A plane E_{1} makes intercepts 1, -3, 4 on th | vedclass

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

Question

(C) The intercept form of the plane $E_{1}$ with intercepts $a=1, b=-3, c=4$ is $\frac{x}{1} + \frac{y}{-3} + \frac{z}{4} = 1$. Multiplying by $12$,we

Vedclass · Accepted Answer

$\frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0$

Vedclass · Answer

(C) The intercept form of the plane $E_{1}$ with intercepts $a=1, b=-3, c=4$ is $\frac{x}{1} + \frac{y}{-3} + \frac{z}{4} = 1$. Multiplying by $12$,we get $12x - 4y + 3z = 12$,or $12x - 4y + 3z - 12 = 0$. The normal vector to this plane is $\vec{n} = 12\hat{i} - 4\hat{j} + 3\hat{k}$. Since the required plane is parallel to $E_{1}$,its equation is of the form $12x - 4y + 3z + k = 0$. Since it passes through $(2, 6, -8)$,we substitute these coordinates: $12(2) - 4(6) + 3(-8) + k = 0$. $24 - 24 - 24 + k = 0 \Rightarrow k = 24$. Thus,the equation is $12x - 4y + 3z + 24 = 0$. Dividing by $12$,we get $x - \frac{y}{3} + \frac{z}{4} + 2 = 0$.

$A$ plane $E_{1}$ makes intercepts $1, -3, 4$ on the coordinate axes. The equation of a plane parallel to plane $E_{1}$ and passing through $(2, 6, -8)$ is

Similar Questions

If the distance between the planes $2x + y + z + 1 = 0$ and $2x + y + z + \alpha = 0$ is $3$ units,then the product of all possible values of $\alpha$ is

The equation of the plane perpendicular to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and passing through the point $(2, 3, 4)$ is:

Let $P$ be a plane passing through the points $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$,and $R$ be the point $(2,1,6)$. Then the image of $R$ in the plane $P$ is:

If for some $\alpha$ and $\beta$ in $\mathbb{R},$ the intersection of the following three planes $x+4y-2z=1$,$x+7y-5z=\beta$,and $x+5y+\alpha z=5$ is a line in $\mathbb{R}^{3},$ then $\alpha+\beta$ is equal to

Find the angle between the planes $x+2y+2z-5=0$ and $3x+3y+2z-8=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module