Question

The image of the line $\frac{x-2}{3}=\frac{y+1}{4} = \frac{z-2}{12}$ in the plane $2x - y + z + 3 = 0$ is the line

• A. $\frac{x+3}{3}=\frac{y-5}{1} = \frac{z-2}{-5}$
• B. $\frac{x+3}{-3}=\frac{y-5}{-1} = \frac{z-2}{5}$
• C. $\frac{x-3}{3}=\frac{y+5}{1} = \frac{z-2}{-5}$
• D. $\frac{x-3}{-3}=\frac{y+5}{-1} = \frac{z-2}{5}$

Answer 1

The Correct Option is A

To solve this problem, we need to find the image of the given line in the plane. The line is given in symmetric form:

$\frac{x-2}{3}=\frac{y+1}{4} = \frac{z-2}{12}$

and the plane is:

$2x - y + z + 3 = 0$

Step-by-Step Solution

1. The direction ratios (d.r.) of the line are $(3, 4, 12)$.
2. The normal vector to the plane $2x - y + z + 3 = 0$ is $(2, -1, 1)$.
3. To find a point on the line, we can set $t = 0$. This gives the point $(2, -1, 2)$.
4. The shortest distance (d) from this point to the plane can be calculated using the formula for the distance from a point to a plane: $$ d = \frac{|2(2) - 1(−1) + 1(2) + 3|}{\sqrt{2^2 + (-1)^2 + 1^2}} = \frac{|4 + 1 + 2 + 3|}{\sqrt{6}} = \frac{10}{\sqrt{6}} $$.
5. The image of the point in the plane can be found using the formula: $ P_{\text{image}} = P + 2\left(\frac{\text{point-plane distance} \times \text{normal}}{|\text{normal}|^2}\right) $.
6. The coordinates of the image are then substituted into the line's equation in symmetric form to get the image line.

Conclusion

The image of the line in the given plane is $\frac{x+3}{3}=\frac{y-5}{1} = \frac{z-2}{-5}$.