Mathematics - Vector Question with Solution | TestHub

Given: 

$\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$

$\Rightarrow \frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}=k\left(\text{let}\right)$

$\Rightarrow x=2, y=2k, z=-k+3 ...\left(i\right)$

Also, $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}=\lambda \left(\text{let}\right)$

$\Rightarrow x=\lambda +1, y=-3\lambda -3. z=\lambda -1 ...\left(ii\right)$

Using $\left(i\right) \text{and} \left(ii\right)$,

$\Rightarrow 2=\lambda +1, 2k=-3\lambda -3$

$\Rightarrow \lambda =1, 2k=-3-3$

$\Rightarrow \lambda =1, k=-3$

$\Rightarrow x=2, y=-6, z=0$

So, the intersection point of two given lines is $\left(2,-6,0\right)$.

Now, we know that centroid $G$ divides $MR$ (median) in the ratio $1 : 2$.

So, by using section formula, $G\equiv \left(\frac{1\times -1+2\times 2}{3},\frac{1\times 4+2\times 1}{3},\frac{1\times 2+2\times 2}{3}\right)$

So, centroid is $G(1, 2, 2)$.

Let $A$ be the point of intersection of given lines.

$\Rightarrow AG=\sqrt{{\left(2-1\right)}^2+{\left(2+6\right)}^2+{\left(2-0\right)}^2}$

$\Rightarrow AG=\sqrt{1+64+4}$

$\Rightarrow AG=\sqrt{69}$