Question

Let \( (\alpha, \beta, \gamma) \) be the mirror image of the point \( (2, 3, 5) \) in the line \(\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}\). Then \( 2\alpha + 3\beta + 4\gamma \) is equal to

• A. 32
• B. 33
• C. 31
• D. 34

Answer 1

The Correct Option is B

To find the mirror image of the point \((2, 3, 5)\) in the given line, we will first determine a point on this line, find the perpendicular from the given point to the line, and use it to find the reflection.

The given line in symmetric form is:

\(\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = t\)

We can write parametric equations for the line:

• \(x = 1 + 2t\)
• \(y = 2 + 3t\)
• \(z = 3 + 4t\)

Let's designate the point \((2 + 2\lambda, 3 + 3\lambda, 5 + 4\lambda)\) as the foot of the perpendicular from the point \((2,3,5)\) on the line. To find this point, we need the direction vector of the line, which is \((2, 3, 4)\).

For the foot of the perpendicular, the vector \((2\lambda, 3\lambda, 4\lambda)\) should be perpendicular to the direction vector \((2, 3, 4)\). Therefore, the dot product is zero:

\([(2 + 2\lambda) - 2, (3 + 3\lambda) - 3, (5 + 4\lambda) - 5] \cdot (2, 3, 4) = 0\)

This simplifies to:

\(2\lambda \cdot 2 + 3\lambda \cdot 3 + 4\lambda \cdot 4 = 0\)

\(4\lambda + 9\lambda + 16\lambda = 0\)

\(29\lambda = 0\)

Therefore, \(\lambda = 0\). This results in the point on the line being \((2, 3, 5)\).

For mirroring a point over a line, the midpoint of the line segment (original point and its image) lies on the line. As \(\lambda = 0\), the point and its image coincide at the point \((2, 3, 5)\).

Finally, we need to calculate \(2\alpha + 3\beta + 4\gamma\) for the point \((\alpha, \beta, \gamma)\).

With \((\alpha, \beta, \gamma) = (2, 3, 5)\), we have:

\(2\alpha + 3\beta + 4\gamma = 2(2) + 3(3) + 4(5)\)

\(= 4 + 9 + 20 = 33\)

Thus, the value of \(2\alpha + 3\beta + 4\gamma\) is 33.

Answer 2

Let \( P(2, 3, 5) \) be the point and \( R(\alpha, \beta, \gamma) \) its mirror image in the line

\[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}. \]

Since \( R \) is the mirror image of \( P \), the line segment \( PR \) is perpendicular to the direction ratios of the line \( (2, 3, 4) \).

Therefore, \( \overrightarrow{PR} \perp (2, 3, 4) \).

So, \( \overrightarrow{PR} \cdot (2, 3, 4) = 0 \).

Let \( \overrightarrow{PR} = (\alpha - 2, \beta - 3, \gamma - 5) \).

Now,

\[ (\alpha - 2, \beta - 3, \gamma - 5) \cdot (2, 3, 4) = 0 \]

which gives:

\[ 2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0 \] \[ \implies 2\alpha + 3\beta + 4\gamma = 4 + 9 + 20 = 33 \]

Thus, the answer is:

33