The distance of the point $-\hat{i}+2 \hat{j}+6 \hat{k}$ from the straight line that passes through the…
The distance of the point $-\hat{i}+2 \hat{j}+6 \hat{k}$ from the straight line that passes through the point $2 \hat{i}+3 \hat{j}-4 \hat{k}$ and is parallel to the vector $6 \hat{i}+3 \hat{j}-4 \hat{k}$ is
9
8
7
10
Solution
Point is $(-1,2,6)$
Line passes through the point $(2,3,-4)$ parallel to vector whose direction ratios is $6,3,-4$.
Equation is $\frac{x-2}{6}=\frac{y-3}{3}=\frac{z+4}{-4}=\lambda$
Any point on this line is given by $x=6$ $\lambda+2, y=3 \lambda+3, z=-4 \lambda-4$
Now, d. Rs of line passing through $(-1,2,6)$ and $\perp$ to this line is $\{(x+1),(y-2),(z-6)\}$
So, $6(x+1)+3(y-2)-4(z-6)=0$ $\Rightarrow 6 x+3 y-4 z+24=0$
Now, $6(6 \lambda+2)+3(3 \lambda+3)+4(4 \lambda+4)$ $+24=0$
$\Rightarrow 61 \lambda+61=0 \Rightarrow \lambda=-1$
So, $x=-4, y=0, z=0$
Now, distance between $(-1,2,6)$ and $(-4,0,0)$ is
$\sqrt{9+4+36}=\sqrt{49}=7$