AP EAMCET · Maths · Vector Algebra
Angle between a diagonal of a cube and a diagonal of its face which are coterminus is
- A \(\frac{\pi}{2}\)
- B \(\operatorname{Cos}^{-1}\left(\sqrt{\frac{2}{3}}\right)\)
- C \(\operatorname{Cos}^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
- D \(\operatorname{Cos}^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(\operatorname{Cos}^{-1}\left(\sqrt{\frac{2}{3}}\right)\)
Step-by-step Solution
Detailed explanation
\(\vec{d_c} = a\hat{i} + a\hat{j} + a\hat{k}\), \(\vec{d_f} = a\hat{i} + a\hat{j}\) \(\cos\theta = \frac{\vec{d_c} \cdot \vec{d_f}}{|\vec{d_c}| |\vec{d_f}|}\) \(\cos\theta = \frac{(a)(a) + (a)(a) + (a)(0)}{\sqrt{a^2+a^2+a^2} \cdot \sqrt{a^2+a^2}}\)…
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