The square of the distance of the point (-2, | vedclass

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(B) Let the point $P = (-2, -8, 6)$.Let the given line be $L_1: \frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1} = k$. Any point $Q$ on $L_1$ is $(k+1, 2k

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let the point $P = (-2, -8, 6)$.Let the given line be $L_1: \frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1} = k$. Any point $Q$ on $L_1$ is $(k+1, 2k+1, -k)$.The vector $\vec{PQ} = (k+1 - (-2), 2k+1 - (-8), -k - 6) = (k+3, 2k+9, -k-6)$.Since the distance is measured along the line with direction vector $\vec{v} = (1, -1, 2)$,the vector $\vec{PQ}$ must be parallel to $\vec{v}$.Thus,$\frac{k+3}{1} = \frac{2k+9}{-1} = \frac{-k-6}{2} = \lambda$.From $\frac{k+3}{1} = \frac{2k+9}{-1}$,we get $-k-3 = 2k+9$,which implies $3k = -12$,so $k = -4$.Substituting $k = -4$ into the coordinates of $Q$,we get $Q = (-4+1, 2(-4)+1, -(-4)) = (-3, -7, 4)$.The distance $PQ$ is the magnitude of vector $\vec{PQ} = (-3 - (-2), -7 - (-8), 4 - 6) = (-1, 1, -2)$.The square of the distance $PQ^2 = (-1)^2 + (1)^2 + (-2)^2 = 1 + 1 + 4 = 6$.

The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

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