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

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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,

=> l+ m+ 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,

=> l+ m+ n2 = 1

=> 1+ 1+ 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,

=> l+ m+ 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,

=> l+ m+ n2 = 1

=> p+ p+ 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,

=> l+ m+ 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,

=> l+ m+ 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,

=> l+ m+ 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,

=> l+ m+ 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})

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