The length of the perpendicular from the orig | vedclass

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

Question

(C) The plane passes through the point $a$ and contains the line $r = b + \lambda c$. Therefore,the plane is parallel to the vector $c$ and the vector

Vedclass · Accepted Answer

$\frac{[a, b, c]}{|b 	imes c + c 	imes a|}$

Vedclass · Answer

(C) The plane passes through the point $a$ and contains the line $r = b + \lambda c$. Therefore,the plane is parallel to the vector $c$ and the vector $(b - a)$.The normal vector $n$ to the plane is given by the cross product of these two vectors:$n = (b - a) 	imes c = b 	imes c - a 	imes c = b 	imes c + c 	imes a$.The equation of the plane is $(r - a) \cdot n = 0$,which simplifies to $r \cdot n = a \cdot n$.Substituting $n = b 	imes c + c 	imes a$,we get $r \cdot (b 	imes c + c 	imes a) = a \cdot (b 	imes c + c 	imes a)$.Since $a \cdot (b 	imes c + c 	imes a) = a \cdot (b 	imes c) + a \cdot (c 	imes a) = [a, b, c] + 0 = [a, b, c]$,the equation of the plane is $r \cdot (b 	imes c + c 	imes a) = [a, b, c]$.The length of the perpendicular from the origin $(0, 0, 0)$ to the plane $r \cdot n = d$ is given by $\frac{|d|}{|n|}$.Here,$d = [a, b, c]$ and $n = b 	imes c + c 	imes a$.Thus,the length of the perpendicular is $\frac{[a, b, c]}{|b 	imes c + c 	imes a|}$.

The length of the perpendicular from the origin to the plane passing through the point $a$ and containing the line $r = b + \lambda c$ is

Similar Questions

The length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $\vec r = (\hat i + \hat j) + \lambda (\hat i + 2\hat j - \hat k)$ and $\vec r = (\hat i + \hat j) + \mu (-\hat i + \hat j - 2\hat k)$ is

$\pi_1$ is a plane passing through the point $(1, 2, 3)$ and perpendicular to the planes $x+2y+3z-6=0$ and $x+2y+2z-5=0$. If $(-1, 2, -3)$ is the foot of the perpendicular drawn from the point $(1, 3, 2)$ onto a plane $\pi_2$,then the angle between the planes $\pi_1$ and $\pi_2$ is

Which of the following points lie on the plane passing through $2 \hat{i}+3 \hat{j}-\hat{k}$,$3 \hat{i}+2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+4 \hat{k}$?

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to

Find the equation of the plane that passes through the three points $(1,1,0), (1,2,1),$ and $(-2,2,-1)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module