Question

The distance of the point $\left(1,1,9\right)$ from the point of intersection of the line $\left(\frac{\mathrm{x}-3}{1}\right)=\left(\frac{\mathrm{y}-4}{2}\right)=\left(\frac{\mathrm{z}-5}{2}\right)$ and the plane $\mathrm{x}+\mathrm{y}+\mathrm{z}=17$ is:

• A. $\sqrt{38}$
• B. $19\sqrt{2}$
• C. $2\sqrt{19}$
• D. $38$

Answer 1

The correct option is A

$\sqrt{38}$

Explanation for the correct option:

Finding the distance:

Given,

Point $A\left(1,1,9\right)$

Line $\left(\frac{\mathrm{x}-3}{1}\right)=\left(\frac{\mathrm{y}-4}{2}\right)=\left(\frac{\mathrm{z}-5}{2}\right)$

Plane $\mathrm{x}+\mathrm{y}+\mathrm{z}=17$

Let, $\left(\frac{\mathrm{x}-3}{1}\right)=\left(\frac{\mathrm{y}-4}{2}\right)=\left(\frac{\mathrm{z}-5}{2}\right)=\mathrm{t}$

Equating the above expression, we get

$\mathrm{x}=3+\mathrm{t}\dots \dots \left(1\right)$

$\mathrm{y}=4+2\mathrm{t}\dots \dots \left(2\right)$

$\mathrm{z}=5+2\mathrm{t}\dots \dots \left(3\right)$

Substituting the values of x, y, z in plane equation

$\mathrm{x}+\mathrm{y}+\mathrm{z}=17$

$\mathrm{t}+3+2\mathrm{t}+4+2\mathrm{t}+5=17$

$5\mathrm{t}+12=17$

$\mathrm{t}=1$

Substituting the value of $t$ in equations $\left(1\right),\left(2\right)\text{and}\left(3\right)$ we get

$\mathrm{x}=4,\mathrm{y}=6,\mathrm{z}=7$

Hence, $\mathrm{Point}\mathrm{B}=\left(4,6,7\right)$

Therefore, the distance between AB $=\sqrt{(x_1-x{)}^2+(y_2-y{)}^2+(z_2-z{)}^2}$

$=\sqrt{(4-1{)}^2+(6-1{)}^2+(9-7{)}^2}$

$=\sqrt{3^2+5^2+2^2}$

$=\sqrt{38}$

Hence, option (A) is the correct answer.