If A and B are the feet of the perpendiculars | vedclass

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

Question

(B) The point $Q$ is $(a, b, c)$. The foot of the perpendicular from $Q$ to the $YZ$-plane $(x=0)$ is $A(0, b, c)$. The foot of the perpendicular from

Vedclass · Accepted Answer

$\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$

Vedclass · Answer

(B) The point $Q$ is $(a, b, c)$. The foot of the perpendicular from $Q$ to the $YZ$-plane $(x=0)$ is $A(0, b, c)$. The foot of the perpendicular from $Q$ to the $ZX$-plane $(y=0)$ is $B(a, 0, c)$. The origin $O$ is $(0, 0, 0)$. The equation of the plane passing through $(0, 0, 0)$,$(0, b, c)$,and $(a, 0, c)$ is given by the determinant equation: $\begin{vmatrix} x & y & z \ 0 & b & c \ a & 0 & c \end{vmatrix} = 0$ Expanding along the first row: $x(bc - 0) - y(0 - ac) + z(0 - ab) = 0$ $bcx + acy - abz = 0$ Dividing the entire equation by $abc$ (assuming $a, b, c 
eq 0$): $\frac{bcx}{abc} + \frac{acy}{abc} - \frac{abz}{abc} = 0$ $\frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$

If $A$ and $B$ are the feet of the perpendiculars drawn from the point $Q(a, b, c)$ to the planes $YZ$ and $ZX$ respectively,then the equation of the plane passing through the points $A, B$,and $O$ is (where $O$ is the origin).

Similar Questions

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

The equation of the plane passing through the point $(1,2,2)$ and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$ is:

The Cartesian equation of the plane passing through the point $A(7, 8, 6)$ and parallel to the $XY$-plane is:

Find the equation of a plane that is at a unit distance from the origin and is parallel to the plane $x - 2y + 2z - 5 = 0$.

The direction ratios of the normal to the plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle of $\pi/4$ with the plane $2x + 3y = 7$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module