RD Sharma Class 12 Ex 23.9 Solutions Chapter 23 Algebra of Vectors

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Here we provide RD Sharma Class 12 Ex 23.9 Solutions Chapter 23 Algebra of Vectors for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 23.9 Solutions Chapter 23 Algebra of Vectors book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter23.9
Exercise23.9
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 23.9 Solutions Chapter 23 Algebra of Vectors

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1: Can a vector have direction angles 45°, 60°, and 120°.

Solution:

We know that if l, m and n are the direction cosines and \alpha , \beta  and \gamma  are the direction angles then,

=> l = \cos \alpha = \cos 45 \degree = \dfrac{1}{\sqrt{2}}

=> m = \cos \beta = \cos 60 \degree = \dfrac{1}{2}

=> n = \cos \gamma = \cos 120 \degree = \dfrac{1}{2}

Also,

=> l2 + m2 + n2 = 1

=> (\dfrac{1}{\sqrt{2}})^2 + (\dfrac{1}{2})^2+ (\dfrac{1}{2})^2 = 1

=> \dfrac{1}{2}+\dfrac{1}{4} +\dfrac{1}{4} = 1

=> As LHS = RHS, the vector can have these direction angles.

Question 2: Prove that 1,1 and 1 can not be the direction cosines of a straight line.

Solution:

Given that, l=1, m=1 and n=1.

We know that,

=> l2 + m2 + n2 = 1

=> 12 + 12 + 12 = 1

=> 3 ≠ 1

Thus, 1, 1 and 1 can never be the direction cosines of a straight line.

=> Hence proved.

Question 3: A vector makes an angle of \dfrac{\pi}{4}  with each of x-axis and y-axis. Find the angle made by it with the z-axis.

Solution:

We know that if l, m and n are the direction cosines and \alpha , \beta  and \gamma  are the direction angles then,

=> l = \cos \alpha = \cos 45 \degree = \dfrac{1}{\sqrt{2}}

=> m = \cos \beta = \cos 45 \degree = \dfrac{1}{\sqrt{2}}

Let \gamma be the angle we have to calculate.

We know that,

=> l2 + m2 + n2 = 1

=> n^2 = 1- [(\dfrac{1}{\sqrt{2}})^2+(\dfrac{1}{\sqrt{2}})^2]

=> n2 = 1 – 1

=> n2 = 0

=> \cos^2 \gamma = 0

=> \cos \gamma = 0

=> \gamma = \cos^{-1}0

=> \gamma = \dfrac{\pi}{2}

Question 4: A vector \vec{r}  is inclined at equal acute angles to x-axis, y-axis and z-axis. If |\vec{r}|  = 6 units, find \vec{r} .

Solution:

Given that \alpha=\beta=\gamma

=> \cos \alpha = \cos \beta = \cos \gamma

=> l = m = n = p (say)

We know that,

=> l2 + m2 + n2 = 1

=> p2 + p2 + p2 = 1

=> 3p2 = 1

=> p = \pm \dfrac{1}{\sqrt{3}}

The vector \vec{r}  can be described as,

=> \vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})

=> \vec{r} = 6(\pm \dfrac{1}{\sqrt{3}} \hat{i}+ \pm \dfrac{1}{\sqrt{3}}\hat{j} +\pm \dfrac{1}{\sqrt{3}}\hat{k})

=> \vec{r} = \pm2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})

Question 5: A vector \vec{r}  is inclined to the x-axis at 45° and y-axis at 60°. If |\vec{r}|=8  units, find \vec{r} .

Solution:

Given that \alpha = 45 \degree  and \beta = 60 \degree

We know that,

=> l2 + m2 + n2 = 1

=> \cos ^2 \alpha + \cos ^2 \beta + \cos^2 \gamma = 1

=> (\dfrac{1}{\sqrt{2}})^2+ (\dfrac{1}{2})^2 + \cos^2 \gamma = 1

=> \cos^2 \gamma = 1- [\dfrac{1}{2}+\dfrac{1}{4}]

=> \cos ^2 \gamma = 1- \dfrac{3}{4}

=> \cos ^2 \gamma = \dfrac{1}{4}

=> \cos \gamma = \pm \dfrac{1}{2}

The vector \vec{r}  can be described as,

=> \vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})

=> \vec{r} = 8(\dfrac{1}{\sqrt{2}} \hat{i}+ \dfrac{1}{2}\hat{j} +\pm \dfrac{1}{2}\hat{k})

=> \vec{r} = 4(\sqrt{2}\hat{i}+\hat{j}+\pm \hat{k})

Question 6: Find the direction cosines of the following vectors:

(i): 2\hat{i}+ 2\hat{j}-\hat{k}

Solution:

The direction ratios are given as 2, 2 and -1.

