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 with direction ratios $(-3, 2, 1)$,we have $-3a

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 with direction ratios $(-3, 2, 1)$,we have $-3a + 2b + c = 0$. The plane passes through $(-1, 3, -2)$ and $(0, 7, -7)$,so the vector connecting these points is $(0 - (-1), 7 - 3, -7 - (-2)) = (1, 4, -5)$. Since this vector lies on the plane,$a + 4b - 5c = 0$. Solving the system $-3a + 2b + c = 0$ and $a + 4b - 5c = 0$ using cross product: $a : b : c = |(2, 1), (4, -5)| : -|( -3, 1), (1, -5)| : |(-3, 2), (1, 4)| = (-10-4) : -(15-1) : (-12-2) = -14 : -14 : -14 = 1 : 1 : 1$. Thus,the plane equation is $1(x+1) + 1(y-3) + 1(z+2) = 0$,which simplifies to $x + y + z = 0$. The line perpendicular to the plane passing through $(1, 2, -1)$ 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 $x+y+z=0$,so $(\lambda+1) + (\lambda+2) + (\lambda-1) = 0$,which gives $3\lambda + 2 = 0$,so $\lambda = -\frac{2}{3}$. The foot of the perpendicular is $(1 - \frac{2}{3}, 2 - \frac{2}{3}, -1 - \frac{2}{3}) = (\frac{1}{3}, \frac{4}{3}, -\frac{5}{3})$. Let the image be $(x', y', z')$. Then $\frac{x'+1}{2} = \frac{1}{3}$,$\frac{y'+2}{2} = \frac{4}{3}$,$\frac{z'-1}{2} = -\frac{5}{3}$. Solving these gives $x' = -\frac{1}{3}$,$y' = \frac{2}{3}$,$z' = -\frac{7}{3}$. Thus,the image is $(\frac{-1}{3}, \frac{2}{3}, \frac{-7}{3})$.

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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