Browse Questions

# Find the equation of the plane passing through the points $(3,4,1)$ and $(0,1,0)$ and parallel to the line $\large\frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}$

Toolbox:
• Eqn. of a line is $\large\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}=\lambda$ where $(a_1,a_2,a_3)$ are the point on the line and $(b_1,b_2,b_3)$ is d.r. of the line.
Step 1:
The given points are $P(3,4,1)$ and $Q(0,1,0)$
Equation of the plane passing through $(3,4,1)$ is
$a(x-3)+b(y-4)+c(2-1)=0$-------(1)
This plane passes through $(0,1,0)$
$\therefore a(0-3)+b(1-4)+c(0-1)=0$
$-3a-3b-c=0$
(i.e) $3a+3b+c=0$-----(2)
Step 2:
The plane (equ(1)) is parallel to the given line
$\large\frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}$
$\therefore$ Normal to the plane is perpendicular to the line.
$\therefore a(2)+7(b)+c(5)=0$
$\Rightarrow 2a+7b+5c=0$------(3)
Step 3:
Now let us solve equ(2) and equ(3)
$\large\frac{a}{\begin{vmatrix}3 & 1\\7 & 5\end{vmatrix}}=\large\frac{b}{\begin{vmatrix}1 & 3\\5& 2\end{vmatrix}}=\large\frac{c}{\begin{vmatrix}3 & 3\\2 & 7\end{vmatrix}}$
(i.e) $\large\frac{a}{8}=\frac{b}{-13}=\frac{c}{15}$$=k$
$a=8k$
$b=-13k$
$c=15k$
Step 4:
Substituting these values in equ(1) we get,
$8(x-3)-13(y-4)+15(2-1)=0$
$\Rightarrow 8x-13y+15z+12=0$
This is the required equation of the plane.