Question

The distance of origin from the point of intersection of the line $\left(\frac{\mathrm{x}}{2}\right)=\frac{\left[\mathrm{y}-2\right]}{3}=\frac{\left[\mathrm{z}-3\right]}{4}$ and the plane $2\mathrm{x}+\mathrm{y}-\mathrm{z}=2$ is

• A. $\sqrt{120}$
• B. $\sqrt{83}$
• C. $2\sqrt{19}$
• D. $\sqrt{78}$

Answer 1

The correct option is D

$\sqrt{78}$

Explanation for the correct option:

Step-1 : Find the value of x,y and z

Given, the point of intersection is given by $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{x}-3}{4}$

Equation of the plane is $2\mathrm{x}+\mathrm{y}-\mathrm{z}=2\dots \dots \left(1\right)$

Let, the points on this line be$\frac{\mathrm{x}}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{x}-3}{4}=\mathrm{t}$

Equating the above expression we get,

$x=2\mathrm{t},y=3\mathrm{t}+2,z=4\mathrm{t}+3$

Substituting the values of $x,y,z$ in equation $\left(1\right)$

$2\left(2\mathrm{t}\right)+3\mathrm{t}+2-(4\mathrm{t}+3)=2$

$4\mathrm{t}+3\mathrm{t}-4\mathrm{t}-1=2$

$$\begin{gathered} 3\mathrm{t}=3 \\ \mathrm{t}=1 \end{gathered}$$

Substituting the value of $t$ in the equations for $x,y,z$

$\begin{array}{rcl}x& =& 2t\\ & =& 2\times 1=2\\ y& =& 3t+2\\ & =& \left(3\times 1\right)+2=5\\ z& =& 4t+3\\ & =& \left(4\times 1\right)+3=7\end{array}$

Thus the point is $(2,5,7)$

Step-2 :Find the distance of origin from the point of intersection.

Distance from origin$$\begin{gathered} =\sqrt{x^2+y^2+z^2} \\ =\sqrt{2^2+5^2+7^2} \end{gathered}$$

$=\sqrt{4+25+49}$

$=\sqrt{\mathrm{78}}$

Hence, option (D) is the correct answer.