Question

The base of the pyramid AOBC is an equilateral triangle OBC with each side equal to $4\sqrt{2}$, 'O' is the origin of reference, AO is perpendicular to the plane of $\triangle OBC$ and $|\overrightarrow{AO}| = 2$. Then the cosine of the angle between the skew straight lines one passing through A and the mid point of OB and the other passing through O and the mid point of BC is :

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{3}{\sqrt{20}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A), (D)

$\frac{1}{\sqrt{2}}$

Answer 2

Solution Outline

1. Coordinate Assignment

   • Let $O=(0,0,0)$.
   • Place $B=(4\sqrt2,0,0)$.
   • Place $C=\bigl(2\sqrt2,2\sqrt6,0\bigr)$ so that $\triangle OBC$ is equilateral of side $4\sqrt2$.
   • Since $AO\perp$ plane $OBC$ and $|AO|=2$, take $A=(0,0,2)$.

2. Mid-points

   • Mid-point of $OB$, $M_1 =\bigl(2\sqrt2,0,0\bigr)$.
   • Mid-point of $BC$, $M_2 =\bigl(3\sqrt2,\sqrt6,0\bigr)$.

3. Direction Vectors

   • For the line through $A$ and $M_1$:
     $\displaystyle \mathbf{v} = M_1 - A = (2\sqrt2,0,-2) = 2(\sqrt2,0,-1)$.
   • For the line through $O$ and $M_2$:
     $\displaystyle \mathbf{w} = M_2 - O = (3\sqrt2,\sqrt6,0)$.

4. Compute Cosine

   $$\mathbf{v}\cdot\mathbf{w} = 2\bigl(\sqrt2\cdot3\sqrt2 +0\cdot\sqrt6 +(-1)\cdot0\bigr) = 12,$$ $$\|\mathbf{v}\| = 2\sqrt{(\sqrt2)^2+(-1)^2} = 2\sqrt3,\quad \|\mathbf{w}\| = \sqrt{(3\sqrt2)^2+(\sqrt6)^2} = 2\sqrt6.$$

   Hence

   $$\cos\theta = \frac{|\mathbf{v}\cdot\mathbf{w}|}{\|\mathbf{v}\|\,\|\mathbf{w}\|} = \frac{12}{(2\sqrt3)(2\sqrt6)} = \frac{12}{4\sqrt{18}} = \frac{3}{\sqrt{18}} = \frac{1}{\sqrt2}.$$