If the equation of the plane passing through | vedclass

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

Question

(A) The equation of the family of planes passing through the intersection of the planes $P_1: 2x - y + z - 3 = 0$ and $P_2: 4x - 3y + 5z + 9 = 0$ is g

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) The equation of the family of planes passing through the intersection of the planes $P_1: 2x - y + z - 3 = 0$ and $P_2: 4x - 3y + 5z + 9 = 0$ is given by $(2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0$.Rearranging the terms,we get $x(2 + 4\lambda) + y(-1 - 3\lambda) + z(1 + 5\lambda) + (-3 + 9\lambda) = 0$.Since this plane is parallel to the line with direction ratios $(-2, 4, 5)$,the normal vector of the plane must be perpendicular to the line. Thus,the dot product of the normal vector $(2 + 4\lambda, -1 - 3\lambda, 1 + 5\lambda)$ and the direction vector $(-2, 4, 5)$ must be zero:$-2(2 + 4\lambda) + 4(-1 - 3\lambda) + 5(1 + 5\lambda) = 0$.$-4 - 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$.$5\lambda - 3 = 0 \implies \lambda = \frac{3}{5}$.Substituting $\lambda = \frac{3}{5}$ into the plane equation:$(2x - y + z - 3) + \frac{3}{5}(4x - 3y + 5z + 9) = 0$.$5(2x - y + z - 3) + 3(4x - 3y + 5z + 9) = 0$.$10x - 5y + 5z - 15 + 12x - 9y + 15z + 27 = 0$.$22x - 14y + 20z + 12 = 0$.Dividing by $2$,we get $11x - 7y + 10z + 6 = 0$.Comparing this with $ax + by + cz + 6 = 0$,we get $a = 11, b = -7, c = 10$.Therefore,$a + b + c = 11 - 7 + 10 = 14$.

If the equation of the plane passing through the line of intersection of the planes $2x - y + z = 3$ and $4x - 3y + 5z + 9 = 0$ and parallel to the line $\frac{x + 1}{-2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $ax + by + cz + 6 = 0$,then $a + b + c$ is equal to $.............$.

Similar Questions

The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ passes through the point

Let $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}$ lie on the plane $px-qy+z=5$,for some $p, q \in R$. The shortest distance of the plane from the origin is

If $P(3, 2, 6)$ is a point in space and $Q$ is a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$,then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is:

The line of intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$ is parallel to which vector?

Find the length and the foot of the perpendicular from the point $\left(1, \frac{3}{2}, 2\right)$ to the plane $2x - 2y + 4z + 5 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module