Finding the Image of a Point in a Plane

The problem asks for the image of a point $P(-3, 8, 4)$ in the plane given by the equation $6x - 3y - 2z + 1 = 0$. Finding the image of a point in a plane is equivalent to finding its reflection across the plane.

Let the image of the point $P(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ be $P'(x', y', z')$. The line segment $PP'$ is perpendicular to the plane, and the midpoint of $PP'$ lies on the plane. The direction ratios of the line $PP'$ are the same as the direction ratios of the normal vector to the plane, which are $(a, b, c)$.

For the given plane $6x - 3y - 2z + 1 = 0$, the normal vector has direction ratios $(6, -3, -2)$. The point is $P(-3, 8, 4)$.

The equation of the line passing through $P(-3, 8, 4)$ and perpendicular to the plane is given by:

\begin{equation*} \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda \end{equation*}

Substituting the coordinates of point $P$ and the direction ratios of the normal vector, we get:

\begin{equation*} \frac{x - (-3)}{6} = \frac{y - 8}{-3} = \frac{z - 4}{-2} = \lambda \end{equation*}

Any point $Q$ on this line can be represented by its coordinates in terms of $\lambda$: $x = -3 + 6\lambda$ $y = 8 - 3\lambda$ $z = 4 - 2\lambda$

The foot of the perpendicular from point $P$ to the plane is a point $Q$ on this line that also lies on the plane. Substituting the coordinates of $Q$ into the plane equation $6x - 3y - 2z + 1 = 0$:

\begin{align*} 6(-3 + 6\lambda) - 3(8 - 3\lambda) - 2(4 - 2\lambda) + 1 &= 0 \\ -18 + 36\lambda - 24 + 9\lambda - 8 + 4\lambda + 1 &= 0 \\ (36 + 9 + 4)\lambda + (-18 - 24 - 8 + 1) &= 0 \\ 49\lambda - 49 &= 0 \\ 49\lambda &= 49 \\ \lambda &= 1 \end{align*}

So, the foot of the perpendicular $Q$ corresponds to $\lambda = 1$. The coordinates of $Q$ are:

\begin{align*} x &= -3 + 6(1) = -3 + 6 = 3 \\ y &= 8 - 3(1) = 8 - 3 = 5 \\ z &= 4 - 2(1) = 4 - 2 = 2 \end{align*}

The foot of the perpendicular is $Q(3, 5, 2)$.

The foot of the perpendicular $Q$ is the midpoint of the segment connecting the original point $P(-3, 8, 4)$ and its image $P'(x', y', z')$. Using the midpoint formula:

\begin{equation*} Q = \left(\frac{x_1 + x'}{2}, \frac{y_1 + y'}{2}, \frac{z_1 + z'}{2}\right) \end{equation*}

So, we have:

• $\frac{-3 + x'}{2} = 3 \implies -3 + x' = 6 \implies x' = 9$
• $\frac{8 + y'}{2} = 5 \implies 8 + y' = 10 \implies y' = 2$
• $\frac{4 + z'}{2} = 2 \implies 4 + z' = 4 \implies z' = 0$

Thus, the coordinates of the image point $P'$ are $(9, 2, 0)$.

Let's check this with the given options.

• Option 1: (0, 2, 9)
• Option 2: (9, 2, 0)
• Option 3: (1, 2, –2)
• Option 4: (–2, 1, 3)

Our calculated image point $(9, 2, 0)$ matches Option 2.

Image of Point in Plane Calculation Steps

To summarize the steps for finding the image of a point $P(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$:

1. Write the equation of the line passing through $P$ and perpendicular to the plane: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$.
2. Find the coordinates of any point $Q$ on this line in terms of $\lambda$: $x = x_1 + a\lambda$, $y = y_1 + b\lambda$, $z = z_1 + c\lambda$.
3. Substitute the coordinates of $Q$ into the plane equation $ax + by + cz + d = 0$ to find the value of $\lambda$ for which $Q$ is the foot of the perpendicular.
4. Calculate the coordinates of the foot of the perpendicular $Q$ using the value of $\lambda$.
5. The foot of the perpendicular $Q$ is the midpoint of $PP'$. If $P'(x', y', z')$ is the image, then $Q = \left(\frac{x_1 + x'}{2}, \frac{y_1 + y'}{2}, \frac{z_1 + z'}{2}\right)$. Use this to find $x', y', z'$. Alternatively, $Q = \frac{P + P'}{2} \implies P' = 2Q - P$.

Using the alternative method $P' = 2Q - P$ with $P(-3, 8, 4)$ and $Q(3, 5, 2)$: $x' = 2(3) - (-3) = 6 + 3 = 9$ $y' = 2(5) - 8 = 10 - 8 = 2$ $z' = 2(2) - 4 = 4 - 4 = 0$ This gives the image point $(9, 2, 0)$, confirming the previous result.