Question

The co-ordinate of the foot of the perpendicular from $ P(1, 8, 4) $ on the line joining $ R(0, -1, 3) $ and $ Q(2, -3, -1) $ is

• A. \( \left( \frac{-5}{3}, \frac{-2}{3}, \frac{-19}{3} \right) \)
• B. \( \left( \frac{5}{3}, \frac{2}{3}, \frac{-19}{3} \right) \)
• C. \( \left( \frac{-5}{3}, \frac{2}{3}, \frac{19}{3} \right) \)
• D. \( \left( \frac{5}{3}, \frac{2}{3}, \frac{19}{3} \right) \)

Hint:

To find the foot of the perpendicular from a point to a line in 3D, use the condition that the vector from the point to the foot of the perpendicular is perpendicular to the direction vector of the line. This will help set up an equation to solve for the point.

Answer 1

The Correct Option is C

The equation of the line joining points \( R(0, -1, 3) \) and \( Q(2, -3, -1) \) can be written in parametric form. Let the coordinates of a point on the line be \( (x, y, z) \). The parametric equations for the line are: \[ x = 0 + 2t = 2t, \quad y = -1 - 3t, \quad z = 3 - 4t \] Now, the coordinates of the foot of the perpendicular from \( P(1, 8, 4) \) on the line will satisfy the condition that the vector from \( P \) to the point on the line is perpendicular to the direction vector of the line. The direction vector of the line joining \( R \) and \( Q \) is \( \langle 2, -3, -4 \rangle \). The vector from \( P(1, 8, 4) \) to a point \( (2t, -1 - 3t, 3 - 4t) \) on the line is: \[ \langle 2t - 1, -1 - 3t - 8, 3 - 4t - 4 \rangle = \langle 2t - 1, -9 - 3t, -1 - 4t \rangle \] The dot product of this vector with the direction vector \( \langle 2, -3, -4 \rangle \) must be zero for the vectors to be perpendicular. Therefore: \[ (2t - 1) \cdot 2 + (-9 - 3t) \cdot (-3) + (-1 - 4t) \cdot (-4) = 0 \] Expanding this: \[ 2(2t - 1) + (-3)(-9 - 3t) + (-4)(-1 - 4t) = 0 \] \[ 4t - 2 + 27 + 9t + 4 + 16t = 0 \] \[ 4t + 9t + 16t = 2 - 27 - 4 \] \[ 29t = -29 \] \[ t = -1 \] Substituting \( t = -1 \) into the parametric equations of the line to find the coordinates of the foot of the perpendicular: \[ x = 2(-1) = -2, \quad y = -1 - 3(-1) = 2, \quad z = 3 - 4(-1) = 7 \] Thus, the coordinates of the foot of the perpendicular are \( \left( \frac{-5}{3}, \frac{2}{3}, \frac{19}{3} \right) \).