Direction cosines are given as,

=> \cos \alpha, \cos \beta, \cos \gamma = \dfrac{2}{|\vec{r}|}, \dfrac{2}{|\vec{r}|}, \dfrac{-1}{|\vec{r}|}

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{2}{\sqrt{2^2+2^2+(-1)^2}}, \dfrac{2}{\sqrt{2^2+2^2+(-1)^2}}, \dfrac{-1}{\sqrt{2^2+2^2+(-1)^2}}

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{-1}{3}

(ii): 6\hat{i}-2\hat{j}-3\hat{k}

Solution:

The direction ratios are given as 6, -2 and -3.

Direction cosines are given as,

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{6}{|\vec{r}|}, \dfrac{-2}{|\vec{r}|}, \dfrac{-3}{|\vec{r}|}

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{6}{\sqrt{6^2+(-2)^2+(-3)^2}}, \dfrac{-2}{\sqrt{6^2+(-2)^2+(-3)^2}}, \dfrac{-3}{\sqrt{6^2+(-2)^2+(-3)^2}}

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{6}{7}, \dfrac{-2}{7}, \dfrac{-3}{7}

(iii): 3\hat{i}-4\hat{k}

Solution:

The direction ratios are given as 3, 0 and -4.

Direction cosines are given as,

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{3}{|\vec{r}|}, \dfrac{0}{|\vec{r}|}, \dfrac{-4}{|\vec{r}|}

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{3}{\sqrt{3^2+0^2+(-4)^2}}, \dfrac{0}{\sqrt{3^2+0^2+(-4)^2}}, \dfrac{-4}{\sqrt{3^2+0^2+(-4)^2}}

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{3}{5}, 0 , \dfrac{-4}{5}

Question 7: Find the angles at which the following vectors are inclined to each of the coordinates axes.

(i): \hat{i}-\hat{j}+\hat{k}

Solution:

The given direction ratios are: 1,-1,1.

Thus,

=> l, m, n = \dfrac{1}{|\vec{r}|}, \dfrac{-1}{|\vec{r}|}, \dfrac{1}{|\vec{r}|}

=> l, m, n = \dfrac{1}{\sqrt{1^2+(-1)^2+1^2}}, \dfrac{-1}{\sqrt{1^2+(-1)^2+1^2}}, \dfrac{1}{\sqrt{1^2+(-1)^2+1^2}}

=> l, m, n = \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}} , \dfrac{1}{\sqrt{3}}

=> \cos \alpha , \cos \beta , \cos \gamma = \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}} , \dfrac{1}{\sqrt{3}}

=> \alpha , \beta , \gamma = \cos ^{-1} (\dfrac{1}{\sqrt{3}}), \cos ^{-1} (\dfrac{-1}{\sqrt{3}}), \cos ^{-1} (\dfrac{1}{\sqrt{3}})

(ii): \hat{j}-\hat{k}

Solution:

The given direction ratios are: 0,1,-1.

Thus,

=> l, m, n = \dfrac{0}{|\vec{r}|}, \dfrac{1}{|\vec{r}|}, \dfrac{-1}{|\vec{r}|}

=> l, m, n = \dfrac{0}{\sqrt{0^2+1^2+(-1)^2}}, \dfrac{1}{\sqrt{0^2+1^2+(-1)^2}}, \dfrac{-1}{\sqrt{0^2+1^2+(-1)^2}}

=> l, m, n = 0 , \dfrac{1}{\sqrt{2}} , \dfrac{-1}{\sqrt{2}}

=> \cos \alpha , \cos \beta , \cos \gamma = 0, \dfrac{1}{\sqrt{2}} , \dfrac{-1}{\sqrt{2}}

=> \alpha , \beta , \gamma = \cos ^{-1}(0), \cos ^{-1} (\dfrac{1}{\sqrt{2}}), \cos ^{-1} (\dfrac{-1}{\sqrt{2}})

=> \alpha , \beta , \gamma = \dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}

(iii): 4\hat{i}+8\hat{j}+\hat{k}

Solution:

The given direction ratios are: 4, 8, 1.

Thus,

=> l, m, n = \dfrac{4}{|\vec{r}|}, \dfrac{8}{|\vec{r}|}, \dfrac{1}{|\vec{r}|}

=> l, m, n = \dfrac{0}{\sqrt{4^2+8^2+1^2}}, \dfrac{8}{\sqrt{4^2+8^2+1^2}}, \dfrac{1}{\sqrt{4^2+8^2+1^2}}

=> l, m, n = \dfrac{4}{9}, \dfrac{8}{9} , \dfrac{1}{9}

=> \cos \alpha , \cos \beta , \cos \gamma = \dfrac{4}{9}, \dfrac{8}{9} , \dfrac{1}{9}

=> \alpha , \beta , \gamma = \cos ^{-1}(\dfrac{4}{9}), \cos ^{-1} (\dfrac{8}{9}), \cos ^{-1} (\dfrac{1}{9})

Question 8: Show that the vector \hat{i}+\hat{j}+\hat{k}  is equally inclined with the axes OX, OY and OZ.

