Consider the line L given by the equation fra | vedclass

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(A) The line $L$ is given by $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1} = k$. Any point on $L$ is $(2k+3, k+1, k+2)$.Let $P_0 = (2,3,-1)$. The vector

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$(1,2,2)$

Vedclass · Answer

(A) The line $L$ is given by $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1} = k$. Any point on $L$ is $(2k+3, k+1, k+2)$.Let $P_0 = (2,3,-1)$. The vector $\vec{P_0M}$ where $M$ is the foot of the perpendicular from $P_0$ to $L$ is $(2k+3-2, k+1-3, k+2+1) = (2k+1, k-2, k+3)$.Since $\vec{P_0M}$ is perpendicular to the direction vector of $L$,$\vec{v} = (2,1,1)$,we have $2(2k+1) + 1(k-2) + 1(k+3) = 0$.$4k+2 + k-2 + k+3 = 0 \implies 6k+3 = 0 \implies k = -1/2$.Thus,$M = (2(-1/2)+3, -1/2+1, -1/2+2) = (2, 1/2, 3/2)$.Since $M$ is the midpoint of $P_0Q$,if $Q = (x,y,z)$,then $\frac{x+2}{2} = 2, \frac{y+3}{2} = 1/2, \frac{z-1}{2} = 3/2$.$x+2 = 4 \implies x=2$; $y+3 = 1 \implies y=-2$; $z-1 = 3 \implies z=4$. So $Q = (2, -2, 4)$.The plane $P$ is perpendicular to $L$,so its normal vector is $\vec{n} = (2,1,1)$.The equation of plane $P$ passing through $Q(2, -2, 4)$ is $2(x-2) + 1(y+2) + 1(z-4) = 0$.$2x - 4 + y + 2 + z - 4 = 0 \implies 2x + y + z - 6 = 0$.Checking the options: For $(1,2,2)$,$2(1) + 2 + 2 - 6 = 2+2+2-6 = 0$. Thus,$(1,2,2)$ lies on the plane.

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