The equation of the plane through the interse | 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 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 =

Vedclass · Accepted Answer

$y-3z+6=0$

Vedclass · Answer

(C) 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 $P_1 + \lambda P_2 = 0$. $(x+y+z-1) + \lambda(2x+3y-z+4) = 0$ $(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$. Since the plane is parallel to the $x$-axis,its normal vector $\vec{n} = (1+2\lambda, 1+3\lambda, 1-\lambda)$ must be perpendicular to the $x$-axis direction vector $\vec{i} = (1, 0, 0)$. Therefore,the dot product is zero: $(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$. $1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$. Substituting $\lambda = -\frac{1}{2}$ into the equation: $(1+2(-\frac{1}{2}))x + (1+3(-\frac{1}{2}))y + (1-(-\frac{1}{2}))z + (4(-\frac{1}{2})-1) = 0$. $0x + (1-\frac{3}{2})y + (1+\frac{1}{2})z + (-2-1) = 0$. $-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$. Multiplying by $-2$,we get $y - 3z + 6 = 0$.

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

Similar Questions

Let the points $P, Q$ and $R$ have position vectors $\overrightarrow{r_1} = 3i - 2j - k, \overrightarrow{r_2} = i + 3j + 4k$ and $\overrightarrow{r_3} = 2i + j - 2k$ respectively relative to an origin $O$. Then the distance of $P$ from the plane $OQR$ is:

Let $N$ be the foot of the perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $.......$.

If the mirror image of the point $(1, 3, 5)$ with respect to the plane $4x - 5y + 2z = 8$ is $(\alpha, \beta, \gamma)$,then $5(\alpha + \beta + \gamma)$ equals:

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

If the equation of a line is $\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z + 5}{2}$ and the equation of a plane is $4x - 2y - z = 1$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module