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