1. Tardigrade
  2. Question
  3. Mathematics
  4. Consider two planes P1: 2 x-3 y+z=-5 and P2: x-4 y-14 z=5. If a line L which is perpendicular to line of intersection of planes P1 P2 and whose one point A(-7, y1, z1) lies on the plane P1 and points B(x2, y2, z2) and C(x3, y3, z3) lie on plane P2 has the equation (x+7/a)=(y-y1/-b) =( z - z I / c ), where a , b , c ∈ N, then

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Consider two planes $P_1: 2 x-3 y+z=-5$ and $P_2: x-4 y-14 z=5$. If a line $L$ which is perpendicular to line of intersection of planes $P_1 \& P_2$ and whose one point $A\left(-7, y_1, z_1\right)$ lies on the plane $P_1$ and points $B\left(x_2, y_2, z_2\right)$ and $C\left(x_3, y_3, z_3\right)$ lie on plane $P_2$ has the equation $\frac{x+7}{a}=\frac{y-y_1}{-b}$ $=\frac{ z - z _{ I }}{ c }$, where $a , b , c \in N$, then

Vector Algebra

Solution:

Planes $P _1$ and $P _2$ are perpendicular
$L : \frac{ x +7}{ a }=\frac{ y - y _1}{- b }=\frac{ z - z _1}{ c }$
The above line lies in plane $P _2$ and perpendicular to plane $P _1$
$\therefore A \left(-7, y _1, z _1\right)$ lies on both the planes.
$\therefore -14-3 y _1+ z _1 =-5 $ ....(i)
$-7-4 y _1-14 z _1 =5$ ....(ii)
image
Solving (i) and (ii), we get $y _1=-3$ and $z _1=0$
$\therefore L: \frac{x+7}{2}=\frac{y+3}{-3}=\frac{z-0}{1}$