Question

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2 x - y + z+ 3 = 0$ 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 find the image of the given line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x - y + z + 3 = 0$, we proceed with the following steps:

1. Identify a point on the given line. By setting the parameter $t = 0$, we get the point $(1, 3, 4)$ on the line.
2. Determine the direction ratios of the line, which are $(3, 1, -5)$.
3. Find the foot of the perpendicular from the point $(1, 3, 4)$ to the plane $2x - y + z + 3 = 0$. The formula to find the foot of the perpendicular point is given by:
4. $(x_1, y_1, z_1) = \left(x - \frac{(ax + by + cz + d)(a)}{a^2 + b^2 + c^2}, y - \frac{(ax + by + cz + d)(b)}{a^2 + b^2 + c^2}, z - \frac{(ax + by + cz + d)(c)}{a^2 + b^2 + c^2}\right)$
5. Calculate the perpendicular point from the plane for $(1, 3, 4)$:
   • Normal vector of the plane is $(2, -1, 1)$.
   • Substitute in the point-plane distance formula: $\text{Distance} = \frac{2(1) - 1(3) + 1(4) + 3}{2^2 + (-1)^2 + 1^2} = \frac{2 - 3 + 4 + 3}{6} = 1$.
   • So, the foot of the perpendicular line is at point $(0, 3.5, 3.5)$.
6. Calculate the image point using the formula for the reflection of a point:
   • $P(x_1, y_1, z_1) = (1, 3, 4), F(x_2, y_2, z_2) = (0, 3.5, 3.5)$.
   • Use the midpoint formula for reflection across the plane:
   • $(x, y, z) \to (2x_2 - x, 2y_2 - y, 2z_2 - z)$.
   • Substituting, we get the coordinates of the reflected image as $I(-1, 4, 3)$.
7. Find the image of the line using the direction ratios, which remain unchanged as the original line's. The image line equation becomes: $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$.

Therefore, the correct answer is: $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$