Question

If $p$ is the perpendicular distance of an angular point of a cube of each side equal to 47 from a diagonal which does not pass through that angular point, then $3p^2$ is equal to

Answer 1

Given cube of side $47$

Angular point is $P(47,47,47)$

The diagonal which does not pass through angular point passes through $(47,47,0)$ and $(0,0,47)$

Line equation diagonal is

$\frac{x-0}{47}=\frac{y}{47}=\frac{z-47}{-47}\left(1\right)$

To find perpendicular distance from $P$ to line, we take arbitrary point on line

$\frac{x}{47}=\frac{y}{47}=\frac{z-47}{-47}=tx=47t,y=47t,z=-47t+47$

Let $Q(47t,47t,-47t+47)$

d.r's of PQ is perpendicular to line $\left(1\right)$

By perpendicular property

$47(47t-47)+47(47t-47)-47\left(47t\right)=0t=\frac{2}{3}$

Substitute $t=\frac{2}{3}$ in $Q$ point

We get $Q=(\frac{47\times 2}{3},\frac{47\times 2}{3},\frac{+47}{3})$

Perpendicular distance $=PQ=P$

By distance formula

$P=\sqrt{{(47-47\times \frac{2}{3})}^2+{(47-47\times \frac{2}{3})}^2+{(47\times \frac{2}{3})}^2}$

Squaring on both sides

$P^2={\left(\frac{47}{3}\right)}^2+{\left(\frac{47}{3}\right)}^2+4\times {\left(\frac{47}{3}\right)}^2$

Multiply $3$ on both sides

$3P^2$$=3[6\times {\left(\frac{47}{3}\right)}^2]=2\times {47}^2=4418$