The plane passing through the intersection of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of any plane passing through the intersection of the planes $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y + z - 4 = 0$ is given by $P_1

Vedclass · Accepted Answer

$(-3, 1, 1)$

Vedclass · Answer

(B) The equation of any plane passing through the intersection of the planes $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y + z - 4 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(x + y + z - 1) + \lambda(2x + 3y + z - 4) = 0$$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 + \lambda)z - (1 + 4\lambda) = 0$.Since the plane is parallel to the $y$-axis,the coefficient of $y$ must be zero.$1 + 3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$.Substituting $\lambda = -\frac{1}{3}$ into the equation:$(1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 - \frac{1}{3})z - (1 + 4(-\frac{1}{3})) = 0$$(1 - \frac{2}{3})x + 0y + \frac{2}{3}z - (1 - \frac{4}{3}) = 0$$\frac{1}{3}x + \frac{2}{3}z + \frac{1}{3} = 0$$x + 2z + 1 = 0$.Checking the options for the point $(x, z)$ satisfying $x + 2z + 1 = 0$:For option $A$: $(-3) + 2(-1) + 1 = -3 - 2 + 1 = -4 
eq 0$.For option $B$: $(-3) + 2(1) + 1 = -3 + 2 + 1 = 0$.Thus,the plane passes through $(-3, 1, 1)$.

The plane passing through the intersection of the planes $x + y + z = 1$ and $2x + 3y + z - 4 = 0$ and parallel to the $y$-axis also passes through the point:

Similar Questions

Let $PQR$ be a triangle with $R(-1, 4, 2)$. Suppose $M(2, 1, 2)$ is the midpoint of $PQ$. The distance of the centroid of $\triangle PQR$ from the point of intersection of the lines $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ and $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ is

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to the planes $3x - 2y + z = 9$ and $x + y + z = 9$ is $x + by + cz + d = 0$, then $d =$

The equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

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

The equation of the plane passing through the intersection of the planes $x + 2y + 3z + 4 = 0$ and $4x + 3y + 2z + 1 = 0$ and the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module