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

(B) The equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by:$(x + 2y + 3z

Vedclass · Accepted Answer

$51x + 15y - 50z + 173 = 0$

Vedclass · Answer

(B) The equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by:$(x + 2y + 3z - 4) + \lambda (2x + y - z + 5) = 0$Rearranging the terms,we get:$x(1 + 2\lambda) + y(2 + \lambda) + z(3 - \lambda) + (5\lambda - 4) = 0 \quad \dots(i)$Since this plane is perpendicular to the plane $5x + 3y + 6z + 8 = 0$,the dot product of their normal vectors must be zero:$5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0$$5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$$7\lambda + 29 = 0$$\lambda = -\frac{29}{7}$Substituting $\lambda = -\frac{29}{7}$ into equation $(i)$:$(x + 2y + 3z - 4) - \frac{29}{7}(2x + y - z + 5) = 0$$7(x + 2y + 3z - 4) - 29(2x + y - z + 5) = 0$$7x + 14y + 21z - 28 - 58x - 29y + 29z - 145 = 0$$-51x - 15y + 50z - 173 = 0$$51x + 15y - 50z + 173 = 0$

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

Similar Questions

The equation of the locus of a point $P(x, y, z)$ such that its distance from the $X$-axis is equal to its distance from the plane $x+z=1$ is

If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then which of the following is true?

The length (in units) of the projection of the line segment,joining the points $(5,-1,4)$ and $(4,-1,3)$,on the plane $x+y+z=7$ is

$X$ intercept of the plane containing the line of intersection of the planes $x-2y+z+2=0$ and $3x-y-z+1=0$ and also passing through $(1,1,1)$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the equation of the plane containing these lines is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module