubic units
Solution:
(c) 64 cubic units
Hint:
Given volume of the parallelepiped with

Question 12.
Consider the vectors \(\overline { a }\), \(\overline { b }\), \(\overline { c }\), \(\overline { d }\) such that (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { c }\) × \(\overline { d }\)) = \(\overline { 0 }\) Let P₁ and P₂ be the planes determined by the pairs of vectors \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\), \(\overline { d }\) respectively. Then the angle between P₁ and P₂ is
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Solution:
(a) 0°
Hint:
A vector perpendicular to the plane P₁ of a, b is \(\overline { a }\) × \(\overline { b }\),
A vector perpendicular to the plane P₂ of c and d is \(\overline { c }\) × \(\overline { d }\)
∴ (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { c }\) × \(\overline { d }\)) = 0
⇒ (\(\overline { a }\) × \(\overline { b }\)) || \(\overline { c }\) × \(\overline { d }\)
⇒ The angle between the planes is \(\overline { 0 }\)

Question 13.
If \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) = (\(\overline { a }\) × \(\overline { b }\)) × \(\overline { c }\) where \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are any three vectors such that \(\overline { b }\).\(\overline { c }\) ≠ 0 and \(\overline { a }\).\(\overline { b }\) ≠ 0, then \(\overline { a }\) and \(\overline { c }\) are
(a) perpendicular
(b) parallel
(c) inclined at angle \(\frac { π }{ 3 }\)
(d) inclined at an angle \(\frac { π }{ 6 }\)
Solution:
(b) parallel
Hint:

Question 14.
If \(\overline { a }\) = 2\(\hat { i }\) + 3\(\hat { j }\) – \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + 2\(\hat { j }\) – 5\(\hat { k }\), \(\overline { c }\) = 3\(\hat { i }\) + 5\(\hat { j }\) – \(\hat { k }\) then \(\overline { a }\) vector perpendicular to a and lies in the plane containing \(\overline { b }\) and \(\overline { c }\) is
(a) -17\(\hat { i }\) + 21\(\hat { j }\) – 97\(\hat { k }\)
(b) 17\(\hat { i }\) + 21\(\hat { j }\) – 123\(\hat { k }\)
(c) -17\(\hat { i }\) – 21\(\hat { j }\) + 97\(\hat { k }\)
(d) -17\(\hat { i }\) – 21\(\hat { j }\) – 97\(\hat { k }\)
Solution:
(d) -17\(\hat { i }\) – 21\(\hat { j }\) – 97\(\hat { k }\)
Hint:
A vector ⊥r to \(\overline { a }\) and lies in the plane containing \(\overline { b }\) and \(\overline { c }\)

Question 15.
The angle between the lines \(\frac { x-2 }{ 3 }\) = \(\frac { y+1 }{ -2 }\), z = 2 and \(\frac { x-1 }{ 1 }\) = \(\frac { 2y+3 }{ 3 }\) = \(\frac { z+5 }{ 2 }\) is
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 4 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { π }{ 2 }\)
Solution:
(d) \(\frac { π }{ 2 }\)
Hint:

Question 16.
If the line \(\frac { x-2 }{ 3 }\) = \(\frac { y-1 }{ -5 }\) = \(\frac { z+2 }{ 2 }\) lies in the plane x + 3y – αz + ß = 0 then (α + ß) is
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Solution:
(b) (-6, 7)
Hint:
\(\frac { x-2 }{ 3 }\) = \(\frac { y-1 }{ 5 }\) = \(\frac { z+2 }{ 2 }\) = λ ⇒ (3λ + 2, -5λ + 1, 2λ – 2)
which lie in x + 3y – αz + ß = 0
(3λ + 2) + 3(-5λ + 1) – α(2λ – 2) + ß = 0
3λ + 2 – 15λ + 3 – 2αλ + 2α + ß = 0.
(-12λ – 2αλ) + 2α + ß + 5 = 0.
-12λ – 2αλ = 0
2αλ = -12λ
α = -6
2α+ ß +5 = 0
-12 + ß + 5 = 0
ß – 7 = 0
ß = 7
(α, ß) = (-6, 7)

Question 17.
The angle between the line \(\overline { r }\) = (\(\hat { i }\) + 2\(\hat { j }\) – 3\(\hat { k }\)) + t(2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\)) and the plane \(\overline { r }\) (\(\hat { i }\) + \(\hat { j }\)) + 4 = 0 is
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Solution:
(c) 45°
Hint:

Question 18.
The co-ordinates of the point where the line \(\overline { r }\) = (6(\(\hat { i }\) – \(\hat { j }\) – 3\(\hat { k }\)) + t(-\(\hat { i }\) + \(\hat { k }\) meets the plane \(\overline { r }\) ((\(\hat { i }\) + (\(\hat { j }\) – (\(\hat { k }\)) = 3 are
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Solution:
(d) (5, -1, 1)
Hint:
Given \(\overline { r }\) = (6(\(\hat { i }\) – (\(\hat { j }\) – 3(\(\hat { k }\)) + t(-(\(\hat { i }\) + (\(\hat { k }\))
\(\frac { x-6 }{ -1 }\) = \(\frac { y+1 }{ 0 }\) = \(\frac { z+3 }{ 4 }\) = t ⇒ (-t + 6, -1, 4t – 3)
which meets x + y – z = 3
-t + 6 – 1 – 4t + 3 = 3
-5t + 5 = 0
5t = 5
t = 1
∴ Co-ordinate is (5, -1, 1)

Question 19.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:
(x₁, y₁, z₁) = (o, 0, o)
(a, b, c) = (3, -6, 2); d = 7.
d = \(\frac { ax_1+by_1+cz_1+d }{ \sqrt{a^2+b^2+c^2} }\) = \(\frac { 7 }{ \sqrt{9+36+4} }\) = \(\frac { 7 }{ 7 }\) = 1

Question 20.
The distance between the planes
x + 2y + 3z + 7 = 0 and 2x + 4y + 6z + 7 = 0 is
(a) \(\frac { √7 }{ 2√2 }\)
(b) \(\frac { 7 }{ 2 }\)
(c) \(\frac { √7 }{ 2 }\)
(d) \(\frac { 7 }{ 2√2 }\)
Solution:
(a) \(\frac { √7 }{ 2√2 }\)
Hint:
x + 2y + 3z+7 = 0
2x + 4y + 6z + 7 = 0
(÷ 2) x + 2y + 3z + \(\frac { 7 }{ 2 }\) = 0
(1) and (2) are parallel planes

Question 21.
If the direction cosines of a line are \(\frac { 1 }{ c }\), \(\frac { 1 }{ c }\), \(\frac { 1 }{ c }\)
(a) c = ±3
(b) c = ±√3
(c) c > 0
(d) 0 < c < 1
Solution:
(b) c = ±√3
Hint:
cos²α + cos²ß + cos²γ = 1
\(\frac { 1 }{ c^2 }\) + \(\frac { 1 }{ c^2 }\) + \(\frac { 1 }{ c^2 }\) = 1
\(\frac { 3 }{ c ^2}\) = 1
c² = 3
c = ±√3

Question 22.
The vector equation \(\overline { r }\) = (\(\hat { i }\) – 2\(\hat { j }\) – \(\hat { k }\)) + t(6\(\hat { i }\) – \(\hat { k }\)) represents a straight line passing through the points
(a) (0, 6, -1) and (1, -2, -1)
(b) (0, 6, -1) and (-1, -4, -2)
(c) (1, -2, -1) and (1, 4, -2)
(d) (1, -2, -1) and (0, -6, 1)
Solution:
(c) (1, -2, -1) and (1, 4, -2)
Hint:
Given vector equation is

Question 23.
If the distance of the point (1, 1, 1) from the origin is half of its distance from the plane x + y + z + k = Q, then the values of k are
(a) ±3
(b) ±6
(c) -3, 9
(d) 3, -9
Solution:
(d) 3, -9
Hint:

Question 24.
If the planes \(\overline { r }\) (2\(\hat { i }\) – λ\(\hat { j }\) + \(\hat { k }\)) = 3 and \(\overline { r }\) (4\(\hat { i }\) + \(\hat { j }\) – µ\(\hat { k }\)) = 5 are parallel, then the value of λ and µ are
(a) \(\frac { 1 }{ 2 }\), -2
(b) –\(\frac { 1 }{ 2 }\), 2
(c) –\(\frac { 1 }{ 2 }\), -2
(d) \(\frac { 1 }{ 2 }\), 2
Solution:
(c) –\(\frac { 1 }{ 2 }\), -2
Hint:

Question 25.
If the length of the perpendicular from the origin to the plane 2x + 3y + λz = 1, λ > 0 is \(\frac { 1 }{ 5 }\), then the value of λ is
(a) 2√3
(b) 3√2
(c) 0
(d) 1
Solution:
(a) 2√3
Hint:

