Question

Find the distance between point $P(6,5,9)$ and the plane determined by the points $A(3,-1,2);B(5,2,4);C(-1,-1,6)$

• A. $\frac{3\sqrt{34}}{17}$
• B. $\frac{3\sqrt{33}}{17}$
• C. $\frac{3\sqrt{34}}{19}$
• D. $\frac{3\sqrt{35}}{19}$

Answer 1

The correct option is A $\frac{3\sqrt{34}}{17}$

The equation of a plane passing through points $A(x_1,y_1,z_1),B(x_2,y_2,z_2)$ and $C(x_3,y_3,z_3)$ is

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

Given, the three points are,

$A(3,-1,2),B(5,2,4)$ and $C(-1,-1,6)$

So, $x_1=3,y_1=-1,z_1=2,x_2=5,y_2=2,z_2=4$ and $x_3=-1,y_3=-1,z_3=6$

$\left|\begin{array}{ccc}x-3& y-(-1)& z-2\\ 5-3& 2-(-1)& 4-2\\ -1-3& -1-(-1)& 6-2\end{array}\right|=0$

$\left|\begin{array}{ccc}x-3& y+1& z-2\\ 2& 3& 2\\ -4& 0& 4\end{array}\right|=0$

$\Rightarrow$ $(x-3)[12-0]-(y+1)[8+8]+(z-2)[0+12]=0$

$\Rightarrow$ $(x-3)\left(12\right)-16(y+1)+12(z-2)=0$

$\Rightarrow$ $3(x-3)-4(y+1)+3(z-2)=0$

$\Rightarrow$ $3x-9-4y-4+3z-6=0$

$\Rightarrow$ $3x-4y+3z=19$

$\therefore$ Equation of plane is $3x-4y+3z=19$

Comparing with $Ax+By+Cz=D$ we get,

$A=3,B=-4,C=3$ and $D=19$

Distance of point form plane $=\left|\frac{A{x}_1+B{y}_1+C{z}_1+-D}{\sqrt{A^2+B^2+C^2}}\right|$

$=\left|\frac{(3\times 6)+(-4\times 5)+(3\times 9)-19}{\sqrt{3^2+4^2+3^2}}\right|$

$=\left|\frac{18-20+27-19}{\sqrt{9+16+9}}\right|$

$=\left|\frac{6}{\sqrt{34}}\right|$

$=\frac{6}{\sqrt{34}}$

$=\frac{6}{\sqrt{34}}\times \frac{\sqrt{34}}{\sqrt{34}}$

$=\frac{6\sqrt{34}}{34}$

$=\frac{3\sqrt{34}}{17}$