The length of the perpendicular from the orig | vedclass

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

Question

(C) The equation of the plane is given in the form $\bar{r} \cdot \bar{n} = d$,where $\bar{n} = 3 \hat{i} - 4 \hat{j} + 12 \hat{k}$ and $d = 8$. The l

Vedclass · Accepted Answer

$\frac{8}{13}$ units

Vedclass · Answer

(C) The equation of the plane is given in the form $\bar{r} \cdot \bar{n} = d$,where $\bar{n} = 3 \hat{i} - 4 \hat{j} + 12 \hat{k}$ and $d = 8$. The length of the perpendicular from the origin to the plane $\bar{r} \cdot \bar{n} = d$ is given by the formula $p = \frac{|d|}{|\bar{n}|}$. Here,$|\bar{n}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Therefore,the length of the perpendicular is $p = \frac{8}{13}$ units.

The length of the perpendicular from the origin to the plane $\bar{r} \cdot (3 \hat{i} - 4 \hat{j} + 12 \hat{k}) = 8$ is

Similar Questions

The line drawn from $(4, -1, 2)$ to the point $(-3, 2, 3)$ meets a plane at right angles at the point $(-10, 5, 4)$. Find the equation of the plane.

If the foot of the perpendicular drawn from the point $(1,0,-2)$ to the plane $\pi$ is $(2,0,-1)$ and the equation of the plane $\pi$ is $ax+by+cz=2$, then $a^2+b^2+c^2=$

The unit vector perpendicular to the plane $4x - 3y + 12z = 15$ is

Find the equation of the plane with intercept $3$ on the $y$-axis and parallel to the $ZOX$ plane.

$A$ plane passes through the point $A(2, 1, -3)$. If the distance of this plane from the origin is maximum,then its equation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module