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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

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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