Consider the line L given by the equation (x-3/2)=(y-1/1)=(z-2/1). Let Q be the mirror image of the point (2,3,-1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P ?

Question

Consider the line L given by the equation (x-3/2)=(y-1/1)=(z-2/1). Let Q be the mirror image of the point (2,3,-1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P ? (A) (-1,1,2) (B) (1,1,1) (C) (1,1,2) (D) (1,2,2)

Tardigrade Admin · Accepted Answer

Plane mathrm p is perpr to line (x-3/2)=(y-1/1)=(z-2/1) passes through pt. (2,3) equation of plane p 2(x-2)+1(y-3)+1(z+1)=0 2 x+y+z-6=0 pt (1,2,2) satisfies above equation

$(- 1, 1, 2)$

$(1, 1, 1)$

$(1, 1, 2)$

$(1, 2, 2)$

Plane mathrm $p$ is $⊥^{r}$ to line
$\frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z - 2}{1}$
\& passes through pt. $(2, 3)$ equation of plane $p$
$2 (x - 2) + 1 (y - 3) + 1 (z + 1) = 0$
$2 x + y + z - 6 = 0$
pt $(1, 2, 2)$ satisfies above equation

Q. Consider the line L given by the equation 2x−3​=1y−1​=1z−2​. Let Q be the mirror image of the point (2,3,−1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P ?

(−1,1,2)

(1,1,1)

(1,1,2)

(1,2,2)

Plane mathrm p is ⊥r to line 2x−3​=1y−1​=1z−2​ \& passes through pt. (2,3) equation of plane p 2(x−2)+1(y−3)+1(z+1)=0 2x+y+z−6=0 pt (1,2,2) satisfies above equation

$(- 1, 1, 2)$

$(1, 1, 1)$

$(1, 1, 2)$

$(1, 2, 2)$

Plane mathrm $p$ is $⊥^{r}$ to line
$\frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z - 2}{1}$
\& passes through pt. $(2, 3)$ equation of plane $p$
$2 (x - 2) + 1 (y - 3) + 1 (z + 1) = 0$
$2 x + y + z - 6 = 0$
pt $(1, 2, 2)$ satisfies above equation