Find the distance from the origin to the plan | vedclass

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

Question

(A) The equation of a plane passing through a point $\vec{a} = (2, 3, -1)$ and perpendicular to a normal vector $\vec{n} = 3\hat{i} - 4\hat{j} + 7\hat

Vedclass · Accepted Answer

$\frac{13}{\sqrt{74}}$

Vedclass · Answer

(A) The equation of a plane passing through a point $\vec{a} = (2, 3, -1)$ and perpendicular to a normal vector $\vec{n} = 3\hat{i} - 4\hat{j} + 7\hat{k}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Substituting the values,we get $(x - 2)(3) + (y - 3)(-4) + (z + 1)(7) = 0$.$3x - 6 - 4y + 12 + 7z + 7 = 0$.$3x - 4y + 7z + 13 = 0$.The distance $d$ of the plane $Ax + By + Cz + D = 0$ from the origin $(0, 0, 0)$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A = 3, B = -4, C = 7, D = 13$.$d = \frac{|13|}{\sqrt{3^2 + (-4)^2 + 7^2}} = \frac{13}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}}$.

Find the distance from the origin to the plane passing through the point $(2, 3, -1)$ and perpendicular to the vector $3\hat{i} - 4\hat{j} + 7\hat{k}$.

Similar Questions

If the angle between the planes $\bar{r} \cdot(11 \hat{i}-2 \hat{j}+\alpha \hat{k})=7$ and $\bar{r} \cdot(2 \hat{i}+4 \hat{j}-2 \hat{k})=5$ is $\frac{\pi}{2}$, then $\alpha=$

If the points $A(2-x, 2, 2)$,$B(2, 2-y, 2)$,$C(2, 2, 2-z)$,and $D(1, 1, 1)$ are coplanar,then the locus of point $P(x, y, z)$ is

If the equation of the plane passing through the point $(3,2,5)$ and perpendicular to the planes $2x-3y+5z=7$ and $5x+2y-3z=11$ is $x+by+cz+d=0$,then $2b+3c+d=$

Assertion: The points $(2, 1, 5)$ and $(3, 4, 3)$ lie on opposite sides of the plane $2x + 2y - 2z - 1 = 0$.
Reason: The algebraic perpendicular distances from the given points to the plane have opposite signs.

The equation of a plane parallel to the plane $x - 2y + 2z - 5 = 0$ and at a unit distance from the origin is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module