Question

The square of the distance of the image of the point \( (6, 1, 5) \) in the line \[ \frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}, \] from the origin is _____ .

Answer 1

Correct Answer: 62

To find the square of the distance of the image of a point \( (6, 1, 5) \) on the given line \(\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}\) from the origin, follow these steps:

1. Identify the direction ratios of the line as \( \langle 3, 2, 4 \rangle \).
2. Find the equation of the line in vector form: \( \mathbf{r} = \langle 1, 0, 2 \rangle + \lambda \langle 3, 2, 4 \rangle \).
3. Calculate the point of contact of the perpendicular from \( (6, 1, 5) \) on the line. Let this point be denoted by \( \mathbf{Q} = \langle 1 + 3\lambda, 2\lambda, 2 + 4\lambda \rangle \).
4. The vector \( \mathbf{PQ} \), where \( \mathbf{P} = \langle 6, 1, 5 \rangle \), is given by \( \langle 1 + 3\lambda - 6, 2\lambda - 1, 2 + 4\lambda - 5 \rangle = \langle 3\lambda - 5, 2\lambda - 1, 4\lambda - 3 \rangle \).
5. Use the condition \( \mathbf{PQ} \cdot \langle 3, 2, 4 \rangle = 0 \) to find \( \lambda \):
   • \((3\lambda - 5) \cdot 3 + (2\lambda - 1) \cdot 2 + (4\lambda - 3) \cdot 4 = 0\)
   • Simplifying, \(9\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0\); hence, \(29\lambda = 29\).
   • Solving gives \(\lambda = 1\).
6. Substitute \(\lambda = 1\) into \( \mathbf{Q} \):
   • \(\mathbf{Q} = \langle 4, 2, 6 \rangle\).
7. The image point \( \mathbf{Q'} \) is given by reflecting \(\mathbf{P}\) around \(\mathbf{Q}\):
   • \( \mathbf{Q'} = 2\mathbf{Q} - \mathbf{P} = 2 \langle 4, 2, 6 \rangle - \langle 6, 1, 5 \rangle = \langle 8, 4, 12 \rangle - \langle 6, 1, 5 \rangle = \langle 2, 3, 7 \rangle\).
8. Calculate the square of the distance from the origin to \(\mathbf{Q'}\):
   • \([(2)^2 + (3)^2 + (7)^2] = 4 + 9 + 49 = 62\).
9. Verify the solution fits the range [62, 62]. Indeed, \(62\) is within this range.

Thus, the square of the distance is \(62\).

Answer 2

Let \( M(3\lambda + 1, 2\lambda, 4\lambda + 2) \)

\(\overrightarrow{AM} \cdot \mathbf{b} = 0\)
\(\implies 9\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0\)
\(\implies 29\lambda = 29  \implies \lambda = 1\)

So,
\(M(4, 2, 6), \quad I = (2, 3, 7)\)

Hence Distance =
\(\sqrt{4 + 9 + 49} = \sqrt{62}\)

Therefore, the square of the distance of the image of the point is 62.