5 = \(\sqrt { 4+9+λ^2 }\)
25 = 4 + 9 + λ²
25 = 13 + λ²
λ² = 12
λ = 2√3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 1.
Find the equation of the plane passing through the line of intersection of the planes \(\overline { r }\) = (2\(\hat { i }\) – 7\(\hat { j }\) + 4\(\hat { k }\)) = 3 and 3x – 5y + 4z + 11 = 0 and the point (- 2, 1, 3).
Solution:
Given planes are
\(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) = 3
2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0
Equation of a plane which passes through the line of intersection of the planes
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0 …………… (1)
This passes through the point (-2, 1, 3).
(1) ⇒ (-4 – 7 + 12 – 3) + λ(-6 – 5 + 12 + 11) = 0
-2 + λ(12) = 0 ⇒ 12λ = 2
λ = \(\frac{2}{12}\) ⇒ λ = \(\frac{1}{6}\)
The required equation is
(1) ⇒ (2x – 7y + 4z – 3) + \(\frac{1}{6}\) (3x – 5y + 4z + 11) = 0
12x – 42y + 24z – 18 + 3x – 5y + 4z + 11 = 0
15x – 47y + 28z – 7 = 0

Question 2.
Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance \(\frac { 2 }{ √3 }\) from the point (3, 1, -1).
Solution:
Required equation of the plane
(x + 2y + 3z – 2) + λ(x – y + z – 3) = 0 ………. (1)
(1 + λ)x + (2 – λ)y + (3 + λ)z + (-2 – 3λ) = 0 ………(2)
Distance from (2) to the point (3, 1, -1) is \(\frac { 2 }{ √3 }\)

putting
λ = \(\frac { -7 }{ 2 }\) in (1)
The required equation
(x + 2y + 3z – 2) – \(\frac { 7 }{ 2 }\) (x – y + z – 3) = 0
2x + 4y + 6z – 4 – 7x + 7y – 7z + 21 = 0
-5x + 11y – z + 17 = 0
5x – 11y + z – 17 = 0

Question 3.
Find the angle between the line
\(\overline { r }\) = (2\(\hat { i }\) – \(\hat { j }\) + \(\hat { k }\)) + (\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\)) and the plane \(\overline { r }\) (6\(\hat { i }\) + 3\(\hat { j }\) + 2\(\hat { k }\)) = 8
Solution:

Question 4.
Find the angle between the planes \(\overline { r }\) (\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\)) = 3 and 2x – 2y + z = 2.
Solution:

Question 5.
Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also, find the distance between the two planes.
Solution:
The required equation parallel to the plane
2x – 3y + 5z + 7 = 0 ………….. (1)
2x – 3y + 5z + λ = 0 ………….. (2)
This passes through (3, 4, -1)
(2) ⇒ 2(3) – 3(4) + 5(-1) + λ = 0
6 – 12 – 5 + 1 = 0
λ = 11
(2) ⇒ The required equation is 2x – 3y + 5z + 11 =0 …………… (3)
∴ Now, distance between the above parallel lines (1) and (3)
a = 2, b = -3, c = 5, d₁ = 7, d₂ = 11

Question 6.
Find the length of the perpendicular from the point (1, -2, 3) to the plane x – y + z = 5.
Solution:

Question 7.
Find the point of intersection of the line with the plane (x – 1) = \(\frac { y }{ 2 }\) = z + 1 with the plane 2x – y – 2z = 2. Also, the angle between the line and the plane.
Solution:
Any point on the line x – 1 = \(\frac{y}{2}\) = z + 1 is
x – 1 = \(\frac{y}{2}\) = z + 1 = λ,(say)
(λ + 1, 2λ, λ – 1)
This passes through the plane 2x – y + 2z = 2
2(λ + 1) – 2λ + 2(λ – 1) = 2
2λ + 2 – 2λ + 2λ – 2 = 2
λ = 1
∴ The required point of intersection is (2, 2, 0)

Question 8.
Find the co-ordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2.
Solution:
Let us take the point P(4, 3, 2) and Q(x₁, y₁, z₁)
⇒ (x₁ – 4, y₁ – 3, z₁ – 2)
Plane x + 2y + 3z = 2 ………. (1)
Cartesian equation of PQ, \(\frac { x_1-4 }{ 1 }\) = \(\frac { y_1-3 }{ 2 }\) = \(\frac { z_3-2 }{ 3 }\) = λ
(λ + 4, 2λ + 3, 3λ + 2) lies in (1)
(λ + 4) + 2(2λ + 3) + 3(3λ + 2) – 2 = 0
λ + 4 + 4λ + 6 + 9λ + 6 – 2 = 0
14λ + 14 = 0
14λ = -14 .
λ = -1
Co-ordinates of the foot of the ⊥r is (3, 1, -1).
Distance PQ = \(\sqrt{(4-3)^2 + (3-1)^2 + (2+1)^2}\)
= \(\sqrt{1^2 + 2^2 + 3^2}\) = \(\sqrt{1 + 4 + 9}\)
= \(\sqrt{14}\) units.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8

Question 1.
Show that the straight lines
\(\overline { r }\) = (5\(\hat { i }\) + 7\(\hat { j }\) – 3\(\hat { k }\)) + s(4\(\hat { i }\) + 4\(\hat { j }\) – 5\(\hat { k }\)) and
\(\overline { r }\) = (8\(\hat { i }\) + 4\(\hat { j }\) + 5\(\hat { k }\)) + t(7\(\hat { i }\) + \(\hat { j }\) + 3\(\hat { k }\)) are coplanar. Find the vector equation of the, plane in which they lie.
Solution:

Question 2.
Show that the lines \(\frac { x-2 }{ 1 }\) = \(\frac { y-3 }{ 1 }\) = \(\frac { z -4}{ 3 }\) and \(\frac { x-1 }{ -3 }\) = \(\frac { y-4 }{ 2 }\) = \(\frac { z-5 }{ 1 }\) are coplanar. Also, find the plane containing these lines.
Solution:

Cartesian equation
x + 2y – 2z = 4
x + 2y – 2z – 4 = 0

Question 3.
If the straight lines \(\frac { x-1 }{ 1 }\) = \(\frac { y-2 }{ 2 }\) = \(\frac { z-3}{ m^2 }\) and \(\frac { x-3 }{ 1 }\) = \(\frac { y-2 }{ m^2 }\) = \(\frac { z-1 }{ 2 }\) are coplanar, find the distinct real values of m
Solution:

2(4 – m⁴) – 2(m² – 2) = 0
8 – 2m⁴ – 2m² + 4 = 0
12 – 2m⁴ – 2m² = 0
(÷ -2) -6 + m⁴ + m² = 0
m⁴ + m² – 6 = 0
(m² – 2)(m² + 3) = 0
m² – 2 = 2; m² = -3 (not possible)
m² = 2
m = ±√2

Question 4.
If the straight lines \(\frac { x-1 }{ 2 }\) = \(\frac { y+1 }{ λ }\) = \(\frac { z }{ 2 }\) and \(\frac { x+1 }{ 5 }\) = \(\frac { y+1 }{ 2 }\) = \(\frac { z }{ λ }\) are coplanar, find λ and equations of the planes containing these two lines.
Solution:
If the two lines are coplanar

When λ = 2
(x₁, y₁, z₁) = (1, -1, 0)
(b₁, b₂, b₃) = (2, 2, 2)
(d₁, d₂, d₃) = (5, 2, 2)
\(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
b_{1} & b_{2} & b_{3} \\
d_{1} & d_{2} & d_{3}
\end{array}\right|\) = 0
⇒ \(\left|\begin{array}{ccc}
x-1 & y+1 & z-0 \\
2 & 2 & 2 \\
5 & 2 & 2
\end{array}\right|\) = 0
⇒ (x – 1)(0) – (y + 1)(-6) + z(6) = 0
⇒6(y + 1) – 6z = 0
⇒ 6y + 6 – 6z = 0
⇒ y – z + 1 = 0
When λ = 2
(b₁, b₂, b₃) = (2, -2, 2)
(d₁, d₂, d₃) = (5, 2, -2)
⇒ \(\left|\begin{array}{ccc}
x-1 & y+1 & z-0 \\
2 & -2 & 2 \\
5 & 2 & -2
\end{array}\right|\) = 0
⇒ (x – 1)(0) – (y + 1)(-14) + z(4 + 10) = 0
⇒ 14(y + 1) + 14z = 0
⇒ 14y + 14 + 14z = 0
⇒ y + z + 1 = 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.7

Question 1.
Find the non-parametric form of vector equation and Cartesian equation of the plane passing through the point (2, 3, 6) and parallel to the .straight lines.
\(\frac { x-1 }{ 2 }\) = \(\frac { y+1 }{ 3 }\) = \(\frac { z-3 }{ 1 }\) and \(\frac { x+3 }{ 2 }\) = \(\frac { y-3 }{ -5 }\) = \(\frac { z+1 }{ -3 }\)
Solution:

Question 2.
Find the non-parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 9.
Solution:

Cartesian equation
3x + 4y – 5z = 9
3x + 4y – 5z – 9 = 0

Question 3.
Find parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (1, -2, 3) and parallel to the straight line passing through the points (2, 1, -3) and (-1, 5, -8)
Solution:

Cartesian equation
-12x + 11y + 16z = 14
12x – 11y – 16z = -14
12x – 11y – 16z + 14 = 0

Question 4.
Find the non-parametric form of vector equation and Cartesian equation of the plane passing through the point (1, -2, 4) and perpendicular to the plane x + 2y – 3z = 11 and parallel to the line \(\frac { x+7 }{ 3 }\) = \(\frac { y+3 }{ -1 }\) = \(\frac { z }{ 1 }\)
Solution:

Non parametric form

Which is the required Cartesian equation of the place.

