Question

If the image of the point \( (4, 4, 3) \) in the line \( \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3} \) is \( (a, \beta, \gamma) \), then \( a + \beta + \gamma \) is equal to:

• A. 12
• B. 7
• C. 8
• D. 9

Hint:

Remember the formula for the reflection of a point across a line in 3D: it involves the directional cosines of the line.

Answer 1

The Correct Option is D

To find the image of the point \((4, 4, 3)\) in the given line \(\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}\), we follow these steps:

Step 1: Parametric form of the line
The line's equation can be rewritten in parametric form as:

• \(x = 1 + 2t\)
• \(y = 2 + t\)
• \(z = 1 + 3t\)

Step 2: Finding the projection of the point onto the line
First, find the direction vector of the line \(\vec{d} = \langle 2, 1, 3 \rangle\) and the vector \(\vec{OP} = \langle 4 - 1, 4 - 2, 3 - 1 \rangle = \langle 3, 2, 2 \rangle\).

The projection of \(\vec{OP}\) onto \(\vec{d}\) is:

\(\text{Projection} = \frac{\vec{OP} \cdot \vec{d}}{\vec{d} \cdot \vec{d}}\vec{d}\), where \(\vec{OP} \cdot \vec{d} = 3 \cdot 2 + 2 \cdot 1 + 2 \cdot 3 = 14\) and \(\vec{d} \cdot \vec{d} = 2^2 + 1^2 + 3^2 = 14\).

The projection is:

\(\text{Projection} = \frac{14}{14}\langle 2, 1, 3 \rangle = \langle 2, 1, 3 \rangle\).

Step 3: Finding the point on the line closest to the point \((4, 4, 3)\)
The closest point \((a, \beta, \gamma)\) on the line is: \((1, 2, 1) + \langle 2, 1, 3 \rangle = \langle 3, 3, 4 \rangle\).

Step 4: Calculating the image of the point
The image point \((x', y', z')\) is equidistant from the line as point \((4, 4, 3)\), reflected over the line. Using the midpoint formula:

• \(x' = 2 \times 3 - 4 = 2\)
• \(y' = 2 \times 3 - 4 = 2\)
• \(z' = 2 \times 4 - 3 = 5\)

Thus, the image is \((2, 2, 5)\).

Final Calculation:
The sum \(a + \beta + \gamma = 2 + 2 + 5 = 9\).