Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}7=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right angled triangle $PQR$, where $R$ is the point $(3,-2,1)$, is

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The foot of the perpendicular from $Q$ is $P(10,1,-3)$. Then $\vec{QP}=(0,4,-2)$ and $\vec{QR}=(-7,1,2)$ have $\vec{QP}\cdot\vec{QR}=0$, so the right angle is at $Q$. $QP=\sqrt{20}$, $QR=\sqrt{54}$, and the area is $\frac12\sqrt{20}\sqrt{54}=\frac12\sqrt{1080}=3\sqrt{30}$. Answer (4).

Q. Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}7=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right angled triangle $PQR$, where $R$ is the point $(3,-2,1)$, is

Explanation

Reviewed by Dr. Abubacker

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