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Provide solution for RD Sharma Maths Class 12 Chapter 26 Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 13.

Answers (1)

Final Answer:

(c) \frac{4}{3}

Hint: Use direction cosines

Given: Line makes angles

\alpha ,\beta ,\gamma ,\delta

with diagonals of cube.

To Find:

\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+\cos ^{2} \delta

Solution: Let us consider a cube as shown

Direction ratio of OP are (a-0,a-0,a-0)

=(a,a,a)

Direction cosine of OP are

\begin{aligned} &\left(\frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}\right) \\ &\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \end{aligned}

Similarly, direction cosine of all diagonals

\begin{aligned} &O P=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \\ &B N=\left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \\ &A M=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right) \\ &C L=\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \end{aligned}

Let (l,m,n) are the direction cosine of line which is inclined at angle

\alpha ,\beta ,\gamma ,\delta with diagonals. Now, using dot product

\begin{aligned} &\cos \alpha=\frac{l+m+n}{\sqrt{3}} \\ &\cos \beta=\frac{-l+m+n}{\sqrt{3}} \\ &\cos \gamma=\frac{l+m-n}{\sqrt{3}} \\ &\cos \delta=\frac{l-m+n}{\sqrt{3}} \end{aligned}

Squaring and adding above equations, we get

\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+\cos ^{2} \delta=\frac{1}{3}\left[(l+m+n)^{2}+(-l+m+n)^{2}+(l+m-n)^{2}+(l-m+n)^{2}\right]\

\begin{aligned} &=\frac{1}{3}\left[4\left(l^{2}+m^{2}+n^{2}\right)\right] \\ &=\frac{4}{3} \end{aligned}

Therefore, value of

\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+\cos ^{2} \delta=\frac{4}{3}

Hence option(c) is correct.

Posted by

infoexpert24

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