Question

Find the coordinates of the foot of perpendicular drawn from $(0,0,0)$ to the line \[ \frac{x}{1}=\frac{y+1}{-1}=\frac{z-3}{-2} \]

Hint:

To find the foot of the perpendicular from a point to a line in 3D, first write the line in parametric form, then use the fact that the vector joining the point to the foot is perpendicular to the direction vector of the line.

Answer 1

Step 1: Convert the line into parametric form.
Let \[ \frac{x}{1}=\frac{y+1}{-1}=\frac{z-3}{-2}=t \] Then \[ x=t \] \[ y=-t-1 \] \[ z=-2t+3 \] Thus the coordinates of any point on the line are \[ P(t,-t-1,-2t+3) \] Step 2: Use perpendicular condition.
The foot of the perpendicular from the origin will satisfy that the vector from origin to this point is perpendicular to the direction of the line.
Direction ratios of the line are \[ (1,-1,-2) \] Vector \[ \overrightarrow{OP}=(t,-t-1,-2t+3) \] Perpendicular condition: \[ \overrightarrow{OP}\cdot(1,-1,-2)=0 \] Step 3: Compute the dot product.
\[ t(1)+(-t-1)(-1)+(-2t+3)(-2)=0 \] \[ t+t+1+4t-6=0 \] \[ 6t-5=0 \] \[ t=\frac{5}{6} \] Step 4: Substitute the value of $t$.
\[ x=\frac{5}{6} \] \[ y=-\frac{5}{6}-1=-\frac{11}{6} \] \[ z=-2\left(\frac{5}{6}\right)+3 \] \[ =-\frac{10}{6}+\frac{18}{6} \] \[ =\frac{8}{6}=\frac{4}{3} \] Step 5: Final conclusion.
Thus the coordinates of the foot of perpendicular are \[ \left(\frac{5}{6},-\frac{11}{6},\frac{4}{3}\right) \] Final Answer: \[ \boxed{\left(\frac{5}{6},-\frac{11}{6},\frac{4}{3}\right)} \]