Question 5.
Find the parametric form of vector equation and Cartesian equations of the plane containing
the line \(\overline { r }\) = (\(\hat { i }\) – \(\hat { j }\) + 3\(\hat { k }\)) + t(2\(\hat { i }\) – \(\hat { j }\) + 4\(\hat { k }\) ) and perpendicular to plane \(\overline { r }\) (\(\hat { i }\) + 2\(\hat { j }\) + \(\hat { k }\)) = 8
Solution:

Cartesian equation
9x – 2y – 5z = -4
9x – 2y – 5z + 4 = 0

Question 6.
Find the parametric vector non-parametric vector and Cartesian form of the equations of the plane passing through the three non- collinear points (3, 6, -2), (-1, -2, 6) and (6, 4, -2).
Solution:


Cartesian equation
⇒ \(\overline { r }\)(2\(\hat { i }\) + 3\(\hat { j }\) + 4\(\hat { k }\)) = 16
⇒ 2x + 3y + 4z – 16 = 0

Question 7.
Find the non-parametric form of vector equation and Cartesian equations of the plane
\(\overline { r }\) = (6\(\hat { i }\) – \(\hat { j }\) + \(\hat { k }\)) + s(\(\hat { -i }\) + 2\(\hat { j }\) + \(\hat { k }\)) + t(\(\hat { -5i }\) – 4\(\hat { j }\) – 5\(\hat { k }\))
Solution:

Cartesian equation:
3x + Sy – 7z = 6
3x + 5y – 7z – 6 = 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.6

Question 1.
Find a parametric form of vector equation of a plane which is at a distance of 7 units from t the origin having 3, -4, 5 as direction ratios of a normal to it.
Solution:
Let \(\overline { d }\) = \(\hat { i }\) – 4\(\hat { j }\) + 5\(\hat { k }\)
p = 7

Question 2.
Find the direction cosines of the normal to the plane 12x + 3y – 4z = 65. Also find the non-parametric form of vector equation of a plane and the length of the perpendicular to the plane from the origin.
Solution:

Length of the ⊥r from the origin = 5 units.

Question 3.
Find the vector and Cartesian equations of the plane passing through the point with position vector 2\(\hat { i }\) + 6\(\hat { j }\) + 3\(\hat { k }\) and normal to the vector \(\hat { i }\) + 3\(\hat { j }\) + 5\(\hat { k }\).
Solution:

Question 4.
A plane passes through the point (-1, 1, 2) and the normal to the plane of magnitude 3√3 makes equal acute angles with the co-ordinate axis. Find the equation of the plane.
Solution:
Magnitude = 3√2
Makes equal acute angles with co-ordinate axes l² + m² + n² = 1
cos²A + cos²A + cos²A = 1 ⇒ 3cos²A = 1
⇒ cos²A = \(\frac { 1 }{ 3 }\)
cos A = \(\frac { 1 }{ √3 }\)
\(\overline { n }\) (normal) = ±3√3(\(\frac { \hat{i} }{ √3 }\)+\(\frac { \hat{j} }{ √3 }\)+\(\frac { \hat{k} }{ √3 }\))
= ±(3\(\hat { i }\) + 3\(\hat { j }\) + 3\(\hat { k }\))
Equation of plane passing through (-1, 1, 2) normal (3, 3, 3)
⇒ 3(x + 1) + 3(y – 1) + 3(z – 2) = 0
⇒ 3x + 3 + 3y – 3 + 3z – 6 = 0
⇒ 3x + 3y + 3z – 6 = 0
⇒ x + y + z = 2
Vector equation \(\overline { r }\) (\(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\)) = 2

Question 5.
Find the intercepts cut off by the plane f \(\overline { r }\) (6\(\hat { i }\) + 4\(\hat { j }\) – 3\(\hat { k }\) ) = 12 on the co-ordinate axes
Solution:

x-intercept = 2, y-intercept = 3, z-intercept = -4

Question 6.
If a plane meets the co-ordinate axes at A, B, C such that the centroid of the triangle ABC is the point (u, v, w), find the equation of the plane.
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.5

Question 1.
Find the parametric form of vector equation and Cartesian equations of straight line passing through (5, 2, 8) and is perpendicular to the straight lines
\(\overline { r }\) = (\(\hat { i }\) + \(\hat { j }\) – \(\hat { k }\)) + s(2\(\hat { i }\) – 2\(\hat { j }\) + \(\hat { k }\)) and
\(\overline { r }\) = (2\(\hat { i }\) – \(\hat { j }\) – 3\(\hat { k }\)) + t(\(\hat { i }\) + 2\(\hat { j }\) + 2\(\hat { k }\)).
Solution:
Given points (5, 2, 8)

2\(\hat { i }\) + \(\hat { j }\) + 3\(\hat { k }\) is a vector perpendicular to both the given straight lines it passes through (5, 2, 8)
The equation is \(\overline { r }\) = (5\(\hat { i }\) + 2\(\hat { j }\) + 8\(\hat { k }\)) + t(2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\))
Cartesian form = \(\frac { x-5 }{ 2 }\) = \(\frac { y-2 }{ 1 }\) = \(\frac { z-8 }{ -2 }\)

Question 2.
Show that the lines
\(\overline { r }\) = (6\(\hat { i }\) + \(\hat { j }\) + 2\(\hat { k }\)) + s(\(\hat { i }\) + 2\(\hat { j }\) – 3\(\hat { k }\)) and
\(\overline { r }\) = (3\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\)) + t(2\(\hat { i }\) + 4\(\hat { j }\) – 5\(\hat { k }\)) are skew lines and hence find the shortest distance between them.
Solution:

\(\overline { b }\) is not a scalar multiple of \(\overline { d }\)
∴ They are not parallel.
∴ The given lines are skew lines.

Question 3.
If the two lines \(\frac { x-1 }{ 2 }\) = \(\frac { y+1 }{ 3 }\) = \(\frac { z-1 }{ 4 }\) and \(\frac { x-3 }{ 1 }\) = \(\frac { y-m }{ 2 }\) = z intersect at a point, find the value of m.
Solution:
(x₁, y₁, z₁) = (1, -1, 1), (x₂, y₂, z₂) = (3, m, 0)
(₁, ₂, b₃) = (2, 3, 4), (d₁, d₂, d₃) = (1, 2, 1).

2(3 – 8) – (m + 1) (2 – 4) – 1(4 – 3) = 0
-10 – (m + 1) (-2) – 1(1) = 0
-10 + 2m + 2 – 1 = 0
2m – 9 = 0
2m = 9
m = \(\frac { 9 }{ 2 }\)

Question 4.
Show that the lines \(\frac { x-3 }{ 3 }\) = \(\frac { y-3 }{ -1 }\), z – 1 = 0 and \(\frac { x-6 }{ 2 }\) = \(\frac { z-1 }{ 3 }\), y – 2 = 0 intersect. Also find the point of intersection.
Solution:

(2s + 6, 2, 3s + 1) ……… (2)
(1) = (2)
((3t + 3), -t + 3, 1) = (2s + 6, 2, 3s + 1)
Compare on both sides
-t + 3 = 2; 3s + 1 = 1
t = 3 – 2; 3s = 0
t = 1; s = 0
(1) ⇒ Point of intersect (6, 2, 1)

Question 5.
Show that the straight lines x + 1 = 2y = -12z and x = y + 2 = 6z – 6 are skew and hence find the shortest distance between them.
Solution:
Given x + 1 = 2y = -12z

\(\overline { b }\) is not a scalar multiple of \(\overline { d }\). So, the two vectors are not parallel.
∴ The given lines are skew lines.

Question 6.
Find the parametric form of vector equation of the straight line passing through (-1, 2, 1) and parallel to the straight line
\(\overline { r }\) = (2\(\hat { i }\) + 3\(\hat { j }\) – \(\hat { k }\)) + t(\(\hat { i }\) – 2\(\hat { j }\) + \(\hat { k }\)) and hence find the shortest distance between the lines.
Solution:

Question 7.
Find the foot of the perpendicular drawn: from the point (5, 4, 2) to the line \(\frac { x+1 }{ 2 }\) = \(\frac { y-3 }{ 3 }\) = \(\frac { z-1 }{ -1 }\). Also, find the equation of the perpendicular.

