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 symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

If the line $\vec{r} = (\hat{i} - 2\hat{j} - \hat{k}) + \lambda (2\hat{i} + \hat{j} + 2\hat{k})$ is parallel to the plane $\vec{r} \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 14$,then the value of $m$ is:

Angle between the lines of intersection of the planes $x-y=0, 2x+y+z=0$ and $2x-z=0, x+y-3z=0$ is (in $^{\circ}$)

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

The distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module