Question

Let N be the foot of perpendicular from the point P (1, –2, 3) on the line passing through the points (4, 5, 8) and (1, –7, 5). Then the distance of N from the plane 2x – 2y + z + 5 = 0 is

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

The Correct Option is B

1. Find the equation of the line passing through \( A(4, 5, 8) \) and \( B(1, -7, 5) \): The direction vector \( \vec{AB} \) is: \[ \vec{AB} = B - A = (1 - 4, -7 - 5, 5 - 8) = (-3, -12, -3) \] The parametric equations of the line are: \[ x = 4 - 3t, \quad y = 5 - 12t, \quad z = 8 - 3t \] 2. Find the foot of the perpendicular \( N \) from \( P(1, -2, 3) \) to the line: The vector from \( A \) to \( P \) is: \[ \vec{AP} = P - A = (1 - 4, -2 - 5, 3 - 8) = (-3, -7, -5) \] The projection of \( \vec{AP} \) onto \( \vec{AB} \) is: \[ t = \frac{\vec{AP} \cdot \vec{AB}}{\vec{AB} \cdot \vec{AB}} = \frac{(-3)(-3) + (-7)(-12) + (-5)(-3)}{(-3)^2 + (-12)^2 + (-3)^2} = \frac{9 + 84 + 15}{9 + 144 + 9} = \frac{108}{162} = \frac{2}{3} \] Substituting \( t = \frac{2}{3} \) into the parametric equations: \[ x = 4 - 3 \left( \frac{2}{3} \right) = 2, \quad y = 5 - 12 \left( \frac{2}{3} \right) = -3, \quad z = 8 - 3 \left( \frac{2}{3} \right) = 6 \] So, \( N(2, -3, 6) \). 3. Find the distance of \( N(2, -3, 6) \) from the plane \( 2x - 2y + z + 5 = 0 \): The distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is: \[ D = \frac{|A x_0 + B y_0 + C z_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \] Substituting the values: \[ D = \frac{|2(2) - 2(-3) + 1(6) + 5|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|4 + 6 + 6 + 5|}{\sqrt{4 + 4 + 1}} = \frac{21}{3} = 7 \] Answer: The distance of \( N \) from the plane is \(\boxed{7}\).