Question

If distance of point (a, 2, 5) from image of point (1, 2, 7) in the line \( \frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2} \) is 4, then sum of all possible values of a is:

• A. 4
• B. 5
• C. 6
• D. 8

Hint:

To find the foot of the perpendicular quickly, use the projection formula for the vector from a point on the line to the external point onto the line's direction vector.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We first need to find the image \(Q(\alpha, \beta, \gamma)\) of point \(P(1, 2, 7)\) in the given line \(L\).
Let \(M\) be the foot of the perpendicular from \(P\) to the line. Then \(M\) is the midpoint of \(PQ\).
Step 2: Key Formula or Approach:
The line is \(\frac{x}{1} = \frac{y - 1}{1} = \frac{z - 2}{2} = \lambda\).
Any point \(M\) on the line is \((\lambda, \lambda + 1, 2\lambda + 2)\).
Direction ratios of \(PM = (\lambda - 1, \lambda + 1 - 2, 2\lambda + 2 - 7) = (\lambda - 1, \lambda - 1, 2\lambda - 5)\).
Since \(PM \perp L\):
\[ 1(\lambda - 1) + 1(\lambda - 1) + 2(2\lambda - 5) = 0 \]
\[ \lambda - 1 + \lambda - 1 + 4\lambda - 10 = 0 \Rightarrow 6\lambda = 12 \Rightarrow \lambda = 2 \]
So, \(M = (2, 3, 6)\).
Step 3: Detailed Explanation:
\(M\) is midpoint of \(P(1, 2, 7)\) and \(Q(\alpha, \beta, \gamma)\):
\[ \frac{\alpha + 1}{2} = 2 \Rightarrow \alpha = 3 \]
\[ \frac{\beta + 2}{2} = 3 \Rightarrow \beta = 4 \]
\[ \frac{\gamma + 7}{2} = 6 \Rightarrow \gamma = 5 \]
The image point is \(Q(3, 4, 5)\).
Distance from \(Q(3, 4, 5)\) to \((a, 2, 5)\) is given as 4:
\[ \sqrt{(a - 3)^2 + (2 - 4)^2 + (5 - 5)^2} = 4 \]
\[ (a - 3)^2 + 4 + 0 = 16 \]
\[ (a - 3)^2 = 12 \Rightarrow a^2 - 6a + 9 = 12 \Rightarrow a^2 - 6a - 3 = 0 \]
The sum of values of \(a\) is the sum of roots of this quadratic:
\[ \text{Sum} = -\frac{-6}{1} = 6 \]
Step 4: Final Answer:
The sum of all possible values of \(a\) is 6.