Question

lf $O=(0,0,0),P=(4,3,-5),Q=(-2,1,-8),\cos \angle POQ=\frac{a}{\sqrt{b}\sqrt{c}}$ and $b>c$, then $\sqrt{b}-\sqrt{c}$ is

• A. $19$
• B. $\sqrt{69}-\sqrt{50}$
• C. $-19$
• D. $17$

Answer 1

Answer: (B)

$\sqrt{69}-\sqrt{50}$

$\cos \left(POQ\right)=\frac{a}{\sqrt{b}\sqrt{c}}$
$\cos \left(POQ\right)=\frac{-8+3+40}{\sqrt{16+9+25}\sqrt{4+1+64}}$
$\cos \left(POQ\right)=\frac{35}{\sqrt{69}\sqrt{50}}$
$\sqrt{b}-\sqrt{c}=\sqrt{69}-\sqrt{50}$