The equation of the plane passing through the | vedclass

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

Question

(C) The equation of any plane passing through the line of intersection of planes $\pi_1 = 0$ and $\pi_2 = 0$ is given by $\pi_1 + \lambda \pi_2 = 0$.

Vedclass · Accepted Answer

$6x + 2y - 4z - 15 = 0$

Vedclass · Answer

(C) The equation of any plane passing through the line of intersection of planes $\pi_1 = 0$ and $\pi_2 = 0$ is given by $\pi_1 + \lambda \pi_2 = 0$. Substituting the given planes: $(2x + 6y + 4z - 7) + \lambda(x - y - 2z - 2) = 0$ $(2 + \lambda)x + (6 - \lambda)y + (4 - 2\lambda)z - (7 + 2\lambda) = 0 \quad \dots(i)$ Since this plane is perpendicular to the plane $x + y + 2z - 5 = 0$, the dot product of their normal vectors must be zero. The normal vectors are $\vec{n_1} = (2 + \lambda, 6 - \lambda, 4 - 2\lambda)$ and $\vec{n_2} = (1, 1, 2)$. $(2 + \lambda)(1) + (6 - \lambda)(1) + (4 - 2\lambda)(2) = 0$ $2 + \lambda + 6 - \lambda + 8 - 4\lambda = 0$ $16 - 4\lambda = 0 \implies \lambda = 4$. Substituting $\lambda = 4$ into equation $(i)$: $(2 + 4)x + (6 - 4)y + (4 - 8)z - (7 + 8) = 0$ $6x + 2y - 4z - 15 = 0$.

The equation of the plane passing through the line of intersection of planes $\pi_1: 2x + 6y + 4z - 7 = 0$ and $\pi_2: x - y - 2z - 2 = 0$, and perpendicular to the plane $x + y + 2z - 5 = 0$ is:

Similar Questions

The distance of the point $P(3,4,4)$ from the point of intersection of the line joining the points $Q(3,-4,-5)$ and $R(2,-3,1)$ and the plane $2x+y+z=7$ is equal to $.....$

The angle between the line $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3\hat{i}+\hat{j})$ and the plane $\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})=8$ is:

The equation of a plane through the line of intersection of the planes $x + 2y = 3$ and $y - 2z + 1 = 0$,and perpendicular to the first plane $x + 2y = 3$ is:

The equation of the plane containing the line of intersection of the planes $2x - y = 0$ and $y - 3z = 0$ and perpendicular to the plane $4x + 5y - 3z - 8 = 0$ is

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies in the plane $2x-4y+z=7$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module