Find the distance of the point P(3, 8, 2) fro | vedclass

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

Question

(D) Let the point $Q$ on the line be $(2\lambda + 1, 4\lambda + 3, 3\lambda + 2)$.Since the line $PQ$ is parallel to the plane $3x + 2y - 2z + 15 = 0$

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Let the point $Q$ on the line be $(2\lambda + 1, 4\lambda + 3, 3\lambda + 2)$.Since the line $PQ$ is parallel to the plane $3x + 2y - 2z + 15 = 0$,the vector $\vec{PQ}$ must be perpendicular to the normal vector of the plane $\vec{n} = (3, 2, -2)$.$\vec{PQ} = (2\lambda + 1 - 3, 4\lambda + 3 - 8, 3\lambda + 2 - 2) = (2\lambda - 2, 4\lambda - 5, 3\lambda)$.Since $\vec{PQ} \cdot \vec{n} = 0$,we have:$3(2\lambda - 2) + 2(4\lambda - 5) - 2(3\lambda) = 0$$6\lambda - 6 + 8\lambda - 10 - 6\lambda = 0$$8\lambda - 16 = 0 \implies \lambda = 2$.Substituting $\lambda = 2$ in the coordinates of $Q$,we get $Q = (2(2) + 1, 4(2) + 3, 3(2) + 2) = (5, 11, 8)$.The distance $PQ = \sqrt{(5-3)^2 + (11-8)^2 + (8-2)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$.

Similar Questions

The transformed equation of $2x^2 + 3y^2 - z^2 - 8x + 18y + 2z + 9 = 0$ when the axes are translated to the point $(2, -3, 1)$ is

The locus of a point $P$ such that $PA + PB = 4$ where $A(2, 3, 4)$ and $B(-2, 3, 4)$ is

What is the angle between two diagonals of a cube?

Consider the tetrahedron with the vertices $A(3,2,4)$,$B(x_1, y_1, 0)$,$C(x_2, y_2, 0)$,and $D(x_3, y_3, 0)$. If the triangle $BCD$ is formed by the lines $y=x$,$x+y=6$,and $y=1$,then the centroid of the tetrahedron is

If the orthocentre and the centroid of a triangle are at $(5,2,-6)$ and $(9,6,-4)$ respectively,then its circumcentre is

Vedclass Test Series

Exam Paper Generator

Online Exam Module