Find the coordinates of the foot of the perpe | vedclass

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

Question

(A) Let the coordinates of the foot of the perpendicular $P$ from the origin $(0, 0, 0)$ to the plane $3y + 4z - 6 = 0$ be $(x_1, y_1, z_1)$.The equat

Vedclass · Accepted Answer

$\left(0, \frac{18}{25}, \frac{24}{25}\right)$

Vedclass · Answer

(A) Let the coordinates of the foot of the perpendicular $P$ from the origin $(0, 0, 0)$ to the plane $3y + 4z - 6 = 0$ be $(x_1, y_1, z_1)$.The equation of the plane is $0x + 3y + 4z = 6$.The direction ratios of the normal to the plane are $(0, 3, 4)$.The magnitude of the normal vector is $\sqrt{0^2 + 3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.Dividing the equation of the plane by $5$,we get the normal form $lx + my + nz = d$:$0x + \frac{3}{5}y + \frac{4}{5}z = \frac{6}{5}$.Here,the direction cosines of the normal are $l = 0, m = \frac{3}{5}, n = \frac{4}{5}$ and the distance from the origin is $d = \frac{6}{5}$.The coordinates of the foot of the perpendicular are given by $(ld, md, nd)$:$x_1 = 0 	imes \frac{6}{5} = 0$$y_1 = \frac{3}{5} 	imes \frac{6}{5} = \frac{18}{25}$$z_1 = \frac{4}{5} 	imes \frac{6}{5} = \frac{24}{25}$Thus,the coordinates of the foot of the perpendicular are $\left(0, \frac{18}{25}, \frac{24}{25}ight)$.

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $3y + 4z - 6 = 0$.

Similar Questions

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by:

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the $\Delta ABC$ is the point $(\alpha, \beta, \gamma)$. Show that the equation of the plane is $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$ intersects the coordinate axes at $A, B, C$,then the area of $\Delta ABC$ is:

The equation of the plane passing through the point $(1,2,2)$ and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$ is:

Let the points on the plane $P$ be equidistant from the points $A(-4, 2, 1)$ and $B(2, -2, 3)$. Then the acute angle between the plane $P$ and the plane $2x + y + 3z = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module