The image of the point (1, 2, -1) on the plan | vedclass

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(B) Let the equation of the plane be $a(x+1) + b(y-3) + c(z+2) = 0$.Since the plane contains the line,the normal vector is perpendicular to the direct

Vedclass · Accepted Answer

$\left( \frac{-1}{3}, \frac{2}{3}, \frac{-7}{3} \right)$

Vedclass · Answer

(B) Let the equation of the plane be $a(x+1) + b(y-3) + c(z+2) = 0$.Since the plane contains the line,the normal vector is perpendicular to the direction vector $(-3, 2, 1)$,so $-3a + 2b + c = 0$.The plane passes through $(0, 7, -7)$,so $a(0+1) + b(7-3) + c(-7+2) = 0$,which gives $a + 4b - 5c = 0$.Solving the system $-3a + 2b + c = 0$ and $a + 4b - 5c = 0$ using cross product,we get the normal vector $\vec{n} = (1, 1, 1)$.The equation of the plane is $1(x+1) + 1(y-3) + 1(z+2) = 0$,which simplifies to $x + y + z = 0$.The line passing through $(1, 2, -1)$ and perpendicular to the plane is $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z+1}{1} = \lambda$.Any point on this line is $(\lambda+1, \lambda+2, \lambda-1)$.For the foot of the perpendicular,this point lies on the plane: $(\lambda+1) + (\lambda+2) + (\lambda-1) = 0$,so $3\lambda + 2 = 0$,which gives $\lambda = \frac{-2}{3}$.The foot of the perpendicular is $(\frac{1}{3}, \frac{4}{3}, \frac{-5}{3})$.Let the image be $(x', y', z')$. The midpoint of the point and its image is the foot of the perpendicular: $\frac{x'+1}{2} = \frac{1}{3} \Rightarrow x' = \frac{-1}{3}$,$\frac{y'+2}{2} = \frac{4}{3} \Rightarrow y' = \frac{2}{3}$,and $\frac{z'-1}{2} = \frac{-5}{3} \Rightarrow z' = \frac{-7}{3}$.Thus,the image is $\left( \frac{-1}{3}, \frac{2}{3}, \frac{-7}{3} \right)$.

The image of the point $(1, 2, -1)$ on the plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$ is:

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