Question

Image of point P (1, 2, a) with respect to line mirror \(\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}\) is point Q (5, b, c), then value of \((a^2 + b^2 + c^2)\) is :

• A. 293
• B. 298
• C. 283
• D. 264

Hint:

Finding the image of a point in a line involves two key geometric conditions: the midpoint of the point and its image lies on the line, and the line connecting the point and its image is perpendicular to the mirror line.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a point P, a line, and the image of P in that line, Q. We need to find the values of the unknown coordinates a, b, and c, and then calculate \(a^2 + b^2 + c^2\). There are two key properties of an image with respect to a line mirror.

Step 2: Applying Geometric Properties:
Property 1: The midpoint of PQ lies on the line.
The midpoint M of the segment PQ is given by: \[ M = \left( \frac{1+5}{2}, \frac{2+b}{2}, \frac{a+c}{2} \right) = \left( 3, \frac{2+b}{2}, \frac{a+c}{2} \right) \] Since M lies on the given line, its coordinates must satisfy the line's equation: \(\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}\). Substituting the coordinates of M: \[ \frac{3-6}{3} = \frac{\frac{2+b}{2}-7}{2} = \frac{\frac{a+c}{2}-7}{-2} \] \[ \frac{-3}{3} = \frac{\frac{2+b-14}{2}}{2} = \frac{\frac{a+c-14}{2}}{-2} \] \[ -1 = \frac{b-12}{4} = \frac{a+c-14}{-4} \] From \(-1 = \frac{b-12}{4}\), we get \(-4 = b-12\), which gives \(\mathbf{b = 8}\).
From \(-1 = \frac{a+c-14}{-4}\), we get \(4 = a+c-14\), which gives \(\mathbf{a+c = 18}\) (Equation I).
Property 2: The line segment PQ is perpendicular to the mirror line.
The direction ratios of the line segment PQ are \((5-1, b-2, c-a) = (4, b-2, c-a)\).
Since we found \(b=8\), the direction ratios of PQ are \((4, 6, c-a)\).
The direction ratios of the mirror line are given by the denominators in its equation: \((3, 2, -2)\).
For two lines to be perpendicular, the dot product of their direction ratios must be zero: \(a_1a_2 + b_1b_2 + c_1c_2 = 0\). \[ (4)(3) + (6)(2) + (c-a)(-2) = 0 \] \[ 12 + 12 - 2(c-a) = 0 \] \[ 24 = 2(c-a) \] \( \mathbf{c-a = 12}\) (Equation II).

Step 3: Solving for a, b, and c:
We have a system of two linear equations for a and c:
\(a + c = 18\)
\(-a + c = 12\) Adding the two equations gives: \(2c = 30 \Rightarrow \mathbf{c = 15}\).
Substituting \(c=15\) into the first equation: \(a + 15 = 18 \Rightarrow \mathbf{a = 3}\).
So we have found \(a=3\), \(b=8\), and \(c=15\).

Step 4: Final Answer:
We need to calculate the value of \(a^2 + b^2 + c^2\). \[ a^2 + b^2 + c^2 = 3^2 + 8^2 + 15^2 \] \[ = 9 + 64 + 225 \] \[ = 73 + 225 = 298 \] The final value is 298.