Q81 / 125 · BITSAT 2013 · Mathematics
Find the angle between the two planes $2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}=5$ and $3 x-6 y-2 z=7$
- Correct answerA$\cos ^{-1}(4 / 21)$
- B$\cos ^{-1}(2 / 21)$
- C$\cos ^{-1}(1 / 21)$
- D$\cos ^{-1}(5 / 21)$
Solution
The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are $\overrightarrow{\mathrm{N}}_{1}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{N}}_{2}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
Therefore, $\cos \theta=\left|\frac{\overrightarrow{\mathrm{N}}_{1} \cdot \overrightarrow{\mathrm{N}}_{2}}{\left|\overrightarrow{\mathrm{N}}_{1}\right|\left|\overrightarrow{\mathrm{N}}_{2}\right|}\right|$
$$
=\left|\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{\sqrt{4+1+4} \sqrt{9+36+4}}\right|=\left(\frac{4}{21}\right)
$$
Hence, $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$
Hence, the correct answer is option: 'A'.