The perpendicular distance of the origin from | vedclass

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

Question

(B) The perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to the plane $Ax+By+Cz+D=0$ is given by the formula $d = \frac{|Ax_1+By_1+Cz_1+D|}{\

Vedclass · Accepted Answer

$\frac{6}{\sqrt{26}}$

Vedclass · Answer

(B) The perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to the plane $Ax+By+Cz+D=0$ is given by the formula $d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$.Here,the point is the origin $(0, 0, 0)$ and the plane is $x-3y+4z-6=0$.Substituting the values $A=1, B=-3, C=4, D=-6$ and $x_1=0, y_1=0, z_1=0$ into the formula:$d = \frac{|1(0) - 3(0) + 4(0) - 6|}{\sqrt{1^2 + (-3)^2 + 4^2}}$$d = \frac{|-6|}{\sqrt{1 + 9 + 16}}$$d = \frac{6}{\sqrt{26}}$

The perpendicular distance of the origin from the plane $x-3y+4z-6=0$ is

Similar Questions

Which of the following points lie on the plane passing through $2 \hat{i}+3 \hat{j}-\hat{k}$,$3 \hat{i}+2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+4 \hat{k}$?

The mirror image of the point $(1, 2, 3)$ in a plane is $\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. Which of the following points lies on this plane?

If the foot of the perpendicular drawn from the origin to a plane is $P(2,-1,4)$,then the equation of the plane is

The equation of the plane which contains the $y$-axis and passes through the point $(1, 2, 3)$ is:

If the equation of the plane that contains the point $(-2, 3, 5)$ and is perpendicular to each of the planes $2x + 4y + 5z = 8$ and $3x - 2y + 3z = 5$ is $\alpha x + \beta y + \gamma z + 97 = 0$,then $\alpha + \beta + \gamma = ...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module