The equation of the plane passing through the | vedclass

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

Question

(D) Let the equation of the plane be $a(x - 4) + b(y - 3) + c(z - 2) = 0$. Since the plane contains the lines,the normal vector $(a, b, c)$ is perpend

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let the equation of the plane be $a(x - 4) + b(y - 3) + c(z - 2) = 0$. Since the plane contains the lines,the normal vector $(a, b, c)$ is perpendicular to the direction vectors of both lines,$(1, 1, 2)$ and $(1, -4, 5)$. Thus,$a + b + 2c = 0$ and $a - 4b + 5c = 0$. Using the cross product to find the normal vector $(a, b, c) = (1, 1, 2) 	imes (1, -4, 5)$: $a = (1)(5) - (2)(-4) = 5 + 8 = 13$. $b = (2)(1) - (1)(5) = 2 - 5 = -3$. $c = (1)(-4) - (1)(1) = -4 - 1 = -5$. So the normal vector is $(13, -3, -5)$. The equation of the plane is $13(x - 4) - 3(y - 3) - 5(z - 2) = 0$. $13x - 52 - 3y + 9 - 5z + 10 = 0$. $13x - 3y - 5z - 33 = 0$ or $13x - 3y - 5z = 33$. Since this does not match any of the given options,the correct answer is $(d)$.

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{-4} = \frac{z}{5}$ is

Similar Questions

Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ and the distance of the plane $P$ from the point $(1,27,7)$ be $c$. Then $a^2+b^2+c^2$ is equal to $.............$.

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0$. Then the value of $(\beta-\alpha)$ is equal to

If the straight lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$,with parameters $s$ and $t$ respectively,are coplanar,then $\lambda$ equals

The distance of the point $(3, 8, 2)$ from the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{3}$ measured parallel to the plane $3x + 2y - 2z = 0$ is

The equation of a plane at a distance of $\sqrt{\frac{2}{21}}$ from the origin,which contains the line of intersection of the planes $x-y-z-1=0$ and $2x+y-3z+4=0$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module