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

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

Vedclass · Accepted Answer

$3x-26y+43z+3=0$

Vedclass · Answer

(D) The equation of any plane passing through the intersection of the planes $P_1: x+2y-3z+1=0$ and $P_2: 3x-2y+4z+3=0$ is given by $P_1 + \lambda P_2 = 0$.$(x+2y-3z+1) + \lambda(3x-2y+4z+3) = 0 \quad \dots(1)$Since the plane passes through the point $(2,2,1)$,we substitute $x=2, y=2, z=1$ into equation $(1)$:$(2 + 2(2) - 3(1) + 1) + \lambda(3(2) - 2(2) + 4(1) + 3) = 0$$(2 + 4 - 3 + 1) + \lambda(6 - 4 + 4 + 3) = 0$$4 + 9\lambda = 0$$\lambda = -\frac{4}{9}$Substituting $\lambda = -\frac{4}{9}$ back into equation $(1)$:$(x+2y-3z+1) - \frac{4}{9}(3x-2y+4z+3) = 0$$9(x+2y-3z+1) - 4(3x-2y+4z+3) = 0$$9x + 18y - 27z + 9 - 12x + 8y - 16z - 12 = 0$$-3x + 26y - 43z - 3 = 0$Multiplying by $-1$,we get $3x - 26y + 43z + 3 = 0$.

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

Similar Questions

The distance of the point $A(3, -4, 5)$ from the plane $2x + 5y - 6z = 16$ measured along the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$,$(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

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:

Find the point of intersection of the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{2}$ and the plane $2x + y + z = 6$.

Find the angle between the plane $\bar{r} \cdot (1, 2, 1) = 1$ and the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module