Solution:

Let \vec{r}= \hat{i}+\hat{j}+\hat{k}

Thus, |\vec{r}| = \sqrt{1^2+1^2+1^2}

=> |\vec{r}| = \sqrt{3}

Thus the direction cosines are: \dfrac{1}{|\vec{r}|} , \dfrac{1}{|\vec{r}|}  and \dfrac{1}{|\vec{r}|}

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}

Thus,

=> \cos \alpha, \cos \beta , \cos \gamma = \dfrac{1}{\sqrt{3}}

=> Thus, the vector is equally inclined with the 3 axes.

Question 9: Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}} .

Solution:

Let the vector be equally inclined at an angle of \alpha .

Then the direction cosines of the vector l, m, n are: \cos \alpha , \cos \beta  and \cos \gamma

We know that,

=> l2 + m2 + n2 = 1

=> \cos^2 \alpha + \cos^2 \alpha+ \cos^2 \alpha = 1

=> 3\cos^2 \alpha = 1

=> \cos \alpha = \pm \dfrac{1}{\sqrt{3}}

=> Thus the direction cosines are: \pm \dfrac{1}{\sqrt{3}} , \pm \dfrac{1}{\sqrt{3}}, \pm \dfrac{1}{\sqrt{3}}.

Question 10: If a unit vector \vec{a}  makes an angle \dfrac{\pi}{3}  with \hat{i}, \dfrac{\pi}{4}  with \hat{j}  and an acute angle \theta  with \hat{k} , then find \theta and hence the components of \vec{a}.

Solution:

The unit vector be,

=> \vec{r} = r_1\hat{i}+r_2\hat{j}+r_3\hat{k}

=> r_1 , r_2, r_3 = \cos (\dfrac{\pi}{3}), \cos (\dfrac{\pi}{4}), \cos \theta

Given that \vec{r}  is unit vector,

=> |\vec{r}| = 1

=> \sqrt{r_1^2+r_2^2+ r_3^2 }= 1

=> r_1^2+r_2^2+ r_3^2 = 1

=>  (\cos (\dfrac{\pi}{3}))^2 + (\cos (\dfrac{\pi}{4}))^2 + (\cos \theta)^2 =1

=> (\cos \theta)^2 = 1-[\dfrac{1}{2}+\dfrac{1}{4}]

=> (\cos \theta)^2 = 1- \dfrac{3}{4}

=> (\cos \theta)^2 = \dfrac{1}{4}

=> \cos \theta = \dfrac{1}{2}

=> \theta = \cos ^{-1} (\dfrac{1}{2})

=> \theta = \dfrac{\pi}{3}

Question 11: Find a vector \vec{r} of magnitude 3\sqrt{2}  units which makes an angle of \dfrac{\pi}{4}  and \dfrac{\pi}{2}  with y and z axes respectively.

Solution:

Let l, m, n be the direction cosines of the vector \vec{r} .

We know that,

=> l2 + m2 + n2 = 1

=> \cos ^2 \alpha+ \cos ^2 \beta + \cos ^2 \gamma =1

=> l^2 =1 - [(\dfrac{1}{\sqrt{2}})^2+(0)^2]

=> l^2 = 1- \dfrac{1}{2}

=> l = \pm \dfrac{1}{\sqrt{2}}

Thus vector is,

=> \vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})

=> \vec{r} = 3\sqrt{2}(\pm \dfrac{1}{\sqrt{2}} \hat{i} + \dfrac{1}{\sqrt{2}}\hat{j} + 0\hat{k})

=> \vec{r} = \pm 3\hat{i} + 3\hat{j}

Question 12: A vector \vec{r}  is inclined at equal angles to the 3 axes. If the magnitude of \vec{r} is 2\sqrt{3} , find \vec{r} .

Solution:

Let l, m, n be the direction cosines of the vector \vec{r} .

Given that the vector is inclined at equal angles to the 3 axes.

=> l = m = n = \cos \alpha = \cos \beta = \cos \gamma

We know that,

=> l2 + m2 + n2 = 1

=> 3\cos ^2\alpha = 1

=> \cos \alpha = \pm \dfrac{1}{\sqrt{3}}

Hence, the vector is given as,

=> \vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})

=> \vec{r} = 2\sqrt{3}(\pm \dfrac{1}{\sqrt{3}}\hat{i}+ \pm \dfrac{1}{\sqrt{3}}\hat{j}+ \pm \dfrac{1}{\sqrt{3}}\hat{k})

=> \vec{r} = \pm 2(\hat{i}+\hat{j}+\hat{k})

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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