Question

A cube is placed so that one corner is at the origin and three edges are along the $x-,y-,$ and $z$-axes of a coordinate system. Use vectors to compute
a. The angle between the edge along the $z$-axis (line ab) and the diagonal from the origin to the opposite corner (line ad).
b. The angle between line $ac$ (the diagonal of a face) and line $ad$.

• A. a. ${\cos }^{-1}\frac{1}{\sqrt{2}}$
  b. ${\cos }^{-1}\frac{\sqrt{2}}{\sqrt{3}}$
• B. a. ${\cos }^{-1}\frac{1}{\sqrt{3}}$
  b. ${\cos }^{-1}\frac{\sqrt{2}}{\sqrt{3}}$
• C. a. ${\cos }^{-1}\frac{1}{\sqrt{3}}$
  b. ${\cos }^{-1}\frac{\sqrt{3}}{\sqrt{3}}$
• D. a. ${\cos }^{-1}\frac{2}{\sqrt{3}}$
  b. ${\cos }^{-1}\frac{\sqrt{2}}{\sqrt{3}}$

Answer 1

The correct option is B a. ${\cos }^{-1}\frac{1}{\sqrt{3}}$

b. ${\cos }^{-1}\frac{\sqrt{2}}{\sqrt{3}}$

$\cos \theta =\frac{a}{\sqrt{3}a}$

$\theta ={\cos }^{-1}\frac{1}{\sqrt{3}}$

$\cos \theta =\frac{{\left(\sqrt{2}a\right)}^2+{\left(\sqrt{3}a\right)}^2-a^2}{2\times \sqrt{3}a\times \sqrt{2}a}$

$=\frac{{4a}^2}{2\sqrt{3}\sqrt{2}{a}^2}$

$=\frac{\sqrt{2}}{3}$

$\theta ={\cos }^{-1}\frac{\sqrt{2}}{3}$