Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 10 Vector Algebra Ex 10.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 10 Vector Algebra Ex 10.2

Question 1.

Compute the magnitudes of the following vectors:

\(\vec{a}\) = \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \),

\(\vec{b}\) = 2\(\hat {i} \) – 7\(\hat {j} \) – 3\(\hat {k} \),

\(\vec{c}\) = \(\frac{1}{\sqrt{3}}\)\(\hat {i} \) + \(\frac{1}{\sqrt{3}}\)\(\hat {j} \) – \(\frac{1}{\sqrt{3}}\)\(\hat {k} \)

Solution:

Question 2.

Write two different vectors having the same magnitude.

Solution:

Consider the vectors \(\vec{a}\) = \(\hat {i} \) + 2\(\hat {j} \) + \(\hat {k} \) and \(\vec{b}\) = \(\hat {i} \) + \(\hat {j} \) + 2\(\hat {k} \).

|\(\vec{a}\)| = \(\sqrt{1^{2}+2^{2}+1^{2}}\) = \(\sqrt{6}\), |\(\vec{b}\)| = \(\sqrt{1^{2}+1^{2}+2^{2}}\) = \(\sqrt{6}\)

Thus, two different vectors \(\vec{a}\) and \(\vec{b}\) have the same magnitude.

Question 3.

Write two different vectors having same direction.

Solution:

Let the two vectors be

\(\vec{a}\) = \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \)

\(\vec{b}\) = 3\(\hat {i} \) + 3\(\hat {j} \) + 3\(\hat {k} \)

Direction cosines of \(\vec{a}\) are

(\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\),\(\frac{1}{\sqrt{3}}\)) and direction cosines of \(\vec{b}\)

are (\(\frac{3}{\sqrt{27}}\), \(\frac{3}{\sqrt{27}}\), \(\frac{3}{\sqrt{27}}\)), i.e; (\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\)).

Hence, vectors \(\vec{a}\) and \(\vec{b}\) have the same direction but different magnitude.

Question 4.

Find the values of x and y so that the vectors 2\(\hat {i} \) + 3\(\hat {j} \) and x\(\hat {i} \) + y\(\hat {j} \) are equal.

Solution:

2\(\hat {i} \) + 3\(\hat {j} \) = x\(\hat {i} \) + y\(\hat {j} \)

Equation the coefficients of \(\hat {i} \) and \(\hat {j} \), we get

x = 2, y = 3.

Question 5.

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (- 5, 7).

Solution:

Let A(2, 1) be the initial point and B(- 5, 7) be the terminal point.

âˆ´ \(\overrightarrow{\mathrm{AB}}\) = (x_{2} – x_{1})\(\hat {i} \) + (y_{2} – y_{1})\(\hat {j} \)

= (- 5 – 2)\(\hat {i} \) + (7 – 1)\(\hat {j} \) = – 7\(\hat {i} \) + 6\(\hat {j} \).

âˆ´ The vector components are – 7\(\hat {i} \) and 6\(\hat {j} \) and scalar components are – 7 and 6.

Question 6.

Find the sum of the vectors \(\vec{a}\) = \(\hat {i} \) – 2\(\hat {j} \) + \(\hat {k} \), \(\vec{b}\) = – 2\(\hat {i} \) + 4\(\hat {j} \) + 5\(\hat {k} \) and \(\vec{c}\) = \(\hat {i} \) – 6\(\hat {j} \) – 7\(\hat {k} \).

Solution:

Sum of vectors = \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)

= (\(\hat {i} \) – 2\(\hat {j} \) + \(\hat {k} \)) + (- 2\(\hat {i} \) + 4\(\hat {j} \) + 5\(\hat {k} \)) + (\(\hat {i} \) – 6\(\hat {j} \) – 7\(\hat {k} \))

= (\(\hat {i} \) – 2\(\hat {i} \) + \(\hat {i} \)) + (- 2\(\hat {j} \) + 4\(\hat {j} \) – 6\(\hat {j} \)) + (\(\hat {k} \) + 5\(\hat {k} \) – 7\(\hat {k} \))

= 0\(\hat {i} \) – 4\(\hat {j} \) – \(\hat {k} \) = – 4\(\hat {j} \) – \(\hat {k} \).

Question 7.

Find the unit vector in the direction of the vector \(\vec{a}\) = \(\hat {i} \) + \(\hat {j} \) + 2\(\hat {k} \).

Solution:

\(\vec{a}\) = \(\hat {i} \) + \(\hat {j} \) + 2\(\hat {k} \)

âˆ´ |\(\vec{a}\)| = \(\sqrt{1^{2}+1^{2}+2^{2}}\) = \(\sqrt{6}\).

âˆ´ Unit vector in the direction of vector \(\vec{a}\)

Question 8.

Find the unit vector in the direction of vector \(\overrightarrow{\mathrm{PQ}}\), where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.

Solution:

The points P and Q are (1, 2, 3) and (4, 5, 6) respectively.

âˆ´ The vector \(\overrightarrow{\mathrm{PQ}}\) = (x_{2} – x_{1})\(\hat {i} \) + (y_{2} – y_{1})\(\hat {j} \) + (z_{2} – z_{1})\(\hat {k} \)

âˆ´ \(\overrightarrow{\mathrm{PQ}}\) = (4 – 1)\(\hat {i} \) + (5 – 2)\(\hat {j} \) + (6 – 3)\(\hat {k} \)

= 3\(\hat {i} \) + 3\(\hat {j} \) + 3\(\hat {k} \)

âˆ´ |\(\overrightarrow{\mathrm{PQ}}\)| = \(\sqrt{3^{2}+3^{2}+3^{2}}\) = \(\sqrt{27}\) = 3\(\sqrt{3}\).

âˆ´ Unit vector in the direction of

Question 9.

For given vectors \(\vec{a}\) = 2\(\hat {i} \) – \(\hat {j} \) + 2\(\hat {k} \) and \(\vec{b}\) = – \(\hat {i} \) + \(\hat {j} \) – \(\hat {k} \), find the unit vector in the direction of the vector \(\vec{a}\) + \(\vec{b}\).

Solution:

\(\vec{a}\) = 2\(\hat {i} \) – \(\hat {j} \) + 2\(\hat {k} \) and \(\vec{b}\) = – \(\hat {i} \) + \(\hat {j} \) – \(\hat {k} \)

âˆ´ \(\vec{a}\) + \(\vec{b}\) = (2\(\hat {i} \) – \(\hat {j} \) + 2\(\hat {k} \)) + (-\(\hat {i} \) + \(\hat {j} \) – \(\hat {k} \))

= (2 – 1)\(\hat {i} \) + (-1 + 1)\(\hat {j} \) + (2 – 1)\(\hat {k} \)

= \(\hat {i} \) + 0.\(\hat {j} \) + \(\hat {k} \) = \(\hat {i} \) + \(\hat {k} \)

So, |\(\vec{a}\) + \(\vec{b}\)| = \(\sqrt{1^{2}+1^{2}}\) = \(\sqrt{2}\).

âˆ´ Unit vector in the direction of \(\vec{a}\) + \(\vec{b}\)

= \(\frac{1}{|\vec{a}+\vec{b}|}\)(\(\vec{a}\) + \(\vec{b}\)) = \(\frac{1}{\sqrt{2}}\)(\(\hat {i} \) + \(\hat {k} \)) = \(\frac{1}{\sqrt{2}}\)\(\hat {i} \) + \(\frac{1}{\sqrt{2}}\)\(\hat {k} \)

Question 10.

Find a vector in the direction of 5\(\hat {i} \) – \(\hat {j} \) + 2\(\hat {k} \) which has magnitude 8 units.

Solution:

The given vector is

\(\vec{a}\) = 5\(\hat {i} \) – \(\hat {j} \) + 2\(\hat {k} \).

âˆ´ |\(\vec{a}\)| = \(\sqrt{5^{2}+(-1)^{2}+2^{2}}\) = \(\sqrt{25+1+4}\) = \(\sqrt{30}\).

âˆ´ Unit vector in the direction of vector \(\vec{a}\)

= \(\frac{1}{\sqrt{30}}\)(5\(\hat {i} \) – \(\hat {j} \) + 2\(\hat {k} \).

âˆ´Vector of magnitude 8 in the direction of vector \(\vec{a}\)

Question 11.

