Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segmen

Q: Wait, why is the distance from the $x$-axis $\sqrt{y^2 + z^2}$ and not $\sqrt{x^2 + y^2 + z^2}$?

That is a classic trap! The distance $\sqrt{x^2 + y^2 + z^2}$ is the distance from the origin $(0,0,0)$. The distance from an axis is the length of the perpendicular segment dropped to that axis. For the $x$-axis, the $x$-coordinate of the projection is the same as the point, so the distance is the hypotenuse in the $yz$-plane, which is $\sqrt{y^2 + z^2}$.

Q: Why did we reject the negative root when solving for $z_1$?

The problem explicitly states that point $P$ lies in the first octant. In the first octant, all coordinates $(x, y, z)$ must be positive. Since we found $z_1^2 = 16$, $z_1$ could be $4$ or $-4$. We must choose $z_1 = 4$ to satisfy the first octant condition.

Q: Is there a faster way to find the image of $Q$ without the full formula?

Yes, since the plane $x + y = 3$ is vertical, you can think of it as a line $x + y = 3$ in the $xy$-plane. The image of $(0,0)$ in the line $x + y = 3$ is $(3,3)$. Since the $z$-coordinate is unaffected by this reflection, the image $P$ is simply $(3, 3, z_1)$. The formula is just a rigorous way to confirm this geometric intuition.

Question

Let $P$ be a point in the first octant, whose image $Q$ in the plane $x + y = 3$ (that is, the line segment $PQ$ is perpendicular to the plane $x + y = 3$ and the mid-point of $PQ$ lies in the plane $x + y = 3$) lies on the $z$-axis. Let the distance of $P$ from the $x$-axis be $5$. If $R$ is the image of $P$ in the $xy$-plane, then the length of $PR$ is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Mirror Wall

The plane $x + y = 3$ has no $z$-term, which is a crucial observation. This implies that for any point on the plane, the $z$-coordinate can be arbitrary, meaning the plane is parallel to the $z$-axis.

Consider point $Q$ resting on the $z$-axis. Its coordinates must be of the form $(0, 0, z_1)$.

When we reflect this point across our vertical mirror, the reflection $P$ will maintain the same $z$-coordinate because the mirror is vertical. We are essentially reflecting the point $(0, 0)$ across the line $x + y = 3$ in the $xy$-plane.

### The Image Hunt

To find the coordinates of $P(x_P, y_P, z_P)$, we employ the standard image formula. For a point $(x_1, y_1, z_1)$ and a plane $ax + by + cz + d = 0$, the image $(x, y, z)$ satisfies:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$$

Substituting our values $Q(0, 0, z_1)$ and the plane $x + y - 3 = 0$, we obtain:

$$\frac{x_P - 0}{1} = \frac{y_P - 0}{1} = \frac{z_P - z_1}{0} = \frac{-2(1(0) + 1(0) + 0(z_1) - 3)}{1^2 + 1^2 + 0^2}$$

Simplifying the right-hand side, we have:

$$\frac{-2(-3)}{2} = 3$$

By equating the ratios to $3$, we find $x_P = 3$ and $y_P = 3$. Since the denominator for the $z$-term is zero, we must have $z_P - z_1 = 0$, confirming our earlier intuition that $z_P = z_1$. Thus, the coordinates of the reflected point are $P = (3, 3, z_1)$.

### The Distance Constraint

We are given that the distance of $P$ from the $x$-axis is $5$. The distance of any point $(x, y, z)$ from the $x$-axis is defined by $\sqrt{y^2 + z^2}$.

Plugging in our coordinates for $P$:

$$\sqrt{3^2 + z_1^2} = 5$$

Squaring both sides yields $9 + z_1^2 = 25$, which simplifies to $z_1^2 = 16$. This gives $z_1 = \pm 4$. Assuming $P$ is in the first octant, we select $z_1 = 4$, making our point $P = (3, 3, 4)$.

### The Final Reflection

The final act involves finding the image $R$ of $P$ in the $xy$-plane. The $xy$-plane is defined by the equation $z = 0$.

Reflecting a point across this plane is straightforward: the $x$ and $y$ coordinates remain unchanged, while the $z$-coordinate is negated. Thus, $R = (3, 3, -4)$.

The length of the segment $PR$ is the distance between $(3, 3, 4)$ and $(3, 3, -4)$. Since the $x$ and $y$ coordinates are identical, the distance is simply the absolute difference in the $z$-coordinates:

$$|4 - (-4)| = 8$$

Through systematic application of geometry, we have arrived at the final answer of **8**.

Enter Numerical Value:

Visualized Solution

The Plane $x + y = 3$

Point $Q$ on the $z$ -axis

Image of $Q$ in the Plane

Image of a Point in a Plane

Applying the Formula

Simplifying the Equation

Coordinates of $P$

Distance of $P$ from $x$ -axis

Distance Formula

Finding $z_{1}$

Image in the $x y$ -plane

Coordinates of $R$

Length of $P R$

The Sigma Insight: Equation of a Plane

Analyzing the Mirror Wall

The Image Hunt

The Distance Constraint

The Final Reflection

Similar Questions

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Enter Numerical Value:

Visualized Solution

The Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y=3

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z-axis

Image of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q in the Plane

Image of a Point in a Plane

Applying the Formula

Simplifying the Equation

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Distance of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis