Question

Let P₁ be the plane 3x-y-7z = 11 and P₂ be the plane passing through the points (2,-1,0), (2,0,-1), and (5,1,1). If the foot of the perpendicular drawn from the point (7,4,-1) on the line of intersection of the planes P₁ and P₂ is (α, β, γ), then a + ẞ+ y is equal to

Hint:

To find the foot of the perpendicular from a point to a line, parameterize the line and use the condition that the vector connecting the point to the line is perpendicular to the line's direction vector.

Answer 1

Correct Answer: 11

1. Equation of the plane \( P_2 \):
   • \( P_2 \) passes through the points \( (2, -1, 0) \), \( (2, 0, -1) \), and \( (5, 1, 1) \).
   • Compute the direction vectors:
     • \( \vec{v_1} = (2 - 2, 0 - (-1), -1 - 0) = (0, 1, -1) \)
     • \( \vec{v_2} = (5 - 2, 1 - (-1), 1 - 0) = (3, 2, 1) \)
   • The normal vector to \( P_2 \) is given by the cross product \( \vec{v_1} \times \vec{v_2} \):
     • \( \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & -1 \\ 3 & 2 & 1 \end{vmatrix} = \mathbf{i}(1 \cdot 1 - (-1) \cdot 2) - \mathbf{j}(0 \cdot 1 - (-1) \cdot 3) + \mathbf{k}(0 \cdot 2 - 1 \cdot 3) \)
   • \( \vec{n_2} = \mathbf{i}(1 + 2) - \mathbf{j}(0 + 3) + \mathbf{k}(0 - 3) = 3\mathbf{i} - 3\mathbf{j} - 3\mathbf{k} \)
   • The normal vector is \( \vec{n_2} = (3, -3, -3) \), so the equation of \( P_2 \) is:
     • \( 3x - 3y - 3z = d \)
   • Substitute \( (2, -1, 0) \) into \( P_2 \) to find \( d \):
     • \( 3(2) - 3(-1) - 3(0) = 6 + 3 = 9 \)
   • Hence, \( P_2 \) is:
     • \( 3x - 3y - 3z = 9 \quad \Rightarrow \quad x - y - z = 3 \)
2. Line of intersection of \( P_1 \) and \( P_2 \):
   • \( P_1: 3x - y - 7z = 11 \)
   • \( P_2: x - y - z = 3 \)
   • Subtract \( P_2 \) from \( P_1 \):
     • \( (3x - y - 7z) - (x - y - z) = 11 - 3 \)
   • \( 2x - 6z = 8 \quad \Rightarrow \quad x = 3z + 4 \)
   • Parameterize the line of intersection:
     • Let \( z = t \), so:
       • \( x = 3t + 4 \), \( y = x - z - 3 = (3t + 4) - t - 3 = 2t + 1 \), \( z = t \)
     • The parametric equation of the line is:
       • \( (x, y, z) = (3t + 4, 2t + 1, t) \)
3. Foot of the perpendicular from \( (7, 4, -1) \) to the line:
   • Let the foot of the perpendicular be \( (3t + 4, 2t + 1, t) \).
   • The vector connecting \( (7, 4, -1) \) to \( (3t + 4, 2t + 1, t) \) is:
     • \( \vec{v} = (3t + 4 - 7, 2t + 1 - 4, t - (-1)) = (3t - 3, 2t - 3, t + 1) \)
   • The direction vector of the line is \( \vec{d} = (3, 2, 1) \).
   • For \( \vec{v} \) to be perpendicular to \( \vec{d} \), their dot product must be zero:
     • \( \vec{v} \cdot \vec{d} = (3t - 3) \cdot 3 + (2t - 3) \cdot 2 + (t + 1) \cdot 1 = 0 \)
   • \( 9t - 9 + 4t - 6 + t + 1 = 0 \quad \Rightarrow \quad 14t - 14 = 0 \quad \Rightarrow \quad t = 1 \)
   • Substitute \( t = 1 \) into the parametric equation:
     • \( x = 3(1) + 4 = 7 \), \( y = 2(1) + 1 = 3 \), \( z = 1 \)
   • The foot of the perpendicular is \( (7, 3, 1) \).
4. Calculate \( \alpha + \beta + \gamma \):
   • \( \alpha = 7 \), \( \beta = 3 \), \( \gamma = 1 \).
   • \( \alpha + \beta + \gamma = 7 + 3 + 1 = 11 \)