Solution:
\(\overline { r }\) = \(\overline { a }\) + t\(\overline { b }\)
\(\overline { a }\) = -i + 3j + k, \(\overline { b }\) = 2i + 3j – k
Given points D (5, 4, 2) to the point A. If P is the foot of the perpendicular from to the straight line.
F is of the form
(2t – 1, 3t + 3, -t + 1) and
\(\overline { DF }\) = \(\overline { OF }\) – \(\overline { OD }\) = (2t – 6)i + (3t – 1)j + (-t – l)k
\(\overline { b }\) is perpendicular to \(\overline { DF }\), we have
\(\overline { b }\).\(\overline { DF }\) = 0 => (2t – 6) 2 + 3(3t – 1) – 1(-t – 1) = 0
4t- 12 + 9t – 3 + t + 1 = 0
14t – 14 = 0
14t = 14
t = 1
∴ F (2 – 1, 3 + 3, -1 + 1) = F (1, 6, 0) is foot point. Equation of the perpendicular.
(x₁, y₁, z₁) = (5, 4, 2), (x₂, y₂, z₂) = (1, 6, 0).

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4

Question 1.
Find the non-parametric form of vector equation and Cartesian equations of the straight line passing through the point with position vector 4\(\hat { i }\) + 3\(\hat { j }\) – 7\(\hat { k }\) and parallel to the vector 2\(\hat { i }\) – 6\(\hat { j }\) + 7\(\hat { k }\)
Solution:

Question 2.
Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point (-2, 3, 4) and parallel to the straight line \(\frac { x-1 }{ -4 }\) = \(\frac { y+3 }{ 5 }\) = \(\frac { 8-z }{ 6 }\)
Solution:
\(\overline { a }\) = -2\(\hat { i }\) + 3\(\hat { j }\) + 4\(\hat { k }\) (x₁, y₁, z₁) = (-2, 3, 4)
\(\overline { b }\) = -4\(\hat { i }\) + 5\(\hat { j }\) + 6\(\hat { k }\) (l, m, n) = (-4, 5, 6)
vector equation

Question 3.
Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
Solution:
The straight line passes through the points (6, 7, 4) and (8, 4, 9)
∴ Direction ratios are 2, -3, 5
So the straight line is parallel to 2\(\hat { i }\) – 3\(\hat { j }\) + 5\(\hat { k }\)
∴ Vector equation is

An arbitrary point on the straight line is of the form (2t + 6, -3t + 7, 5t + 4) or (2s + 8, -3s + 4, 5s + 9)
(i) xz plane mean y = 0
-3t + 7 = 0
3t = 7
t = \(\frac { 7 }{ 3 }\)
Point = (2(\(\frac { 7 }{ 3 }\)) + 6, 0, 5(\(\frac { 7 }{ 3 }\)) + 4) = (\(\frac { 32 }{ 3 }\), 0, \(\frac { 47 }{ 3 }\))

(ii) The straight Line cuts yz-plane
So we get x = 0
2t + 6 = 0 ⇒ 2t = -6
t = -3
-3t + 7 = -3 (-3) + 7 = 9 + 7 = 16
5t + 4 = 5(-3) + 4 = -15 + 4 = -11
The required point (0, 16, -11).

Question 4.
Find the direction cosines of the straight line passing through the points (5, 6, 7) and (7, 9, 13). Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points.
Solution:
Given points (5, 6, 7) and (7, 9, 13)
Direction ratios are (2, 3, 6)
So the straight line is parallel to 2\(\hat { i }\) + 3\(\hat { j }\) + 6\(\hat { k }\)

Question 5.
Find the acute angle between the following lines. ]
(i) \(\overline { r }\) = (4\(\hat { i }\) – \(\hat { j }\)) + t(\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\)),
\(\overline { r }\) = (\(\hat { i }\) – 2\(\hat { j }\) + 4\(\hat { k }\)) + s(-\(\hat { i }\) – 2\(\hat { j }\) + 2\(\hat { k }\))
(ii) \(\frac { x+4 }{ 3 }\) = \(\frac { y-7 }{ 4 }\) = \(\frac { z+5 }{ 5 }\), \(\overline { r }\) = 4\(\hat { k }\) + t(2\(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\))
(iii) 2x = 3y = -z and 6x = -y = -4z.
Solution:

Question 6.
The vertices of ΔABC are A(7, 2, 1), 5(6, 0, 3), and C(4, 2, 4). Find ∠ABC.
Solution:

Question 7.
If the straight line joining the points (2, 1, 4) and (a – 1, 4, -1) is parallel to the line joining the points (0, 2, b – 1) and (5, 3, -2) find the values of a and b.
Solution:
(i) Points are (2, 1, 4) (a – 1, 4, -1)
Cartesian equation

(ii) Points are (0, 2, b – 1) (5, 3, -2)
Cartesian equation

Question 8.
If the straight lines \(\frac { x-5 }{ 5m+2 }\) = \(\frac { 2-y }{ 5 }\) = \(\frac { 1-z }{ -1 }\) and x = \(\frac { 2y+1 }{ 4ni }\) = \(\frac { 1-z }{ -3 }\) are perpendicular to each other, find the value of m.
Solution:
If the lines are perpendicular
b₁d₁ + b₂d₂ + b₃d₃ = 0
(5m + 2) – 5(2m) + 1(+3) = 0
5m + 2 – 10m + 3 = 0
-5m = -5
m = \(\frac { -5 }{ -5 }\) = 1
⇒ ∴ m = 1

Question 9.
Show that the points (2, 3, 4), (-1, 4, 5) and (8, 1, 2) are collinear.
Solution:

∴ given points are collinear.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3

Question 1.
If \(\overline { a }\) = \(\hat { i }\) – 2\(\hat { j }\) + 3\(\hat { k }\), \(\overline { b }\) = 2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\), \(\overline { c }\) = 3\(\hat { i }\) + 2\(\hat { j }\) + \(\hat { k }\)
Find (i) (\(\overline { a }\) × \(\overline { b }\) ) × \(\overline { c }\)
(ii) \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\))
Solution:
(i) \(\overline { a }\) × \(\overline { b }\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 1 & -2
\end{array}\right|\)
= \(\hat { i }\)(4 – 3) – \(\hat { j }\)(-2 – 6) + \(\hat { k }\)(1 + 4)
= \(\hat { i }\) + 8\(\hat { j }\) + 5\(\hat { k }\)
\(\overline { a }\) × \(\overline { b }\) × \(\overline { c }\) = \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 8 & 5 \\
3 & 2 & 1
\end{array}\right|\)
= \(\hat { i }\)(8 – 10) – \(\hat { j }\)(1 – 15) + \(\hat { k }\)(2 – 24)
= -2\(\hat { i }\) + 14\(\hat { j }\) – 22\(\hat { k }\)

(ii) \(\overline { b }\) × \(\overline { c }\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & -2 \\
3 & 2 & 1
\end{array}\right|\)
= \(\hat { i }\)(1 + 4) – \(\hat { j }\)(2 + 6) + \(\hat { k }\)(4 – 3)
= 5\(\hat { i }\) – 8\(\hat { j }\) + \(\hat { k }\)
\(\overline { a }\) × \(\overline { b }\) × \(\overline { c }\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
5 & -8 & 1
\end{array}\right|\)
= \(\hat { i }\)(-2 + 24) – \(\hat { j }\)(1 – 15) + \(\hat { k }\)(-8 + 10)
= 22\(\hat { i }\) + 14\(\hat { j }\) + 2\(\hat { k }\)

Question 2.
For any vector a, prove that
\(\hat { i }\)(\(\overline { a }\) × \(\hat { i }\)) + \(\hat { j }\) × (\(\overline { a }\) × \(\hat { j }\)) + \(\hat { k }\) × (\(\overline { a }\) × \(\hat { k }\)) = 2\(\overline { a }\).
Solution:
Let \(\overline { a }\) = a₁\(\hat { i }\) + a₂\(\hat { j }\) + a₃\(\hat { k }\)
\(\overline { a }\) × \(\hat { i }\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_{1} & a_{2} & a_{3} \\
1 & 0 & 0
\end{array}\right|\)
= \(\hat { i }\)(0) – \(\hat { j }\)(-a₃) + \(\hat { k }\)(0 – a₂)
\(\hat { i }\) × (\(\overline { a }\) × \(\hat { i }\)) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 0 \\
0 & a_{3} & -a_{2}
\end{array}\right|\)
= \(\hat { i }\)(0) – \(\hat { j }\)(-a₂) + \(\hat { k }\)(a₃)
= a₂\(\hat { j }\) + a₃\(\hat { k }\)
Similarly \(\hat { j }\) × (\(\overline { a }\) × \(\hat { j }\)) = a₁\(\hat { i }\) + a₃\(\hat { k }\)
\(\hat { k }\) × (\(\overline { a }\) × \(\hat { k }\)) = a₁\(\hat { i }\) + a₂\(\hat { j }\)
\(\hat { i }\) × (\(\overline { a }\) × \(\hat { i }\)) + \(\hat { j }\) × (\(\overline { a }\) × \(\hat { j }\)) + \(\hat { k }\) × (\(\overline { a }\) × \(\hat { k }\))
= 2a₁\(\hat { i }\) + 2a₂\(\hat { j }\) + 2a₃\(\hat { k }\)
= 2(a₁\(\hat { i }\) + a₂\(\hat { j }\) + a₃\(\hat { k }\))= 2 \(\overline { a }\)

Question 3.
Prove that [\(\overline { a }\) – \(\overline { b }\), \(\overline { b }\) – \(\overline { c }\), c – a] = 0.
Solution:

Hence proved.

