Question

Let P be the plane passing through the line \(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}\) and the point (2, 4, –3). If the image of the point (-1, 3, 4) in the plane P is (α, β, γ) then α + β + γ is equal to

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The Correct Option is B

To find the image of a point in a plane, we first need to determine the equation of the plane P that passes through the given line and the point (2, 4, -3). Then, we'll use the concept of image reflection in terms of vectors.

1. **Equation of the Line**: The given line is \frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}. This can be rewritten in parametric form as:

• x = 1 + t
• y = 2 - 3t
• z = -5 + 7t

The direction ratios of the line are (1, -3, 7).

2. **Normal Vector to the Plane**: A plane passing through a line will have the line's direction vector as one of its vectors. We need a normal vector to represent the plane's orientation, which should be perpendicular to this direction vector. Let the normal vector of the plane be \vec{n} = (1, -3, 7).

3. **Equation of the Plane**: The plane passes through the point (2, 4, -3). Thus, the vector equation of the plane becomes:

1(x - 2) - 3(y - 4) + 7(z + 3) = 0

Simplifying, we get:

x - 3y + 7z = -18

4. **Finding the Image of the Point**: The point (-1, 3, 4) has its image in the plane (x - 3y + 7z + 18 = 0). The formula for the reflection of a point about a plane is given by the formula:

• If (x_1, y_1, z_1) is a point and Ax + By + Cz + D = 0 is the plane, then the reflection (x', y', z') is given by:

x' = x_1 - \frac{2A(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2} \\ y' = y_1 - \frac{2B(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2} \\ z' = z_1 - \frac{2C(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2}

Substituting A = 1, B = -3, C = 7, D = 18 and the point (x_1, y_1, z_1) = (-1, 3, 4), we compute:

Ax_1 + By_1 + Cz_1 + D = 1(-1) - 3(3) + 7(4) + 18 = 29

Normalizing factor:

A^2 + B^2 + C^2 = 1^2 + (-3)^2 + 7^2 = 59

Thus the coordinates of the image are:

• x' = -1 - \frac{2 \times 1 \times 29}{59} = -1 - \frac{58}{59} = -\frac{117}{59}
• y' = 3 - \frac{2 \times (-3) \times 29}{59} = 3 + \frac{174}{59} = \frac{351}{59}
• z' = 4 - \frac{2 \times 7 \times 29}{59} = 4 - \frac{406}{59} = \frac{2}{59}

Finally, sum up the coordinates:

\alpha + \beta + \gamma = -\frac{117}{59} + \frac{351}{59} + \frac{2}{59} = 10

Therefore, the sum \alpha + \beta + \gamma is 10.