The equation of a plane passing through (-1, | vedclass

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

Question

(C) Let the normal vector to the plane be $\vec{n} = (a, b, c)$. Since the normal makes equal angles with the coordinate axes,the direction cosines ar

Vedclass · Accepted Answer

$x+y+z-4=0$

Vedclass · Answer

(C) Let the normal vector to the plane be $\vec{n} = (a, b, c)$. Since the normal makes equal angles with the coordinate axes,the direction cosines are equal,i.e.,$\cos \alpha = \cos \beta = \cos \gamma$. This implies $a = b = c$. Thus,the equation of the plane is of the form $x + y + z = d$. Since the plane passes through the point $(-1, 2, 3)$,we substitute these coordinates into the equation: $-1 + 2 + 3 = d \Rightarrow d = 4$. Substituting the value of $d$ back into the equation,we get: $x + y + z = 4$ $x + y + z - 4 = 0$.

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

Similar Questions

Find the Cartesian equation of the following plane:
$\vec{r} \cdot [(s-2t) \hat{i} + (3-t) \hat{j} + (2s+t) \hat{k}] = 15$

If the plane $56x + 4y + 9z = 2016$ meets the coordinate axes at points $A$,$B$,and $C$,then the centroid of the $\triangle ABC$ is

The equation of the plane passing through the points $(2, 3, 1)$ and $(4, -5, 3)$ and parallel to the $X$-axis is:

If $L_1$ and $L_2$ are two lines which pass through the origin and have direction ratios $(3, 1, -5)$ and $(2, 3, -1)$ respectively,then the equation of the plane containing $L_1$ and $L_2$ is

What are the direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module