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

For which values of $a$ do the two points $(1, a, 1)$ and $(-3, 0, a)$ lie on opposite sides of the plane $3x + 4y - 12z + 13 = 0$?

The equation of the plane passing through the point $(1, 2, 1)$ and perpendicular to the planes $x + 2y + 2z - 7 = 0$ and $3x + 3y + 2z - 5 = 0$ is:

Find the angle between the planes $x + y + 2z = 9$ and $2x - y + z = 15$.

$A$ plane $\pi$ makes intercepts $3$ and $4$ respectively on $Z$-axis and $X$-axis. If $\pi$ is parallel to $Y$-axis, then its equation is:

If from a point $P(a, b, c)$, perpendiculars $PA$ and $PB$ are drawn to $YZ$ and $ZX$ planes respectively, then the equation of the plane $OAB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module