Question

The image of the point \(P(1,3,4)\) in the plane \(2x - y + z + 3 = 0\), is

• A. \((3,5,-2)\)
• B. \((-3,5,2)\)
• C. \((3,-5,2)\)
• D. \((3,5,2)\)

Hint:

The formula for image is derived from the foot of perpendicular and midpoint.

Answer 1

The Correct Option is B

 To find the image of the point \( P(1,3,4) \) in the plane \( 2x - y + z + 3 = 0 \), we need to follow these steps:

1. First, find the normal vector of the plane. The equation of the plane is given by \( 2x - y + z + 3 = 0 \). The normal vector \(n\\) to the plane can be extracted from the coefficients of \(x\\), \(y\\), and \(z\\), which is \(\mathbf{n} = (2, -1, 1)\).
2. We use the formula to find the perpendicular distance from a point \( (x_1, y_1, z_1) \) to the plane \(ax+by+cz+d=0\):

D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

1. Substituting \( (x_1, y_1, z_1) = (1,3,4) \) and \( a = 2, b = -1, c = 1, d = 3 \), we calculate:

D = \frac{|2(1) - 1(3) + 1(4) + 3|}{\sqrt{2^2 + (-1)^2 + 1^2}} = \frac{|2 - 3 + 4 + 3|}{\sqrt{4 + 1 + 1}} = \frac{|6|}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}

1. Find the coordinates of the foot of the perpendicular from point \( P(1,3,4) \) on the plane. The coordinates of the foot of the perpendicular \( (x_2, y_2, z_2) \) are computed using:

(x_2, y_2, z_2) = (x_1, y_1, z_1) - D \cdot \mathbf{n}

1. where \( D \) is the perpendicular distance calculated previously, and \( \mathbf{n} \) is a unit normal vector to the plane:

\mathbf{n}_{unit} = \left( \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)

1. Compute the coordinates:

(x_2, y_2, z_2) = \left( 1 - \sqrt{6} \cdot \frac{2}{\sqrt{6}}, 3 - \sqrt{6} \cdot \frac{-1}{\sqrt{6}}, 4 - \sqrt{6} \cdot \frac{1}{\sqrt{6}} \right) = (1 - 2, 3 + 1, 4 - 1) = (-1,4,3)

1. The image of the point \( P \) with respect to the plane is the symmetric point about the foot of the perpendicular. Use the midpoint formula between \( P \) and its image \( M \) such that:

(x_2, y_2, z_2) = \frac{(x + 1)}{2}, \frac{(y + 3)}{2}, \frac{(z + 4)}{2}

1. Solving the above:

\begin{align*} x + 1 & = -2 \times (-1), \\ y + 3 & = 2 \times 4, \\ z + 4 & = 2 \times 3, \\ \end{align*}

(x, y, z) = \left ( -3, 5, 2 \right )

Thus, the image of point \( P(1,3,4) \) in the given plane is \((-3,5,2)\), which matches the correct option.