Question 4.
If \(\overline { a }\) = 2\(\hat { i }\) + 3\(\hat { j }\) – \(\hat { k }\), \(\overline { b }\) = 3\(\hat { i }\) + 5\(\hat { j }\) + 2\(\hat { k }\), \(\overline { c }\) = –\(\hat { i }\) – 2\(\hat { j }\) + 3\(\hat { k }\)
(i) (\(\overline { a }\) × \(\overline { b }\)) × \(\overline { c }\) = (\(\overline { a }\). \(\overline { c }\))\(\overline { b }\) – (\(\overline { b }\).\(\overline { c }\))\(\overline { a }\)
(ii) \(\overline { a }\) (\(\overline { b }\) × \(\overline { c }\)) = (\(\overline { a }\). \(\overline { c }\))\(\overline { b }\) – (\(\overline { a }\).\(\overline { b }\))\(\overline { c }\)
Solution:

Question 5.
\(\overline { a }\) = 2\(\hat { i }\) + 3\(\hat { j }\) – \(\hat { k }\), \(\overline { b }\) = –\(\hat { i }\) + 2\(\hat { j }\) – 4\(\hat { k }\), \(\overline { c }\) = \(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\) then find the value of (\(\overline { a }\) × \(\overline { b }\)) – (\(\overline { a }\) × \(\overline { c }\))
Solution:

Question 6.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) and \(\overline { d }\) are coplanar vectors, show that (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { c }\) × \(\overline { d }\)) = \(\overline { 0 }\)
Solution:

Question 7.
If \(\overline { a }\) = \(\hat { i }\) + 2\(\hat { j }\) + 3\(\hat { k }\), \(\overline { b }\) = 2\(\hat { i }\) – \(\hat { j }\) + \(\hat { k }\), \(\overline { c }\) = 3\(\hat { i }\) + 2\(\hat { j }\) + \(\hat { k }\) and \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) = l\(\overline { a }\) + m\(\overline { b }\) + n\(\overline { c }\), find the values of l, m, n.
Solution:

m = 3 + 4 + 3; n = -(2 – 2 + 3)
m = 10; n = -3
l = 0; m = 10; n = -3

Question 8.
If \(\hat { a }\), \(\hat { b }\), \(\hat { c }\) are three unit vectors such that \(\hat { b }\) and \(\hat { c }\) are non-parallel and \(\hat { a }\) × (\(\hat { b }\) × \(\hat { c }\)) = \(\frac { 1 }{ 2 }\) \(\hat { b }\) find the angle between \(\hat { a }\) and \(\hat { c }\).
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

Question 1.
Prove by vector method that if a line is drawn from the centre of a circle to the midpoint of a chord; then the line is perpendicular to the chord.
Solution:
A circle with centre at O. AB is chord of the circle and OP bisects AB (ie) AP = PB
To prove \(\overline { OP }\) ⊥ \(\overline { AB }\) O is the position vector
∴ \(\overline { OA }\) = \(\overline { OB }\) = Radius
Position vector of P

Hence proved

Question 2.
Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
Solution:
In isosceles ΔABC
Let AB = AC and AD is the median
D is the mid point of BC

\(\overline { DA }\).\(\overline { DB }\) = 0
\(\overline { DA }\) ⊥ \(\overline { DB }\)

Question 3.
Prove by vector method that an angle in a semi-circle is a right angle.
Solution:

Let us consider a circle with centre O and diameter AB.
Let P be any point on the semi-circle.
Let us prove that ∠APB = 90°
W.K.T OA = OB = OP (∵ radius)
\(\overline { PA }\) = \(\overline { PO }\) + \(\overline { OA }\)
\(\overline { PB }\) = \(\overline { PO }\) + \(\overline { OB }\)
= \(\overline { PO }\) – \(\overline { OA }\)
\(\overline { PA }\). \(\overline { PB }\) = (\(\overline { PO }\) + \(\overline { OA }\))(\(\overline { PO }\) – \(\overline { OA }\))
= \(\overline { PO }\)² – \(\overline { OA }\)² = 0
\(\overline { PA }\) ⊥ \(\overline { PB }\)
⇒ ∠APB = 90°. Hence proved.

Question 4.
Prove by vector method that the diagonals of a rhombus bisect each other at right angles.
Solution:
ABCD is a rhombus

\(\overline { AB }\) = \(\overline { a }\) and \(\overline { AD }\) = \(\overline { b }\)
Here AB = BC = CD = DA
(ie) |\(\overline { a }\)| = |\(\overline { b }\)|
\(\overline { AC }\) = \(\overline { AB }\) + \(\overline { BC }\)
= \(\overline { a }\) + \(\overline { b }\)
\(\overline { BD }\) = \(\overline { BC }\) + \(\overline { CD }\)
= \(\overline { AD }\) – \(\overline { AB }\)
= \(\overline { b }\) – \(\overline { a }\)
\(\overline { AC }\).\(\overline { BD }\) = (\(\overline { a }\) + \(\overline { b }\)).(\(\overline { b }\) – \(\overline { a }\))
= (\(\overline { b }\) + \(\overline { a }\)).(\(\overline { b }\) – \(\overline { a }\))
= (\(\overline { b }\))² – (\(\overline { a }\))² = 0 (∵ |\(\overline { a }\)| = |\(\overline { b }\)|)
\(\overline { AC }\).\(\overline { BD }\) = 0 ⇒ \(\overline { AC }\) ⊥ \(\overline { BD }\)
Hence proved.

Question 5.
Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.
Solution:

ABCD is a parallelogram with sides
\(\overline { AB }\) = \(\overline { a }\), \(\overline { AD }\) = \(\overline { b }\) and the diagonals are \(\overline { AC }\) and \(\overline { BD }\)
\(\overline { AC }\) = \(\overline { AB }\) + \(\overline { BC }\) = \(\overline { AB }\) + \(\overline { AD }\) = \(\overline { a }\) + \(\overline { b }\)
\(\overline { BD }\) = \(\overline { BA }\) + \(\overline { AD }\) = \(\overline { AD }\) – \(\overline { AB }\) = \(\overline { b }\) – \(\overline { a }\)
Since the diagonals are equal
|\(\overline { AC }\)| = |\(\overline { BD }\)|
|\(\overline { a }\) + \(\overline { b }\)| = |\(\overline { b }\) – \(\overline { a }\)|
(\(\overline { a }\) + \(\overline { b }\))² = (\(\overline { b }\) – \(\overline { a }\))²
\(\overline { a }\)² + 2\(\overline { a }\).\(\overline { b }\) + \(\overline { b }\)² = \(\overline { b }\)² – 2\(\overline { b }\) \(\overline { a }\) + \(\overline { a }\)²
4\(\overline { a }\) \(\overline { b }\) = 0
\(\overline { a }\).\(\overline { b }\) = 0
\(\overline { a }\) ⊥\(\overline { b }\) ⇒ ABCD is a rectangle. Since ∠A = 90°.

Question 6.
Prove by vector method that the area of the quadrilateral ABCD having diagonals AC and BD is \(\frac { 1 }{ 2 }\) |\(\overline { AC }\)| = |\(\overline { BD }\)|
Solution:
Vector area of a quadrilateral ABCD
= Vector area of ΔABC + Vector area of ΔACD

Area of quadrilateral
= \(\frac { 1 }{ 2 }\) |\(\overline { AC }\)| = |\(\overline { BD }\)|

Question 7.
Prove by vector method that the parallelograms on the same base and between the same parallels are equal in area.
Solution:
\(\overline { AB }\) = \(\overline { a }\), \(\overline { AD }\) = \(\overline { b }\)

Vector area of the parallelogram is \(\overline { b }\) × \(\overline { a }\) ……(1)
Consider the parallelogram ABB’A’
\(\overline { AB }\) = \(\overline { a }\), \(\overline { AB }\) = m\(\overline { a }\)
Because \(\overline { A’B }\) is parallel to \(\overline { AB }\)
Consider the triangle ADA’
By law of vectors AA’ = m\(\overline { a }\) + \(\overline { b }\)
Hence the vector area of parallelogram
ABB’A = \(\overline { a }\) × (m\(\overline { a }\) + \(\overline { b }\))
= m(\(\overline { a }\) × \(\overline { a }\)) + (\(\overline { a }\) × \(\overline { b }\))
= 0 + (\(\overline { a }\) × \(\overline { b }\))
= \(\overline { a }\) × \(\overline { b }\) ……..(2)
By (1) and (2) Area of ABCD = Area of ABB’A’
Hence proved.

Question 8.
If G is the centroid of a ΔABC, prove that (area of ΔGAB) = (area of ΔGBC) = (area of ΔGCA) = \(\frac { 1 }{ 3 }\) (area of ΔABC).
Solution:
W.K.T the median of a triangle divides it into two triangles of equal area.

