If the points (1, 1, p) and (-3, 0, 1) are eq | vedclass

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

Question

(A) The perpendicular distance $D$ from a point with position vector $\vec{a}$ to the plane $\vec{r} \cdot \vec{n} + d = 0$ is given by $D = \frac{|\v

Vedclass · Accepted Answer

$1, \frac{7}{3}$

Vedclass · Answer

(A) The perpendicular distance $D$ from a point with position vector $\vec{a}$ to the plane $\vec{r} \cdot \vec{n} + d = 0$ is given by $D = \frac{|\vec{a} \cdot \vec{n} + d|}{|\vec{n}|}$.For the given plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0$,we have $\vec{n} = 3 \hat{i} + 4 \hat{j} - 12 \hat{k}$ and $d = 13$.The magnitude $|\vec{n}| = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.Let $D_1$ be the distance from $(1, 1, p)$ to the plane:$D_1 = \frac{|(1)(3) + (1)(4) + (p)(-12) + 13|}{13} = \frac{|3 + 4 - 12p + 13|}{13} = \frac{|20 - 12p|}{13}$.Let $D_2$ be the distance from $(-3, 0, 1)$ to the plane:$D_2 = \frac{|(-3)(3) + (0)(4) + (1)(-12) + 13|}{13} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.Since the points are equidistant,$D_1 = D_2$:$\frac{|20 - 12p|}{13} = \frac{8}{13} \implies |20 - 12p| = 8$.Case $1$: $20 - 12p = 8 \implies 12p = 12 \implies p = 1$.Case $2$: $20 - 12p = -8 \implies 12p = 28 \implies p = \frac{28}{12} = \frac{7}{3}$.Thus,the values of $p$ are $1$ and $\frac{7}{3}$.

If the points $(1, 1, p)$ and $(-3, 0, 1)$ are equidistant from the plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0$,then find the value of $p$.

Similar Questions

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is . . . . . . units.

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

The equation of a plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$ is

The line passing through the points $(1, 1, -1)$ and $(3, -1, 0)$ makes an angle of $\operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}\right)$ with the plane $\sqrt{\lambda} x + 3y + 6z = 17$. Then $\lambda =$

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and passing through the point $(2,1,3).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module