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

If $(3,4,-7)$ is the foot of the perpendicular drawn from the point $(-2,3,6)$ to the plane $\pi$,then the sum of the intercepts made by the plane $\pi$ on the $X$ and $Y$-axes is

Direction ratios of the normal to a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle of $\frac{\pi}{4}$ with the plane $x + y - 3 = 0$ are:

The equation of a plane,containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(2, 1, 0)$,is

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them.
$2x + y + 3z - 2 = 0$ and $x - 2y + 5 = 0$

The equation of the plane passing through the points $(2,1,0)$,$(3,2,-2)$,and $(3,1,7)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module