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If $\theta$ is the acute angle between the diagonals of a cube, then which one of the following is correct?
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Verified Answer
The correct answer is:
$3 \cos \theta=1$
diagonals are OP and AD and Acute anngle $=\theta$
$\cos \theta=\left|\frac{\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}}{\sqrt{\mathrm{a}_{1}^{2}+\mathrm{b}_{1}^{2}+\mathrm{c}_{1}^{2}} \sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}\right|$
$=\left|\frac{\mathrm{a}(-\mathrm{a})+(\mathrm{a})(\mathrm{a})+(\mathrm{a})(\mathrm{a})}{\sqrt{\mathrm{a}^{2}+\mathrm{a}^{2}+\mathrm{a}^{2}} \sqrt{\mathrm{a}^{2}+\mathrm{a}^{2}+\mathrm{a}^{2}}}\right|$
$=\left|\frac{-\mathrm{a}^{2}+\mathrm{a}^{2}+\mathrm{a}^{2}}{\sqrt{3 \mathrm{a}^{2}} \sqrt{3 \mathrm{a}^{2}}}\right|=\left|\frac{\mathrm{a}^{2}}{3 \mathrm{a}^{2}}\right|=\frac{1}{3}$
$\Rightarrow 3 \cos \theta=1$
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