Show that the vectors 2\(\hat {i} \) – 3\(\hat {j} \) + 4\(\hat {k} \) and – 4\(\hat {i} \) + 6\(\hat {j} \) – 8\(\hat {k} \) are collinear.

Solution:

\(\vec{a}\) = – 2\(\hat {i} \) – 3\(\hat {j} \) + 4\(\hat {k} \)

\(\vec{b}\) = – 4\(\hat {i} \) + 6\(\hat {j} \) – 8\(\hat {k} \) = – 2(2\(\hat {i} \) – 3\(\hat {j} \) + 4\(\hat {k} \))

Vectors \(\vec{a}\) and \(\vec{b}\) have the same direction. Therefore, they are collinear.

Question 12.

Find the direction cosines of the vectors \(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \).

Solution:

Directions cosines of vector x\(\hat {i} \) + y\(\hat {j} \) + z\(\hat {k} \) are

Here, for the vector \(\vec{a}\) = \(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \)

x = 1, y = 2, z = 3

âˆ´ \(\sqrt{x^{2}+y^{2}+z^{2}}\) = \(\sqrt{1^{2}+2^{2}+3^{2}}\) = \(\sqrt{1+4+9}\) = \(\sqrt{14}\).

âˆ´ Direction of cosines are given by

Question 13.

Find the direction cosines of the vector joining the points A(1, 2, – 3) and B(- 1, – 2, 1), directed from A to B.

Solution:

Vector joining the points A and B is

(x_{2} – x_{1})\(\hat {i} \) + (y_{2} – y_{1})\(\hat {j} \) + (z_{2} – z_{1})\(\hat {k} \)

Here, A is (1, 2, – 3)

âˆ´ x_{1} = 1, y_{1} = 2, z_{1} = – 3.

B is (- 1, – 2, 1)

âˆ´ x_{2} = – 1, y_{2} = – 2, z_{2} = 1.

âˆ´ Vector joining the points A and B

= \(\overrightarrow{\mathrm{AB}}\) = (- 1 – 1)\(\hat {i} \) + (- 2 – 2)\(\hat {j} \) + [1 – (- 3)]\(\hat {k} \)

= – 2\(\hat {i} \) – 4\(\hat {j} \) + 4\(\hat {k} \).

Direction cosines of vectors x\(\hat {i} \) + y\(\hat {j} \) + z\(\hat {k} \) are

Question 14.

Show that the vector \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \) is equally inclined to the axes OX, OY and OZ.

Solution:

Let \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \) = a.

Direction cosines of vector x\(\hat {i} \) + y\(\hat {j} \) + z\(\hat {k} \) are

which shows that the vector \(\vec{a}\) is equally inclined to the axes OX, OY and OZ.

Question 15.

Find the position vector of a point R which divides the line segment joining the points, whose positive vectors are

P (\(\hat {i} \) + 2\(\hat {j} \) – \(\hat {k} \)) and Q(- \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \)), in the ratio 2 : 1

(i) internally,

(ii) externally.

Solution:

(i) Internal Division:

The point R which divides the line segment joining the points P(\(\vec{a}\)) and Q(\(\vec{b}\)) in the ratio m : n is given by

(ii) External Division:

Similarly, the point R’ which divides PQ externally in the ratio m : n is given by

Question 16.

Find the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, – 2).

Solution:

Mid point of the vector joining the points P(\(\vec{a}\)) and Q(\(\vec{b}\)) is \(\frac{\vec{a}+\vec{b}}{2}\).

Here, \(\vec{a}\) = 2\(\hat {i} \) + 3\(\hat {j} \) + 4\(\hat {k} \) and \(\vec{b}\) = 4\(\hat {i} \) + \(\hat {j} \) – 2\(\hat {k} \).

âˆ´ Required mid-point is

Question 17.

Show that the points A, B and C with position vectors \(\vec{a}\) = 3\(\hat {i} \) – 4\(\hat {j} \) – 4\(\hat {k} \),

\(\vec{b}\) = 2\(\hat {i} \) – \(\hat {j} \) + \(\hat {k} \) and \(\vec{c}\) = \(\hat {i} \) – 3\(\hat {j} \) – 5\(\hat {k} \), respectively form the vertices of a right angled traiangle.

