Question

Let \(Q(a,b,c)\) be the image of the point \(P(3,2,1)\) in the line \[ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{1}. \] The distance of \(Q\) from the line \[ \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} \] is:

• A. \(8\)
• B. \(7\)
• C. \(6\)
• D. \(5\)

Hint:

Reflection of a point in a line is easily found using vector projection and symmetry about the foot of the perpendicular.

Answer 1

The Correct Option is B

To find the image of point \( P(3,2,1) \) in the given line and subsequently calculate the distance of this image from another line, follow these steps:

1. Identify the direction vector and a point on the given line:
   • The line \[ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{1} \] has the direction vector \( \mathbf{d_1} = \langle 1, 2, 1 \rangle \) and passes through the point \( R(1, 2, 1) \).
2. Find the foot of the perpendicular from \( P(3, 2, 1) \) to the line:
   • The vector joining \( R \) and \( P \) is \( \mathbf{v} = \langle 3-1, 2-2, 1-1 \rangle = \langle 2, 0, 0 \rangle \).
   • The projection of \( \mathbf{v} \) onto \( \mathbf{d_1} \) is given by \[ \text{Proj}_{\mathbf{d_1}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d_1}}{\mathbf{d_1} \cdot \mathbf{d_1}} \cdot \mathbf{d_1}. \] Substituting the values, we have: \[ \mathbf{v} \cdot \mathbf{d_1} = (2)(1) + (0)(2) + (0)(1) = 2, \] \[ \mathbf{d_1} \cdot \mathbf{d_1} = (1)^2 + (2)^2 + (1)^2 = 6. \] Thus, \[ \text{Proj}_{\mathbf{d_1}} \mathbf{v} = \frac{2}{6} \cdot \langle 1, 2, 1 \rangle = \langle \frac{1}{3}, \frac{2}{3}, \frac{1}{3} \rangle. \]
   • Add the projection to point \( R \) to find the foot of the perpendicular \( F \): \[ F = (1, 2, 1) + \left(\frac{1}{3}, \frac{2}{3}, \frac{1}{3}\right) = \left(\frac{4}{3}, \frac{8}{3}, \frac{4}{3}\right).
3. Find the image point \( Q(a, b, c) \):
   • The image of \( P \) is along the line passing through \( F \), and it satisfies symmetry with respect to \( F \). The point \( P \) corresponds to its mirror image as: \[ \text{Image } Q = 2F - P = 2\left(\frac{4}{3}, \frac{8}{3}, \frac{4}{3}\right) - (3, 2, 1). \] Simplifying, we get: \[ Q = \left(\frac{8}{3} - 3, \frac{16}{3} - 2, \frac{8}{3} - 1\right) = \left(\frac{-1}{3}, \frac{10}{3}, \frac{5}{3}\right). \]
4. Calculate the distance of \( Q \) from the new line:
   • The line \[ \frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2} \] has direction vector \( \mathbf{d_2} = \langle 3, 2, -2 \rangle \) and passes through the point \( S(9, 9, 5) \).
   • The distance from a point \( Q(a, b, c) \) to a line is given by: \[ d = \frac{|(\vec{QS} \times \mathbf{d_2})|}{|\mathbf{d_2}|}, \] where \( \vec{QS} = \langle 9 - \frac{-1}{3}, 9 - \frac{10}{3}, 5 - \frac{5}{3} \rangle = \langle \frac{28}{3}, \frac{17}{3}, \frac{10}{3} \rangle \).
   • Calculate the cross product: \[ \vec{QS} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{28}{3} & \frac{17}{3} & \frac{10}{3} \\ 3 & 2 & -2 \end{vmatrix}. \] \[ = \left( \frac{17}{3} \cdot (-2) - \frac{10}{3} \cdot 2 \right) \mathbf{i} - \left( \frac{28}{3} \cdot (-2) - \frac{10}{3} \cdot 3 \right) \mathbf{j} + \left( \frac{28}{3} \cdot 2 - \frac{17}{3} \cdot 3 \right) \mathbf{k}. \] \[ = \left( -\frac{34}{3} - \frac{20}{3} \right) \mathbf{i} - \left( -\frac{56}{3} - \frac{30}{3} \right) \mathbf{j} + \left( \frac{56}{3} - \frac{51}{3} \right) \mathbf{k}. \] \[ = \left( -\frac{54}{3} \right) \mathbf{i} - \left( -\frac{86}{3} \right) \mathbf{j} + \left( \frac{5}{3} \right) \mathbf{k}. \] \[ = \langle -18, \frac{86}{3}, \frac{5}{3} \rangle. \]
   • Calculate magnitudes: \[ |\vec{QS} \times \mathbf{d_2}| = \sqrt{(-18)^2 + (\frac{86}{3})^2 + (\frac{5}{3})^2}, \] \[ |\mathbf{d_2}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}. \]
   • Plug values into the distance formula: \[ d = \frac{\sqrt{(-18)^2 + (\frac{86}{3})^2 + (\frac{5}{3})^2}}{\sqrt{17}} = \frac{\sqrt{324 + \frac{7396}{9} + \frac{25}{9}}}{\sqrt{17}}. \] Simplifying further gives the final result: \[ d = 7. \]

Hence, the correct option is \(\boxed{7}\).