The equation of the plane which is parallel t | vedclass

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

Question

(B) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, 0, 0)$,we have $a(0) + b(0) + c(0) = d$,so $d = 0$. Thus,the pla

Vedclass · Accepted Answer

$x - 19y - 11z = 0$

Vedclass · Answer

(B) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, 0, 0)$,we have $a(0) + b(0) + c(0) = d$,so $d = 0$. Thus,the plane is $ax + by + cz = 0$. Since it passes through $(3, -1, 2)$,we have $3a - b + 2c = 0$ (Equation $1$). The plane is parallel to the line with direction ratios $(1, -4, 7)$,so the normal to the plane is perpendicular to the line. Thus,$a(1) + b(-4) + c(7) = 0$,which gives $a - 4b + 7c = 0$ (Equation $2$). Solving Equations $1$ and $2$ using cross-multiplication: $\frac{a}{(-1)(7) - (2)(-4)} = \frac{-b}{(3)(7) - (2)(1)} = \frac{c}{(3)(-4) - (-1)(1)}$ $\frac{a}{-7 + 8} = \frac{-b}{21 - 2} = \frac{c}{-12 + 1}$ $\frac{a}{1} = \frac{-b}{19} = \frac{c}{-11}$ So,$a = 1, b = -19, c = -11$. The equation of the plane is $1x - 19y - 11z = 0$.

The equation of the plane which is parallel to the line $\frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7}$ and passes through the points $(0, 0, 0)$ and $(3, -1, 2)$ is

Similar Questions

Point $P$ is the intersection of the line joining points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from point $T(2, 1, 4)$ to the line $QR$,find the length of the line segment $PS$.

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to each of the planes $3x - 2y + z = 8$ and $x + y + z = 6$ is $lx + my + nz = 1$,then $4m + 2n - 31 =$

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,if

The equation of a plane through the line of intersection of the planes $x + 2y = 3$ and $y - 2z + 1 = 0$,and perpendicular to the first plane $x + 2y = 3$ is:

Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=1$. If $L$ makes an angle $\alpha$ with the positive $x$-axis,then $\cos \alpha$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module