The distance of a point vec{a} from the plane | vedclass

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

Question

(D) The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = d$ is given by $\frac{|\vec{a} \cdot \vec{m} - d|}{|\vec{m}|}$.Given the

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = d$ is given by $\frac{|\vec{a} \cdot \vec{m} - d|}{|\vec{m}|}$.Given the plane $\vec{r} \cdot (2\hat{i} + 6\hat{j} - 9\hat{k}) = -1$,we have $\vec{m} = 2\hat{i} + 6\hat{j} - 9\hat{k}$ and $d = -1$.The magnitude $|\vec{m}| = \sqrt{2^2 + 6^2 + (-9)^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11$.For the point $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,the distance $p$ is:$p = \frac{|(1)(2) + (2)(6) + (3)(-9) - (-1)|}{11} = \frac{|2 + 12 - 27 + 1|}{11} = \frac{|-12|}{11} = \frac{12}{11}$.The distance $q$ of the origin $(0, 0, 0)$ from the plane is:$q = \frac{|0 - (-1)|}{11} = \frac{1}{11}$.Therefore,$p - q = \frac{12}{11} - \frac{1}{11} = \frac{11}{11} = 1$.

Similar Questions

The equation of a plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

$A$ plane is parallel to two lines whose direction ratios are $2, 0, -2$ and $-2, 2, 0$ and it contains the point $(2, 2, 2)$. If it cuts the coordinate axes at $A, B, C$,then the volume of the tetrahedron $OABC$ (in cubic units) is

The direction cosines of the normal drawn to the plane passing through the points $(2,-1,5)$,$(1,-3,4)$,and $(5,2,1)$ are

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin.
$Z=2$

If the plane passing through the points $(1, 2, 3)$,$(2, 3, 1)$,and $(3, 1, 2)$ is $a x + b y + c z = 1$,then $a + 2 b + 3 c = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module