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  • Provide solution for RD Sharma Maths Class 12 Chapter 26 Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 12.

Provide solution for RD Sharma Maths Class 12 Chapter 26 Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 12.

Answers (1)

Final Answer:

(d) \cos ^{-1}\frac{1}{3}

Given: A cube

To Find: Angle between its diagonals

Solution: Let us consider a cube OABCDEFG with vertices as shown

O(0,0,0),  A(a,0,0), B(a,a,0), C(0,a,0), D(0,a,a), E(0,0,a), F(a,0,a), G(a,a,a)

There are four diagonals OG, CF, AD, and BE

Let us consider OG and AD

\overrightarrow{OG} = Position vector of G- Position vector of O

\begin{aligned} &\overrightarrow{O G}=(a-0) \hat{i}+(a-0) \hat{j}+(a-0) \hat{k} \\ &\overrightarrow{O G}=a \hat{i}+a j+a \hat{k} \end{aligned}

Also,

\overrightarrow{AD} =Position vector of D - Position vector of A

\begin{aligned} &\overrightarrow{A D}=(0-a) \hat{i}+(a-0) \hat{j}+(a-0) \hat{k} \\ &\overrightarrow{A D}=-a i+a \hat{j}+a \hat{k} \end{aligned}

Now using dot product

\begin{aligned} &\overrightarrow{O G} \cdot \overrightarrow{A D}=|\overrightarrow{O G} \| \overrightarrow{A D}| \cos \theta \\ &\cos \theta=\frac{\overrightarrow{O G} \cdot \overrightarrow{A D}}{|\overrightarrow{O G}||\overrightarrow{A D}|} \\ &\cos \theta=\frac{(a \hat{i}+a \hat{j}+a \hat{k}) \cdot(-a \hat{i}+a \hat{j}+a k)}{\sqrt{a^{2}+a^{2}+a^{2}} \sqrt{a^{2}+a^{2}+a^{2}}} \end{aligned}

\begin{aligned} &\cos \theta=\frac{-a^{2}+a^{2}+a^{2}}{a \sqrt{3} \times a \sqrt{3}} \\ &\cos \theta=\frac{a^{2}}{a^{2} \times 3} \\ &\theta=\cos ^{-1}\left(\frac{1}{3}\right) \end{aligned}

 

Therefore angle between diagonals of cube is

\cos ^{-1}\frac{1}{3}

Hence option (d) is correct.

Posted by

infoexpert24

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