Question

Find the co-ordinates of those points on the line $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6}$ which is at a distance of $3$ units from point $(1,-2,3)$.

Answer 1

Given line,

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6}$

Hence point of the line satisfy the condition,

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6}=R$

$or,\left(x,y,z\right)=\left(2k+1,3R-2,6R+3\right)......\left(i\right)$

(where $R$ is a constant)

Now, if distance between a point on the line and a point $(1,-2,3)$ is $3$ unit, then we have

$\sqrt{{\left(2R+1-1\right)}^2+{\left(3R-2+2\right)}^2+{\left(6R+3-3\right)}^2}=\mathrm{3......}\left(ii\right)$

Solving the above equation for $R$.

We get $R=\frac{3}{7}$

Now,

From (i) we get the coordinates of required point as $\left(\frac{13}{7},\frac{-5}{7},\frac{39}{7}\right)$