Question

the equation of the plane which passes through the point $(2,1,-1)$ and $(-1,3,4)$ and perpendicular to the plane $x-2y+4z=0$ is

• A. $18x+17y+4z=49$
• B. $18x-17y+4z=49$
• C. $18x+17y-4z+49=0$
• D. $18x+17y+4z+49=0$

Answer 1

The correct option is A $18x+17y+4z=49$

Let the equation of the plane through $(2,1,-1)$ be

$a(x-2)+b(y-1)+c(z+1)=0⟹\left(1\right)$

If it passes through $(-1,3,4)$ then,substitute for x,y and z.we get,

$a(-1-2)+b(3-1)+c(4+1)=0$

$-3a+2b+5c=0⟹\left(2\right)$

If this plane is perpendicular to the plane

$x-2y+4z=10$ then

$a-2b+4c=0⟹\left(3\right)$

Now let us solve the equation $\left(2\right)$ and $\left(3\right)$

$\frac{a}{\left|\begin{array}{cc}2& 5\\ -2& 4\end{array}\right|}=\frac{b}{\left|\begin{array}{cc}5& -3\\ 4& 1\end{array}\right|}=\frac{c}{\left|\begin{array}{cc}-3& 2\\ 1& -2\end{array}\right|}$

$\frac{a}{8+10}=\frac{b}{5+12}=\frac{c}{6-2}$

$\frac{a}{18}=\frac{b}{17}=\frac{c}{4}=\lambda \left(say\right)$

$a=18\lambda ,b=17\lambda ,c=4\lambda$

Substituitng for $a,b,c$ in equ $\left(1\right)$ we get,

$18\lambda (x-2)+17\lambda (y-1)+4\lambda (z+1)=0$

$18x+17y+4z-49=0$

$18x+17y+4z=49$

This is the required equation of the plane.