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

Find the coordinates of the foot of the perpendicular drawn from \( (0,0,0) \) to the line

The equation of the line is given by:
\[ \frac{x}{1} = \frac{y + 1}{-1} = \frac{z - 3}{-2} \] This equation represents a parametric form of the line. Let the parameter be \( t \). We can rewrite the parametric equations of the line as:
\[ x = t, \quad y = -1 - t, \quad z = 3 - 2t \] Thus, the coordinates of any point on the line are \( (t, -1 - t, 3 - 2t) \).

Step 1: Direction Vector of the Line
The direction vector of the line is given by the coefficients of \( t \) in the parametric equations, which is \( \vec{v} = (1, -1, -2) \).

Step 2: Perpendicular from the Origin
The coordinates of the foot of the perpendicular from the point \( P(0, 0, 0) \) to the line will be the point on the line closest to the origin. The vector \( \overrightarrow{OP} \) (from the origin to any point \( (t, -1 - t, 3 - 2t) \) on the line) is given by:
\[ \overrightarrow{OP} = (t, -1 - t, 3 - 2t) \] The foot of the perpendicular lies on the line, so the vector \( \overrightarrow{OP} \) must be perpendicular to the direction vector \( \vec{v} = (1, -1, -2) \).

Step 3: Find the value of \( t \)
For the vectors \( \overrightarrow{OP} \) and \( \vec{v} \) to be perpendicular, their dot product must be zero. The dot product of \( \overrightarrow{OP} = (t, -1 - t, 3 - 2t) \) and \( \vec{v} = (1, -1, -2) \) is:
\[ t \cdot 1 + (-1 - t) \cdot (-1) + (3 - 2t) \cdot (-2) \] Simplifying: \[ t + (1 + t) - 2(3 - 2t) = 0 \] \[ t + 1 + t - 6 + 4t = 0 \] \[ 6t - 5 = 0 \] \[ t = \frac{5}{6} \]

Step 4: Find the Coordinates of the Foot of the Perpendicular
Substituting \( t = \frac{5}{6} \) into the parametric equations of the line: \[ x = \frac{5}{6}, \quad y = -1 - \frac{5}{6} = -\frac{11}{6}, \quad z = 3 - 2 \times \frac{5}{6} = \frac{8}{3} \] Therefore, the coordinates of the foot of the perpendicular are \( \left( \frac{5}{6}, -\frac{11}{6}, \frac{8}{3} \right) \).

Final Answer:
The coordinates of the foot of the perpendicular are \( \boxed{\left( \frac{5}{6}, -\frac{11}{6}, \frac{8}{3} \right)} \).