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 \(\overline { c }\) = c_{1}\(\hat { i }\) + c_{2}\(\hat { j }\) + c_{3}\(\hat { k }\). If c_{1} = 1 and c_{2} = 2, find c_{3} 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(c_{3}) + 1(c_{2}) = 0

-c_{3} + c_{2} = 0

c_{3} = c_{2} = 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