In ΔABC, AD is the median
area (ΔABD) = area (ΔACD) ………. (1)
In ΔGBC, GD is die median
area (ΔGBD) = area (ΔGCD) ………. (2)
Sub (2) from (1) we get
area (ΔABD) – area (ΔGBD)
= area (ΔACD) – area (ΔGCD)
area(ΔAGB) = area(ΔAGC) ………… (3)
Similarly
area (ΔAGB) = area (ΔBGC) …………(4)
From (3) and (4) we get
area (ΔAGB) = area(ΔAGC) = area(ΔBGC) ……….. (5)
Now
area(ΔAGB) + area(ΔAGC) + area(ΔBGC) = area(ΔABC)
⇒ area(ΔAGB) + area(ΔAGB) + area(ΔAGB)
= area (ΔABC) (using 5)
⇒ 3area(ΔAGB) = area(ΔABC)
⇒ axea(ΔAGB) = \(\frac { 1 }{ 3 }\) area(ΔABC) ………..(6)
From (5) and (6) we get
area(ΔAGB) = area(ΔAGB) = area(ΔBGC)
= \(\frac { 1 }{ 3 }\) area(ΔABC)

Question 9.
Using vector method, prove that
cos(α – ß) = cos α cos ß + sin α sin ß.
Solution:
Let \(\overline { a }\) = \(\overline { OA }\), \(\overline { b }\) = \(\overline { OB }\) be the unit vectors and which make angles α and ß respectively with positive x-axis where A and B are as in diagram.

Draw AL and BM perpendicular to the X axis, then
\(\overline { OL }\) = \(\overline { OA }\) = cos α
|\(\overline { OL }\)| = |\(\overline { OA }\)| cos α = cos α
|\(\overline { LA }\)| = |\(\overline { OA }\)| sin α = sin α
\(\overline { OL }\) = |\(\overline { OL }\)|\(\hat { i }\) = cos α \(\hat { i }\)
\(\overline { LA }\) = sin α (+\(\hat { j }\))
\(\overline { a }\) = \(\overline { OA }\) = \(\overline { OL }\) + \(\overline { LA }\)
= cosα \(\hat { i }\) + sinα \(\hat { j }\) ……… (1)
Similarly \(\overline { b }\) = cos ß \(\hat { i }\) + sin ß \(\hat { j }\) ……… (2)
The angle between \(\overline { a }\) and \(\overline { b }\) is α – ß and so \(\overline { a }\).\(\overline { b }\) = |\(\overline { a }\)||\(\overline { b }\)| cos(α – ß) = cos(α – ß) ……..(3)
From (1) and (2)
\(\overline { a }\).\(\overline { b }\) = (cosα\(\hat { i }\) + sinα \(\hat { j }\)).(cosß\(\hat { i }\) + sinß\(\hat { j }\))
= cos α cos ß + sin α sin ß ………. (4)
From (3) and (4)
cos(α – ß) = cos α cos ß + sin α sin ß

Question 10.
Prove by vector method that
sin(α + ß) = sin α cos ß +cos α sin ß.
Solution:
Let \(\overline { a }\) = \(\overline { OA }\), \(\overline { b }\) = \(\overline { OB }\) be the unit vectors making angles α and ß respectively with positive x axis where A and B are as shown in the diagram

Draw AL and BM perpendicular to the X axis, then
\(\overline { OL }\) = \(\overline { OA }\) cos α
|\(\overline { OL }\)| = |\(\overline { OA }\)| cos α = cos α
|\(\overline { LA }\)| = |\(\overline { OA }\)| sin α = sin α
|\(\overline { OL }\)| = |\(\overline { OL }\)| \(\hat { i }\) = cos α \(\hat { i }\)
\(\overline { LA }\) = sin α(-\(\hat { j }\))
\(\overline { a }\) = \(\overline { OA }\) = \(\overline { OL }\) + \(\overline { LA }\)
= cos α \(\hat { i }\) + sin α \(\hat { j }\) ………… (1)
similarly \(\overline { b }\) = cos ß \(\hat { i }\) – sin ß \(\hat { j }\) ………… (2)
The angle between \(\overline { a }\) and \(\overline { b }\) is α + ß and the vectors \(\overline { b }\), \(\overline { a }\), \(\overline { k }\) from a right handed system.
\(\overline { b }\) × \(\overline { a }\) = |\(\overline { b }\)||\(\overline { a }\)| sin(α + ß)\(\hat { k }\) = sin(α + ß)\(\hat { k }\) ………..(1)
\(\overline { b }\) × \(\overline { a }\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
\cos \beta & -\sin \beta & 0 \\
\cos \alpha & \sin \alpha & 0
\end{array}\right|\)
= (sin α cos ß + cos α sin ß)\(\hat { k }\) ……… (2)
From (1) & (2)
sin(α + ß) = sin α cos ß + cos α sin ß

Question 11.
A particle acted on by constant forces 8\(\hat { i }\) + 2\(\hat { j }\) – 6\(\hat { k }\) and 6\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\) is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces.
Solution:
\(\overline { OA }\) = \(\hat { i }\) + 2\(\hat { j }\) + 3\(\hat { k }\)
\(\overline { OB }\) = 5\(\hat { i }\) + 4\(\hat { j }\) + \(\hat { k }\)
\(\overline { d }\) = \(\overline { AB }\) = \(\overline { OB }\) – \(\overline { OA }\)
= 4\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\)
\(\overline { F_1 }\) = 8\(\hat { i }\) + 2\(\hat { j }\) – 6\(\hat { k }\)
and \(\overline { F_2 }\) = 6\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\)
\(\overline { F }\) = \(\overline { F_1 }\) + \(\overline { F_2 }\) = 14\(\hat { i }\) + 4\(\hat { j }\) – 8\(\hat { k }\)
Work done = \(\overline { F }\) \(\overline { d }\)
= (14\(\hat { i }\) + 4\(\hat { j }\) – 8\(\hat { k }\)).(4\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\))
= 56 + 8 + 16
= 80 units

Question 12.
Forces of magnitudes 5√2 and 10√2 units acting in the directions 3\(\hat { i }\) + 4\(\hat { j }\) + 5\(\hat { k }\) and 10\(\hat { i }\) + 6\(\hat { j }\) – 8\(\hat { k }\), respectively, act on a particle which is displaced from the point with position vector 4\(\hat { i }\) – 3\(\hat { j }\) – 2\(\hat { k }\) to the point with position vector 6\(\hat { i }\) – \(\hat { j }\) – 3\(\hat { k }\). Find the work done by the forces.

Resultant force \(\overline { F }\) = \(\overline { F_1 }\) + \(\overline { F_2 }\)
= 5√2\(\overline { F_1 }\) + 10√2 \(\overline { F_2 }\)
= 3\(\hat { i }\) + 4\(\hat { j }\) + 5\(\hat { k }\) + 10\(\hat { i }\) + 6\(\hat { j }\) – 8\(\hat { k }\)
= 13\(\hat { i }\) + 10\(\hat { j }\) – 3\(\hat { k }\)
\(\overline { OA }\) = 4\(\hat { i }\) – 3\(\hat { j }\) – 2\(\hat { k }\)
\(\overline { OB }\) = 6\(\hat { i }\) + \(\hat { j }\) – 3\(\hat { k }\)
\(\overline { d }\) = \(\overline { AB }\) = \(\overline { OB }\) – \(\overline { OA }\) = 2\(\hat { i }\) + 4\(\hat { j }\) – \(\hat { k }\)
\(\overline { F }\).\(\overline { d }\) = (13\(\hat { i }\) + 10\(\hat { j }\) – 3\(\hat { k }\)).(2\(\hat { i }\) + 4\(\hat { j }\) – \(\hat { k }\))
= 26 + 40 + 3
= 69 units

Question 13.
13. Find the magnitude and direction cosines of the torque of a force represented by 3\(\hat { i }\) + 4\(\hat { j }\) – 5\(\hat { k }\) about the point with position vector 2\(\hat { i }\) – 3\(\hat { j }\) + 4\(\hat { k }\) acting through a point whose position vector is 4\(\hat { i }\) + 2\(\hat { j }\) – 3\(\hat { k }\)
Solution:
\(\overline { OA }\) = 2\(\hat { i }\) – 3\(\hat { j }\) + 4\(\hat { k }\)
\(\overline { OB }\) = 4\(\hat { i }\) + 2\(\hat { j }\) – 3\(\hat { k }\)
\(\hat { r }\) = \(\overline { AB }\) = \(\overline { OB }\) – \(\overline { OA }\)
= 2\(\hat { i }\) + 5\(\hat { j }\) – 7\(\hat { k }\)
= 3\(\hat { i }\) + 4\(\hat { j }\) – 5\(\hat { k }\)

