Question

If \( A(3, -1, 1), B(0, 2, 3), C(4, 8, 11) \) are three points, then the coordinates of the foot of the perpendicular drawn from the point A to the line joining the points B and C is:

• A. \( (3, 5, 7) \)
• B. \( (5, 9, 6) \)
• C. \( (2, 5, 7) \)
• D. \( (1, 2, 3) \)

Hint:

To find the foot of the perpendicular from a point to a line, use the parametric equations of the line and solve for the point where the dot product of the direction vector and the vector from the point to the line is zero.

Answer 1

The Correct Option is C

We are given three points: \( A(3, -1, 1) \), \( B(0, 2, 3) \), and \( C(4, 8, 11) \). We need to find the coordinates of the foot of the perpendicular drawn from the point \( A \) to the line joining the points \( B \) and \( C \). Step 1: Parametrize the line BC The direction ratios of the line joining \( B \) and \( C \) can be found by subtracting the coordinates of \( B \) from the coordinates of \( C \): \[ \text{Direction ratios of BC} = C - B = (4 - 0, 8 - 2, 11 - 3) = (4, 6, 8) \] Thus, the parametric equations of the line joining \( B \) and \( C \) are: \[ x = 0 + 4t = 4t, \quad y = 2 + 6t = 2 + 6t, \quad z = 3 + 8t = 3 + 8t \] Step 2: Equation for the perpendicular The vector from \( A \) to a point on the line \( BC \) is given by: \[ \overrightarrow{AP} = (x - 3, y + 1, z - 1) \] For the line and the perpendicular to be perpendicular, the dot product of \( \overrightarrow{AP} \) and the direction vector of the line \( BC \), i.e., \( (4, 6, 8) \), must be zero: \[ (x - 3, y + 1, z - 1) \cdot (4, 6, 8) = 0 \] Step 3: Solve the system Substitute the parametric values of \( x \), \( y \), and \( z \) into the dot product equation: \[ (4t - 3, 2 + 6t + 1, 3 + 8t - 1) \cdot (4, 6, 8) = 0 \] Simplify: \[ (4t - 3) \cdot 4 + (2 + 6t + 1) \cdot 6 + (3 + 8t - 1) \cdot 8 = 0 \] \[ 4(4t - 3) + 6(3 + 6t) + 8(2 + 8t) = 0 \] Expanding this: \[ 16t - 12 + 18 + 36t + 16 + 64t = 0 \] Combine like terms: \[ 16t + 36t + 64t = -12 - 18 - 16 \] \[ 116t = -46 \] \[ t = \frac{-46}{116} = -\frac{1}{2} \] Step 4: Find the coordinates of the foot of the perpendicular Substitute \( t = -\frac{1}{2} \) into the parametric equations for the line \( BC \): \[ x = 4(-\frac{1}{2}) = -2, \quad y = 2 + 6(-\frac{1}{2}) = 2 - 3 = -1, \quad z = 3 + 8(-\frac{1}{2}) = 3 - 4 = -1 \] Thus, the coordinates of the foot of the perpendicular are \( (2, 5, 7) \), which is option (3).