The projection of the line (x+1/-1)=(y/2)=(z-1/3) on the plane x-2 y+z=6 is the line of intersection of this plane with the plane

Question

The projection of the line (x+1/-1)=(y/2)=(z-1/3) on the plane x-2 y+z=6 is the line of intersection of this plane with the plane (A) 2 x+y+2=0 (B) 3 x+y-z=2 (C) 2 x-3 y+8 z=3 (D) None of these

Tardigrade Admin · Accepted Answer

Equation of the plane through (-1,0,1) is a(x+1)+b(y-0)+c(z-1)=0 dots(i) which is parallel to the given line and perpendicular to the given plane -a+2 b+3 c=0 dots(ii) and a-2b+c=0 dots(iii) From Eqs. (ii) and (iii), we get c=0, a=2b From Eq. (i), 2 b(x+1)+b y=0 ⇒ 2 x+y+2=0

Q. The projection of the line $\frac{x + 1}{- 1} = \frac{y}{2} = \frac{z - 1}{3}$ on the plane $x - 2 y + z = 6$ is the line of intersection of this plane with the plane

$2 x + y + 2 = 0$

$3 x + y - z = 2$

$2 x - 3 y + 8 z = 3$

None of these

Q. The projection of the line −1x+1​=2y​=3z−1​ on the plane x−2y+z=6 is the line of intersection of this plane with the plane

2x+y+2=0

3x+y−z=2

2x−3y+8z=3

None of these

Equation of the plane through (−1,0,1) is a(x+1)+b(y−0)+c(z−1)=0…(i) which is parallel to the given line and perpendicular to the given plane −a+2b+3c=0…(ii) and a−2b+c=0…(iii) From Eqs. (ii) and (iii), we get c=0,a=2b From Eq. (i), 2b(x+1)+by=0 ⇒2x+y+2=0

Q. The projection of the line $\frac{x + 1}{- 1} = \frac{y}{2} = \frac{z - 1}{3}$ on the plane $x - 2 y + z = 6$ is the line of intersection of this plane with the plane

$2 x + y + 2 = 0$

$3 x + y - z = 2$

$2 x - 3 y + 8 z = 3$