Question 14.
Find the torque of the resultant of the three forces represented by -3\(\hat { i }\) + 6\(\hat { j }\) – 3\(\hat { k }\), 4\(\hat { i }\) – 10\(\hat { j }\) + 12\(\hat { k }\) and 4\(\hat { i }\) + 7\(\hat { j }\) acting at the point with position vector 8\(\hat { i }\) – 6\(\hat { j }\) – 4\(\hat { k }\) about the point with position vector 18\(\hat { i }\) + 3\(\hat { j }\) – 9\(\hat { k }\)
Solution:
\(\overline { F_1 }\) = -3\(\hat { i }\) + 6\(\hat { j }\) – 3\(\hat { k }\)
\(\overline { F_2 }\) = 4\(\hat { i }\) – 10\(\hat { j }\) + 12\(\hat { k }\)
\(\overline { F_3 }\) = 4\(\hat { i }\) + 7\(\hat { j }\)
\(\overline { F }\) = \(\overline { F_1 }\) + \(\overline { F_2 }\) + \(\overline { F_3 }\)
= 5\(\hat { i }\) + 3\(\hat { j }\) + 9\(\hat { k }\)
\(\overline { OB }\) = 8\(\hat { i }\) – 6\(\hat { j }\) – 4\(\hat { k }\)
\(\overline { OA }\) = 18\(\hat { i }\) + 3\(\hat { j }\) – 9\(\hat { k }\)
\(\overline { AB }\) = \(\overline { OB }\) – \(\overline { OA }\)
= -10\(\hat { i }\) – 9\(\hat { j }\) + 5\(\hat { k }\)
\(\overline { t }\) = \(\overline { r }\) × \(\overline { F }\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-10 & -9 & 5 \\
5 & 3 & 9
\end{array}\right|\)
= \(\hat { i }\)(-81 – 15) – \(\hat { j }\)(-90 – 25) + \(\hat { k }\)(-30 + 45)
= -96\(\hat { i }\) + 115\(\hat { j }\) + 15\(\hat { k }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.6

Choose the most suitable answer from the given four alternatives:

Question 1.
The equation of the circle passing through (1, 5) and (4, 1) and touching y-axis is x² + y² – 5x – 6y + 9 + λ (4x + 3y – 19) = 0 where λ is equal to
(a) 0, –\(\frac {40}{9}\)
(b) 0
(c) \(\frac {40}{9}\)
(d) –\(\frac {40}{9}\)
Solution:
(a) 0, –\(\frac {40}{9}\)
Hint:
x² + y² – 5x – 6y + 9 + λ(4x + 3y – 19) = 0
x² + y² + x (-5 + 4λ) + y (- 6 + 3λ) + 9 – 19λ = 0
It touches the y-axis put x = 0.
y² + (3λ – 6) y + 9 – 19λ = 0
Now, b² – 4ac = 0
⇒ (3λ – 6)² – 4 (1) (9 – 19λ) = 0
Solving this equation we get
λ = 0 or λ = –\(\frac {40}{9}\)

Question 2.
The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distence between the foci is
(a) \(\frac {4}{3}\)
(b) \(\frac {4}{√3}\)
(c) \(\frac {2}{√3}\)
(d) –\(\frac {3}{2}\)
Solution:
(c) \(\frac {2}{√3}\)
Hint:
Length of Latus Rectum \(\frac {2b^2}{a}\) = 8
⇒ b² = 4a …….. (1)
Length of conjugate axes
2b = \(\frac {1}{2}\)(2ae)
⇒ b = \(\frac {1}{2}\) …….. (2)
b² = \(\frac {a^2e^2}{4}\)
Now b² = a²(e² – 1)
\(\frac {a^2e^2}{4}\) = a²(e² – 1)
e² = 4e² – 4
3e² = 4
e² = \(\frac {4}{3}\)
∴ e = \(\frac {2}{√3}\)

Question 3.
The circle x² + y² = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if
(a) 15 < m < 65
(b) 35 < m < 85
(c) -85 < m < -35
(d) -35 < m < 15
Solution:
(d) -35 < m < 15
Hint:

C : x² + y² – 4x – 8y – 5 = 0
(x – 2)² + (y – 4)² = 25
C (2, 4); r = 5
Distance from centre < r

-25 < 10 + m < 25
⇒ -25 – 10 < m < 25 – 10
-35 < m < 15

Question 4.
The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3).
(a) \(\frac {6}{5}\)
(b) \(\frac {5}{3}\)
(c) \(\frac {10}{3}\)
(d) \(\frac {3}{5}\)
Solution:
(c) \(\frac {10}{3}\)
Hint:

CA = CB
CA² = CB²
(1 – 1)² + (h – 0)² = (1 – 2)² + (h – 3)²
h² = 1 + h² + 9 – 6h
6h = 10
h = \(\frac {10}{6}\) = \(\frac {5}{3}\)
Diameter is 2h = 2(\(\frac {5}{3}\)) = \(\frac {10}{3}\)

Question 5.
The radius of the circle 3x² + by² + 4bx – 6by + b² = 0 is
(a) 1
(b) 3
(c) \(\sqrt {10}\)
(d) \(\sqrt {11}\)
Solution:
(c) \(\sqrt {10}\)
Hint:
Equation of circle
3x² + by² + 4 bx – 6by + b² = 0
a = b ⇒ b = 3
3x² + 3y² + 12x – 18y + 9 = 0
÷ by 3 x² + y² + 4x – 6y + 3 = 0
2g = 4; 2f = -6; c = 3
g = 2; f = -3
r = \(\sqrt {g^2+f^2-c}\)
= \(\sqrt {4+9-3}\)
= \(\sqrt {10}\)

Question 6.
The centre of the circle inscribed in a square formed by the lines x² – 8x – 12 = 0 and y² – 14y + 45 = 0 is
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)
Solution:
(a) (4, 7)
Hint:
Equation of lines
x² – 8x – 12 = 0
(x – 6)(x – 2) = 0
x = 2, 6
Another lines
y² – 14y + 45 = 0
(y – 5 )(y – 9) = 0
y = 5, 9
Hence the extremities of the diameter are (6, 9) and (2, 5).
Centre is mid point of (6, 9) and (2, 5)
Centre = (\(\frac {6+2}{2}\),\(\frac {9+5}{2}\))
= (4, 7)

Question 7.
The equation of the normal to the circle x² + y² – 2x – 2y + 1 = 0 which is parallel to the line 2x + 4y = 3 is
(a) x + 2y = 3
(b) x + 2y + 3 = 0
(c) 2x + 4y + 3 = 0
(d) x – 2y + 3 = 0
Solution:
(a) x + 2y = 3
Hint:

x² + y² – 2x – 2y + 1 = 0
2g = -2 2f = -2
g = -1 f = -1
Parallel line be 2x + 4y + λ = 0
Centre be (-g, -f) = (1, 1)
Which lies on line
2 + 4 + λ = 0 ⇒ λ = -6
∴ 2x + 4y – 6 = 0 ⇒ x + 2y = 3

Question 8.
If P(x, y) be any point on 16x² + 25y² = 400 with foci F(3, 0) then PF₁ + PF₂ is
(a) 8
(b) 6
(c) 10
(d) 12
Solution:
(c) 10
Hint:
16x² + 25y² = 400
\(\frac {x^2}{25}\) + \(\frac {y^2}{16}\) = 1
a² = 25
⇒ ∴ a = ±5
PF₁ + PF₂ = major axis = 2a
= 2 × 5 = 10.

Question 9.
The radius of the circle passing through the points (6,2) two of whose diameter are x + y = 6 and x + 2y = 4 is
(a) 10
(b) 2√5
(c) 6
(d) 4
Solution:
(b) 2√5
Hint:
x + y = 6 …….. (1)
x + 2y = 4 ……… (2)
(1) – (2) -y = 2 ⇒ y = -2
(1) ⇒ x – 2 = 6 ⇒ x = 8.
point be (8, -2)
another point (6, 2)
radius = \(\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}\)
= \(\sqrt {(8-6)^2+(2+2)^2}\)
= \(\sqrt {2^2+4^2}\) = \(\sqrt {4+16}\)
= \(\sqrt {20}\) = 2√5.

Question 10.
The area of quadrilateral formed with foci of the hyperbolas \(\frac {x^2}{a^2}\) – \(\frac {y^2}{b^2}\) = 1 and \(\frac {x^2}{a^2}\) – \(\frac {y^2}{b^2}\) = -1 is
(a) 4(a² + b²)
(b) 2(a² + b²)
(c) a² + b²
(d) \(\frac {1}{2}\)(a²+ b²)
Solution:
(b) 2(a² + b²)
Hint:

Question 11.
If the normals of the parabola y² = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)² + (y + 2)² = r², then the value of r² is
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(a) 2
Hint:
y² = 4x
4a = 4
a = 1
End points of latus rectum = (a, ±2a)
= (1, ±2)
Normal equation
xyx + 2ay = x₁y₁ + 2 ay₁
Equation of normal at points (1, ±2)
y = -x + 3, y = x + 3
x + y – 3 = 0, x – y + 3 = 0

Question 12.
If x + y = k is a normal to the parabola y² = 12x, then the value of k is 14.
(a) 3
(b) -1
(c) 1
(d) 9
Solution:
(d) 9
Hint:
y² = 12x ⇒ 4a = 12
⇒ a = 3
y = mx + c ∴ x + y = k
⇒ y = -x + k
∴ m = -1, c = k.
c = -2am – am² ⇒ k = -2a(-1) – a(-1)³
k = -6(-1) – 3(-1) = 6 + 3 = 9
k = 9

