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 point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ is

Let the equation of the plane passing through the line $x-2y-z-5=0=x+y+3z-5$ and parallel to the line $x+y+2z-7=0=2x+3y+z-2$ be $ax+by+cz=65$. Then the distance of the point $(a, b, c)$ from the plane $2x+2y-z+16=0$ is $..........$.

If the straight lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$,with parameters $s$ and $t$ respectively,are coplanar,then $\lambda$ equals

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 point $P$ is the intersection of the straight line joining the points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from the point $T(2, 1, 4)$ to $QR$,then the length of the line segment $PS$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module