Question

The length of the perpendicular from the point \( P(1, -1, 2) \) to the given line \[ \frac{x + 1}{2} = \frac{y - 2}{-3} = \frac{z + 2}{4} \]

• A. \( \sqrt{29} \) units
• B. \( \sqrt{21} \) units
• C. \( \sqrt{6} \) units
• D. 0 units

Hint:

The perpendicular distance from a point to a line in 3D can be found by using the vector cross product method or the dot product condition for perpendicularity.

Answer 1

The Correct Option is D

Step 1: Parametric form of the given line.
The given symmetric equation is:
\[ \frac{x + 1}{2} = \frac{y - 2}{-3} = \frac{z + 2}{4} \]
Let \( t \) be the parameter. The parametric equations of the line are:
\[ x = 2t - 1, \quad y = -3t + 2, \quad z = 4t - 2 \]

Step 2: Find the point on the line closest to \( P(1, -1, 2) \).
To find the perpendicular distance from point \( P(1, -1, 2) \) to the line, we need the point on the line closest to \( P \). Using the vector form of the line, we get:
\[ \vec{L}(t) = (2t - 1, -3t + 2, 4t - 2) \]
The vector from \( P(1, -1, 2) \) to a point on the line is:
\[ \vec{v}(t) = (2t - 1 - 1, -3t + 2 + 1, 4t - 2 - 2) = (2t - 2, -3t + 3, 4t - 4) \]

Step 3: Dot product condition.
For the vector \( \vec{v}(t) \) to be perpendicular to the line, its dot product with the direction vector of the line \( \vec{d} = (2, -3, 4) \) must be zero:
\[ (2t - 2, -3t + 3, 4t - 4) \cdot (2, -3, 4) = 0 \]
Expanding the dot product:
\[ 2(2t - 2) - 3(-3t + 3) + 4(4t - 4) = 0 \]
Simplifying:
\[ 4t - 4 + 9t - 9 + 16t - 16 = 0 \] \[ 29t - 29 = 0 \Rightarrow t = 1 \]

Step 4: Find the coordinates of the foot of the perpendicular.
Substitute \( t = 1 \) into the parametric equations of the line:
\[ x = 2(1) - 1 = 1, \quad y = -3(1) + 2 = -1, \quad z = 4(1) - 2 = 2 \] So, the foot of the perpendicular is the point \( (1, -1, 2) \).

Step 5: Conclusion.
Since the foot of the perpendicular coincides with the point \( P(1, -1, 2) \), the length of the perpendicular is \( 0 \) units.