Question 13.
The ellipse E₁ : \(\frac {x^2}{9}\) + \(\frac {y^2}{4}\) = 1 is inscribed in a rectangle R whose sides are parallel to the co-ordinate axes. Another ellipse E₂ passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse is
(a) \(\frac {√2}{2}\)
(b) \(\frac {√3}{2}\)
(c) \(\frac {1}{2}\)
(d) \(\frac {3}{4}\)
Solution:
(c) \(\frac {1}{2}\)
Hint:

Question 14.
Tangents are drawn to, the, hyperbola \(\frac {x^2}{9}\) – \(\frac {y^2}{4}\) = 1 parallel to the straight line 2x – y – 1. One of the points of contact of tangents on the hyperbola is
(a) (\(\frac {9}{2√2}\), \(\frac {-1}{√2}\))
(b) (\(\frac {-9}{2√2}\), \(\frac {1}{√2}\))
(c) (\(\frac {9}{2√2}\), \(\frac {1}{√2}\))
(d) (3√3, -2√2)
Solution:
(c) (\(\frac {9}{2√2}\), \(\frac {1}{√2}\))
Hint:
a² = 9 b² = 4, 2x – y = 1
y = 2x – 1
m = 2

Question 15.
The equation of the circle passing through the foci of the ellipse \(\frac {x^2}{16}\) + \(\frac {y^2}{9}\) = 1 having centre at (0, 3) is
(a) x² + y² – 6y – 7 = 0
(b) x² + y² – 6y + 7 = 0
(c) x² + y² – 6y – 5 = 0
(d) x² + y² – 6y + 5 = 0
Solution:
(a) x² + y² – 6y – 7 = 0
Hint:
a² = 16, b² = 16
(h, k) = (0, 3)
e = \(\sqrt{1-\frac {b^2}{a^2}}\) = \(\sqrt{1-\frac {9}{16}}\)
= \(\sqrt{\frac {7}{16}}\) = \(\frac {√7}{4}\)
ae = 4\(\frac {√7}{4}\) = √7.
F(√7, 0) lies on circle.
(x – h)² + (y – k)² = r²
(√7 – 0)² + (0 – 3)² = r² ⇒ √7² + 3² = r²
7 + 9 = r²
⇒ r² = 16.
∴ (x – 0)² + (y – 3)² = 16
x² + y² – 6y + 9 = 16
x² + y² – 6y – 7 = 0

Question 16.
Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centered at (0, y) passing through the origin and touching the circle C externally, then the radius of T is equal to
(a) \(\frac {√3}{√2}\)
(b) \(\frac {√3}{2}\)
(c) \(\frac {1}{2}\)
(d) \(\frac {1}{4}\)
Solution:
(d) \(\frac {1}{4}\)
Hint:

ΔOO’A
(1 + y)² = (1 – y)² + 1
1 + y² + 2y = 1 + y² – 2y + 1
4y = 1 ⇒ y = \(\frac {1}{4}\)

Question 17.
Consider an ellipse whose centre is of the origin and its major axis is a long x-axis. If its eccentricity is \(\frac {3}{5}\) and the distance between its foci is 6, then the area of the quadrilateral’ inscribed in the ellipse with diagonals as major and minor axis, of the ellipse is
(a) 8
(b) 32
(c) 80
(d) 40
Solution:
(d) 40
Hint:

e = \(\frac {3}{5}\)
2ae = 6 ⇒ 2a(\(\frac {3}{5}\)) = 6
a = 5; b = 4
Area = 4 × \(\frac {1}{2}\) × ab = 2 ab
= 2 × 5 × 4
= 40

Question 18.
Area of the greatest rectangle inscribed in the ellipse \(\frac {x^2}{16}\) + \(\frac {y^2}{9}\) = 1 is
(a) 2ab
(b) ab
(c) \(\sqrt {ab}\)
(d) \(\frac {a}{b}\)
Solution:
(a) 2ab
Hint:

x = a cosθ; y = b sinθ
length = 2acosθ; breadth = 2bsinθ
A = l × b = 4absinθcosθ
A = 2ab sin2θ
\(\frac {dA}{dt}\) = 2ab cos2θ (2)
= 4ab cos2θ
\(\frac {dA}{dt}\) = 0
cos2θ = 0
2θ = \(\frac {π}{2}\)
θ = \(\frac {π}{4}\)
∴ A = 2ab sin²(\(\frac {π}{4}\))
= 2ab sin \(\frac {π}{2}\)
A = 2ab

Question 19.
An ellipse has OB as semi minor axes, F and F’ its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is
(a) \(\frac {1}{√2}\)
(b) \(\frac {1}{2}\)
(c) \(\frac {1}{4}\)
(d) \(\frac {1}{√3}\)
Solution:
(a) \(\frac {1}{√2}\)
Hint:

Distance between foci and end of minor axes = a
∴ F₁B = F₂ = a
F₁F₂ = 2ae
In right angle F₁BF₂
F₁B² + F₂B² = F₁F₂²
a² + a² = (2ae)²
2a² = 4a²e²
e² = \(\frac {2}{4}\) = \(\frac {1}{2}\)
∴ e = \(\frac {1}{√2}\)

Question 20.
The eccentricity of the ellipse
(x – 3)² + (y – 4)² = \(\frac {y²}{9}\) is
(a) \(\frac {√3}{2}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {1}{3√2}\)
(d) \(\frac {1}{√3}\)
Solution:
(b) \(\frac {1}{3}\)
Hint:
PF = e²p³
(x – h)² + (y – k)² = e²(\(\frac {ax+by+c}{\sqrt{a^2+b^2}}\))
(h, k) = (3, 4),
a = 0, c = 0
e² = \(\frac {1}{3}\)
e = \(\frac {1}{3}\)

Question 21.
If the two tangents drawn from a point P to the parabola y2 = 4r are at right angles then the locus of P is {SEA
(a) 2x + 1 = 0
(b) x = -1
(c) 2x – 1 = 0
(d) x = 1
Solution:
(b) x = -1
Hint:
Locus of P = Directrix of y² = 4x; 4a = 4
∴ a = 1
Equation of directrix x = -a = -1
∴ -x = -1

Question 22.
The circle passing through (1, -2) and touching the axis of x at (3, 0) passing through the point
(a) (-5, 2)
(b) (2, -5)
(c) (5, -2)
(d) (-2, 5)
Solution:
(c) (5, -2)
Hint:
(x – 3)² + (y – 0)² + λy = 0
At (1, -2), (1 – 3)² + (-2 – 0)² + λy = 0
4 + 4 – 2λ = 0
8 = 2λ
λ = 4
x² – 6x + 9 + y² + 4y = 0
Apply all the point which satisfied that passes through the circle At (+5, -2),
25 – 30 + 9 + 4 – 8 = 0

Question 23.
The locus of a point whose distance from (- 2, 0) is \(\frac {2}{3}\) times its distance from the line x = \(\frac {-9}{2}\) is
(a) a parabola
(b) a hyperbola
(c) an ellipse
(d) a circle
Solution:
(c) an ellipse
Hint:
P(h, k) Q(-2, 0)
x = –\(\frac {9}{2}\) PQ = \(\frac {2}{3}\)
2x + 9 = 0
\(\sqrt {(h+2)^2+k^2}\) = \(\frac {2}{3}\)|\(\frac {2h+9}{2}\)|
(h + 2)² + k² = \(\frac {1}{9}\)(2h + 9)²
h² + 4 + 4h + k² = \(\frac {1}{9}\)(4h² + 36h + 81)
9h² + 36 + 36h + 9k² = 4h² + 36h + 81
5h² + 9k² = 45
\(\frac {h^2}{9}\) + \(\frac {k^2}{5}\) = 1
\(\frac {x^2}{9}\) + \(\frac {y^2}{5}\) = 1
Which is ellipse.

Question 24.
The values of m for which the line y = mx + 2√5 touches the hyperbola 16x² – 9y² = 144 are the roots of x² – (a + b)x – 4 = 0, then the value of (a + b) is
(a) 2
(b) 4
(c) 0
(d) -2
Solution:
(c) 0
Hint:
a² = 9; b² = 16
a = 3; b = 4
c² = a²m² – b²
(2√5)² = 9m² – 16
20 + 16 = 9m²; m² = \(\frac {36}{9}\)
∴ m = 2 which is roots of x² -(a + b)x – 4 = 0
2² -(a + b)2 – 4 = 0
a + b = 0

Question 25.
If the coordinates at one end of a diameter of the circle x² + y² – 8x – 4y + c = 0 are (11, 2) the cordinates of the other end are
(a) (-3, 2)
(b) (2, -5)
(c) (5, -2)
(d) (-2, 5)
Solution:
(a) (-3, 2)
Hint:

2g = -g; 2f = -4
g = -4; f = -2
c(-g, -f) = (4, 2)
\(\frac {x_1+x_2}{9}\) = 4; \(\frac {y_1+y_2}{2}\) = 2
\(\frac {x_1+11}{2}\) = 4; \(\frac {y_1+2}{2}\) = 2
x₁ = 8 – 11; y₁ = 4 – 2
x₁ = -3; y₁ = 2
∴ Other end be (-3, 2)