Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3\hat i+2\hat j+2\hat k$. If the distance of the point $Q$ from the point $R(1,3,3)$ is 5, then the square of the area of $ riangle PQR$ is equal to :

Question

Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3\hat i+2\hat j+2\hat k$. If the distance of the point $Q$ from the point $R(1,3,3)$ is 5, then the square of the area of $	riangle PQR$ is equal to :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\vec{PR}=(3,4,0)$ and $\vec d=(3,2,2)$, $|\vec d|=\sqrt{17}$. $\vec{PR}	imes\vec d=(8,-6,-6)$, so the distance of $R$ from the line is $h=\sqrt{136/17}=\sqrt8$. The projection of $PR$ on the line is $\frac{17}{\sqrt{17}}=\sqrt{17}$, and $Q$ is at distance $\sqrt{25-8}=\sqrt{17}$ from the foot, so $PQ=2\sqrt{17}$ ($Q
e P$). Area $=\frac12\cdot2\sqrt{17}\cdot\sqrt8=\sqrt{136}$, so (area)$^2=136$. Answer (2).

Q. Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3\hat i+2\hat j+2\hat k$. If the distance of the point $Q$ from the point $R(1,3,3)$ is 5, then the square of the area of $\triangle PQR$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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