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  • Provide solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise Fill in the blanks question 14

Provide solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise Fill in the blanks question 14

Answers (1)

Final Answer:  \frac{\pi }{3}

Hint:

Dot product of two direction cosine two vectors

Given:

\begin{aligned} &\left(a_{1}, b_{1}, c_{1}\right)=(1,1,2) \text { and } \\ &\left(a_{2}, b_{2}, c_{2}\right)=(\sqrt{3}-1,-\sqrt{3}-1,4) \end{aligned}

To Find: 

Angle between two vectors

Solution:

Angle between two vectors with direction ratios \left(a, b, c_{1}\right) \text { and }\left(a_{2}, b_{2}, c_{2}\right)  is
given by

\cos \theta=\left[\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right]

\text { Now, } \sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}=\sqrt{(1)^{2}+(1)^{2}+(2)^{2}}=\sqrt{1+1+4}=\sqrt{6}

\text { Also } \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}=\sqrt{(\sqrt{3}-1)^{2}+(-\sqrt{3}-1)^{2}+4^{2}}

                                      =\sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}+16}=\sqrt{24}

\text { Hence } \cos \theta=\left[\frac{1(\sqrt{3}-1)+1 \cdot(-\sqrt{3}-1)+2 \cdot 4}{\sqrt{6} \cdot \sqrt{24}}\right]

\begin{aligned} &\cos \theta=\left[\frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6 \times 24}}\right] \\ &\cos \theta=\frac{6}{12} \end{aligned}

\begin{aligned} &\theta=\cos ^{-1}\left[\frac{1}{2}\right] \\ &\theta=\frac{\pi}{3} \end{aligned}

Therefore, the angle between two vectors is \frac{\pi }{3}

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