Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.2

Question 1.
If \(\overline { a }\) = \(\hat { i }\) – 2\(\hat { j }\) + 3\(\hat { k }\), b = 2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\), c = 3\(\hat { i }\) + 2\(\hat { j }\) + \(\hat { k }\) find \(\overline { a }\).(\(\overline { b}\) × \(\overline { c }\)).
Solution:
\(\overline { a }\).(\(\overline { b}\) × \(\overline { c }\)) = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = \(\left|\begin{array}{ccc}
1 & -2 & 3 \\
2 & 1 & -2 \\
3 & 2 & 1
\end{array}\right|\)
= 1(1 + 4) + 2(2 + 6) + 3(4 – 3)
= 5 + 16 + 3 = 24

Question 2.
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors -6\(\hat { i }\) + 14\(\hat { j }\) + 10\(\hat { k }\), 14\(\hat { i }\) – 10\(\hat { j }\) – 6\(\hat { k }\) and 2\(\hat { i }\) + 4\(\hat { j }\) – 2\(\hat { k }\)
Solution:
Volume of the parallelepiped = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ]
= \(\left|\begin{array}{ccc}
-6 & 14 & 10 \\
14 & -10 & -6 \\
2 & 4 & -2
\end{array}\right|\)
= -6(20 + 24) -14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760
= 720 cu. units.

Question 3.
The volume of the parallelepiped whose coterminous edges are 7\(\hat { i }\) + λ\(\hat { j }\) – 3\(\hat { k }\), \(\hat { i }\) + 2\(\hat { j }\) – \(\hat { k }\), -3\(\hat { i }\) + 7\(\hat { j }\) + 5\(\hat { k }\) is 90 cubic units. Find the value of λ
Solution:
volume of the parallelepiped = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ]
\(\left|\begin{array}{ccc}
7 & \lambda & -3 \\
1 & 2 & -1 \\
-3 & 7 & 5
\end{array}\right|\) = 90
7(10 + 7) – λ(5 – 3) – 3(7 + 6) = 90
7(17) – λ(2) – 3(13) = 90
119 – 2λ – 39 = 90
2λ = 119 – 39 – 90
2λ = -10
λ = -5

Question 4.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three non-coplanar vectors represented by concurrent edges of a parallelepiped of volume 4 cubic units, find the value of
(\(\overline { a }\) + \(\overline { b }\)).(\(\overline { b }\) × \(\overline { c }\)) + (\(\overline { b }\) + \(\overline { c }\)).(\(\overline { c }\) × \(\overline { a }\)) + (\(\overline { c }\) + \(\overline { a }\)).(\(\overline { a }\) × \(\overline { b }\)).
Solution:
Given [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = ±4.
{(\(\overline { a }\) + \(\overline { b }\)).(\(\overline { b }\) × \(\overline { c }\)) + (\(\overline { b }\) + \(\overline { c }\)).(\(\overline { c }\) × \(\overline { a }\)) + (\(\overline { c }\) + \(\overline { a }\)).(\(\overline { a }\) × \(\overline { b }\)).}
= \(\overline { a }\)(\(\overline { b }\) × \(\overline { c }\)) + \(\overline { b }\) – (\(\overline { b }\) × \(\overline { c }\)) + \(\overline { b }\) – (\(\overline { c }\) × \(\overline { a }\)) + \(\overline { c }\) – (\(\overline { c }\) × \(\overline { a }\)) + \(\overline { c }\)(\(\overline { a }\) × \(\overline { b }\)) + \(\overline { a }\) – (\(\overline { a }\) × \(\overline { b }\))
= [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] + [ \(\overline { b }\), \(\overline { b }\), \(\overline { c }\) ] + [ \(\overline { p }\), \(\overline { c }\), \(\overline { a }\) ] + [ \(\overline { c }\), \(\overline { c }\), \(\overline { a }\) ] + [ \(\overline { c }\), \(\overline { a }\), \(\overline { b }\) ] + [ \(\overline { a }\), \(\overline { a }\), \(\overline { b }\) ]
= ± 4 + 0 ± 4 + 0 ± 4 + 0 = ± 12

