Find the distance of the point (0, 0, 0) from | vedclass

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

Question

(A) The distance $d$ of a point $P(x_1, y_1, z_1)$ from a plane $Ax + By + Cz - D = 0$ is given by the formula:$d = \left| \frac{Ax_1 + By_1 + Cz_1 -

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The distance $d$ of a point $P(x_1, y_1, z_1)$ from a plane $Ax + By + Cz - D = 0$ is given by the formula:$d = \left| \frac{Ax_1 + By_1 + Cz_1 - D}{\sqrt{A^2 + B^2 + C^2}} \right|$Given the point $(x_1, y_1, z_1) = (0, 0, 0)$ and the plane equation $3x - 4y + 12z - 3 = 0$,we identify $A = 3, B = -4, C = 12$,and $D = 3$.Substituting these values into the formula:$d = \left| \frac{3(0) - 4(0) + 12(0) - 3}{\sqrt{3^2 + (-4)^2 + 12^2}} \right|$$d = \left| \frac{-3}{\sqrt{9 + 16 + 144}} \right|$$d = \left| \frac{-3}{\sqrt{169}} \right|$$d = \frac{3}{13}$

Find the distance of the point $(0, 0, 0)$ from the plane $3x - 4y + 12z = 3$. (in $/13$)

Similar Questions

The equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$ is:

The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$,$(-1, 1, -2)$ and $(3, -4, 2)$ is:

If the distance of the point $(1, 1, 1)$ from the origin is half its distance from the plane $x + y + z + k = 0$,then $k = $

The equation of the plane which bisects the angle between the planes $3x - 6y + 2z + 5 = 0$ and $4x - 12y + 3z - 3 = 0$ which contains the origin is

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $2x + 3y - z = 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module