Question

Find the co-ordinates of a point, which is at a distance of $21$ units from the point $A=(1,-3,4)$ in the direction of vectors $2\hat{i}-3\hat{j}-6\hat{k}$

Answer 1

Let the required point be $P(x_1,y_1,z_1)$

Direvtion ratio of line $x_1-1,y_1+3,z_1-4$

Since AP is parallel to $2\hat{i}-3\hat{j}-6\hat{k}$

So, $\frac{x_1-1}{2},\frac{y_1+3}{-3},\frac{z_1-4}{-6}=k$

where $k$ is any constant

$\Rightarrow x_1=2k+1,y_1=-3k-3,z_1=-6k+4$

But it is given that $AP=21$

so $\sqrt{{\left(x_1-1\right)}^2+{\left(y_1+3\right)}^2+{\left(z_1-4\right)}^2}=21$

$\sqrt{{\left(2k\right)}^2+{\left(-3k\right)}^2+{\left(-6k\right)}^2}=21$

$\sqrt{49k^2}=21$

$k=3$

Therefore required point

$\begin{array}{c}P=\left(2k+1,-3k-3,-6k+4\right)\\ =\left(2\times 3+1,-3\times 3-3,-6\times 3+4\right)\\ =\left(7,-12,-14\right)\end{array}$