"Let Q be the foot of the perpendicular from the point P(7, $-$2, 13) on the plane containing the lines ${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$ and ${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$. Then (PQ)2, is equal to ___________."

Question

Accepted Answer

"Containing the line $$\left| {\matrix{ {x + 1} & {y - 1} & {z - 3} \cr 6 & 7 & 8 \cr 3 & 5 & 7 \cr } } \right| = 0$$

$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$

$x - 2y + z = 0$

$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6$

$P{Q^2} = 96$"

Let Q be the foot of the perpendicular from the point P(7, $-$2, 13) on the plane containing the lines ${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$ and ${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$. Then (PQ)², is equal to ___________.

Solution

About this question

Drill 25 more like these. Every day. Free.

Let Q be the foot of the perpendicular from the point P(7, $-$2, 13) on the plane containing the lines ${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$ and ${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$. Then (PQ)2, is equal to ___________.

Solution

About this question

More Three Dimensional Geometry questions

Drill 25 more like these. Every day. Free.

Let Q be the foot of the perpendicular from the point P(7, $-$2, 13) on the plane containing the lines ${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$ and ${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$. Then (PQ)², is equal to ___________.