Find the equation of the plane passing throug | vedclass

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

Question

(D) The equation of any plane passing through the intersection of planes $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$.$(3x + 2y + 5z + 1) + \la

Vedclass · Accepted Answer

$2x + 5y - 4z + 19 = 0$

Vedclass · Answer

(D) The equation of any plane passing through the intersection of planes $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$.$(3x + 2y + 5z + 1) + \lambda(x + y + z + 2) = 0$$(3 + \lambda)x + (2 + \lambda)y + (5 + \lambda)z + (1 + 2\lambda) = 0$Since this plane is parallel to the line $L$ with direction ratios $(1, 2, 3)$,the normal to the plane must be perpendicular to the line. Thus,the dot product of the normal vector $(3 + \lambda, 2 + \lambda, 5 + \lambda)$ and the direction vector of the line $(1, 2, 3)$ is zero.$1(3 + \lambda) + 2(2 + \lambda) + 3(5 + \lambda) = 0$$3 + \lambda + 4 + 2\lambda + 15 + 3\lambda = 0$$6\lambda + 22 = 0$$6\lambda = -22 \Rightarrow \lambda = -\frac{11}{3}$Substituting $\lambda = -\frac{11}{3}$ into the equation of the plane:$(3 - \frac{11}{3})x + (2 - \frac{11}{3})y + (5 - \frac{11}{3})z + (1 - \frac{22}{3}) = 0$$-\frac{2}{3}x - \frac{5}{3}y + \frac{4}{3}z - \frac{19}{3} = 0$Multiplying by $-3$,we get $2x + 5y - 4z + 19 = 0$.

Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:
$P_1 : 3x + 2y + 5z + 1 = 0$
$P_2 : x + y + z + 2 = 0$
$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$

Similar Questions

The line passing through $(4, -1, 2)$ and $(-3, 2, 3)$ meets the plane at right angles at the point $(-10, 5, 4)$. Then the equation of the plane is:

Find the distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$.

The equation of the plane passing through the point $(2,2,1)$ and the intersection of the planes $x+2y-3z+1=0$ and $3x-2y+4z+3=0$ is

If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,then $k$ can have:

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=17$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:$P_1 : 3x + 2y + 5z + 1 = 0$$P_2 : x + y + z + 2 = 0$$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$