Question 5.
Find the altitude of a parallelepiped determined by the vectors \(\overline { a }\) = -2\(\hat { i }\) + 5\(\hat { j }\) + 3\(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + 3\(\hat { j }\) – 2\(\hat { k }\) and \(\overline { c }\) = -3\(\hat { i }\) + \(\hat { j }\) + 4\(\hat { k }\) if the base is taken as the parallelogram determined by \(\overline { b }\) and \(\overline { c }\).
Solution:
V = \(\overline { a }\) -(\(\overline { b }\) × \(\overline { c }\)) = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ]
= \(\left|\begin{array}{ccc}
-2 & 5 & 3 \\
1 & 3 & -2 \\
-3 & 1 & 4
\end{array}\right|\)
= -2(12 + 2) -5(4 – 6) + 3(1 + 9)
= -2(14) -5(-2) + 3(10)
= -28 + 10 + 30 = 12
Area = |\(\overline { b }\) × \(\overline { c }\)| = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 3 & -2 \\
-3 & 1 & 4
\end{array}\right|\)
= \(\hat { i }\)(12 + 2) – \(\hat { j }\)(4 – 6) + \(\hat { k }\)(1 + 9)
= 14\(\hat { i }\) + 2\(\hat { j }\) + 10\(\hat { k }\)
|\(\overline { b }\) × \(\overline { c }\)| = \(\sqrt { 196+4+100 }\) = \(\sqrt { 300 }\)
= 10√3
Altitude h = \(\frac { V }{ Area }\) = \(\frac { 12 }{ 10√3 }\) = \(\frac { 12×√3 }{ 10×3 }\) = \(\frac { 2√3 }{ 5 }\)

Question 6.
Determine whether the three vectors 2\(\hat { i }\) + 3\(\hat { j }\) + \(\hat { k }\), \(\hat { i }\) – 2\(\hat { j }\) + 2\(\hat { k }\) and 3\(\hat { i }\) + \(\hat { j }\) + 3\(\hat { k }\) are coplanar.
Solution:
If vectors are coplanar, [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = 0
[ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = \(\left|\begin{array}{ccc}
2 & 3 & 1 \\
1 & -2 & 2 \\
3 & 1 & 3
\end{array}\right|\)
= 2(-6 – 2) -3(3 – 6) + 1(1 + 6)
= -2(-8) -3(-3) + 1(7) = -16 + 9 + 7 = 0
∴ The given vectors are coplanar

Question 7.
Let \(\overline { a }\) = \(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) and \(\overline { c }\) = c₁\(\hat { i }\) + c₂\(\hat { j }\) + c₃\(\hat { k }\). If c₁ = 1 and c₂ = 2, find c₃ such that \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) are coplanar.
Solution:
If \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) are coplanar [ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = 0
\(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 0 \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0
1(0) – 1(c₃) + 1(c₂) = 0
-c₃ + c₂ = 0
c₃ = c₂ = 2

Question 8.
If \(\overline { a }\) = \(\hat { i }\) – \(\hat { k }\), \(\overline { b }\) = x\(\hat { i }\) + \(\hat { j }\) + (1 – x)\(\hat { k }\) c = y\(\hat { i }\) + x\(\hat { j }\) + (1 + x – y) \(\hat { k }\) Show that [ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = 0
Solution:
[ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = \(\left|\begin{array}{ccc}
1 & 0 & -1 \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|\)
= 1(1 + x – y – x + x²)-1(x² – y)
(1 + x – y – x + x² – x² + y)
= 1
There is no x and y terms
∴ [ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] depends on neither x nor y.

Question 9.
If the vectors a\(\hat { i }\) + a\(\hat { j }\) + c\(\hat { k }\), \(\hat { i }\) + \(\hat { k }\) and c\(\hat { i }\) + c\(\hat { j }\) + b\(\hat { k }\) are coplanar, prove that c is the geometric mean of a and b.
Solution:
[ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = 0
\(\left|\begin{array}{lll}
a & a & c \\
1 & 0 & 1 \\
c & c & b
\end{array}\right|\) = 0
a(0 – c) – a(b – c) + c(c) = 0
-ac – ab + ac + c² = 0
c² – ab = 0
c² = ab
⇒ c in the geometric mean of a and b.

Question 10.
Let \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) be three non-zero vectors such that \(\overline { c }\) is a unit vector perpendicular to both \(\overline { a }\) and \(\overline { b }\). If the angle between \(\overline { a }\) and \(\overline { b }\) is \(\frac { π }{ 6 }\) show that [\(\overline { a }\), \(\overline { b }\) and \(\overline { c }\)]² = \(\frac { 1 }{ 4 }\) |\(\overline { a }\)|² |\(\overline { b }\)|²
Solution:

Hence proved