Solution:

\(\overrightarrow{\mathrm{AB}}\) = \(\vec{b}\) – \(\vec{a}\)

= (2\(\hat {i} \) – \(\hat {j} \) + \(\hat {k} \)) – (3\(\hat {i} \) – 4\(\hat {j} \) – 4\(\hat {k} \))

= – \(\hat {i} \) + 3\(\hat {j} \) + 5\(\hat {k} \).

So, |\(\overrightarrow{\mathrm{AB}}\)| = (- 1)^{2} + 3^{2} + 5^{2}

= 1 + 9 + 25 = 35

\(\overrightarrow{\mathrm{BC}}\) = \(\vec{c}\) – \(\vec{b}\)

= (\(\hat {i} \) – 3\(\hat {j} \) – 5\(\hat {k} \)) – (2\(\hat {i} \) – \(\hat {j} \) + \(\hat {k} \))

= (1 – 2)\(\hat {i} \) + (- 3 + 1)\(\hat {j} \) + (- 5 – 1)\(\hat {k} \) = – \(\hat {i} \) – 2\(\hat {j} \) – 6\(\hat {k} \).

âˆ´ |\(\overrightarrow{\mathrm{BC}}\)|^{2} = (- 1)^{2} + (- 2)^{2} + (- 6)^{2} = 1 + 4 + 36 = 41.

\(\overrightarrow{\mathrm{CA}}\) = \(\vec{a}\) – \(\vec{c}\) = (3\(\hat {i} \) – 4\(\hat {j} \) – 4\(\hat {k} \)) – (\(\hat {i} \) – 3\(\hat {j} \) – 5\(\hat {k} \))

= (3 – 1)\(\hat {i} \) + (- 4 + 3)\(\hat {j} \) + (- 4 + 5)\(\hat {k} \) = 2\(\hat {i} \) – \(\hat {j} \) + \(\hat {k} \).

|\(\overrightarrow{\mathrm{CA}}\)|^{2} = 2^{2} + (- 1)^{2} = 4 + 1 + 1 = 6.

Now, |\(\overrightarrow{\mathrm{AB}}\)|^{2} + |\(\overrightarrow{\mathrm{AC}}\)|^{2} = 35 + 6 = 41 = |\(\overrightarrow{\mathrm{BC}}\)|^{2}

or |\(\overrightarrow{\mathrm{AB}}\)|^{2} + |\(\overrightarrow{\mathrm{AC}}\)|^{2} = |\(\overrightarrow{\mathrm{BC}}\)|^{2}

Hence, âˆ† ABC is a right angled traiangle, right angled at A.

Question 18.

In triangle ABC, which of the following is not true?

(A) \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) + \(\overrightarrow{\mathrm{CA}}\) = \(\vec{0}\)

(B) \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) – \(\overrightarrow{\mathrm{AC}}\) = \(\vec{0}\)

(C) \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) – \(\overrightarrow{\mathrm{CA}}\) = \(\vec{0}\)

(D) \(\overrightarrow{\mathrm{AB}}\) – \(\overrightarrow{\mathrm{CB}}\) + \(\overrightarrow{\mathrm{CA}}\) = \(\vec{0}\)

Solution:

By traingle law of vector addition,

\(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) = \(\overrightarrow{\mathrm{AC}}\) or \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) = – \(\overrightarrow{\mathrm{CA}}\)

or \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) + \(\overrightarrow{\mathrm{CA}}\) = \(\vec{0}\)

Thus, part (C) \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) – \(\overrightarrow{\mathrm{CA}}\) = \(\vec{0}\) is not true.

Question 19.

If \(\vec{a}\) and \(\vec{b}\) are two collinear vectors, then which of the following is incorrect:

(A) \(\vec{b}\) = Î»\(\vec{a}\), for some scalar

(B) \(\vec{a}\) = Â± \(\vec{b}\)

(C) the respective components of \(\vec{a}\) and \(\vec{b}\) are proportional.

(D) both the vectors \(\vec{a}\) and \(\vec{b}\) have same direction, but different magnitude.

Solution:

Part (D) is incorrect since both the vectors \(\vec{a}\) and \(\vec{b}\), being collinear, are not necessarily in the same direction.

They may have opposite directions. Their magnitudes may be different.