Let (α, β, γ) be the image of the point (8, 5, 7) in the line (x-1)/2 = (y+1)/3 = (z-2)/5. Then α +

Q: Why do we need to find the foot of the perpendicular $M$ first? Can't we just use the distance formula?

The distance formula gives you the magnitude, but it doesn't give you the direction or the specific coordinates of the image. In 3D geometry, the image is defined by the fact that the line segment connecting the object and the image is perpendicular to the mirror line and is bisected by it. Finding the foot of the perpendicular $M$ is the most robust way to locate the exact point of intersection, which then acts as the midpoint to find the image $A'$.

Q: Why is the dot product of $\vec{AM}$ and the direction vector $\vec{d}$ equal to zero?

This is a fundamental property of perpendicular vectors. If two vectors are perpendicular, their dot product is zero. Since $M$ is the foot of the perpendicular from $A$ to the line, the vector $\vec{AM}$ is perpendicular to the line itself. Therefore, $\vec{AM}$ must be perpendicular to the line's direction vector $\vec{d}$. This condition allows us to solve for the unknown parameter $\lambda$.

Q: How do I know if my final image coordinates are correct?

A great way to verify is to check two things: First, ensure the midpoint $M$ of your calculated $A$ and $A'$ actually lies on the given line (substitute the coordinates of $M$ into the line equation). Second, ensure the vector $\vec{AA'}$ is perpendicular to the line's direction vector $\vec{d}$. If both conditions hold, your answer is definitely correct!

Question

Let $(\alpha, \beta, \gamma)$ be the image of the point $(8, 5, 7)$ in the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$. Then $\alpha + \beta + \gamma$ is equal to :

Options:

(A) 16

(B) 20

(C) 14

(D) 18

Sigma Percentile · Accepted Answer

### The Mirror in 3D Space: A Geometric Journey

Imagine you are standing in a room, and before you lies a mirror. But this is not a flat, two-dimensional mirror; this is a mirror that exists as a line in three-dimensional space.

You have a point $A(8, 5, 7)$ floating in this space, and your goal is to find its reflection, $A'(\alpha, \beta, \gamma)$, on the other side of this line. This is the essence of 3D geometry—a beautiful, logical dance of vectors and parameters.

### Phase 1: The Parametric Leap

To find the reflection, we first need to find the point where the perpendicular from $A$ hits the line. Let's call this point $M$, the foot of the perpendicular.

The line is given by the equation:

$$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$$

In 3D geometry, we love parameters. By setting this entire expression equal to a variable $\lambda$, we can describe any point on this line as a function of $\lambda$.

Thus, $M$ becomes $(2\lambda+1, 3\lambda-1, 5\lambda+2)$. This is our anchor, allowing us to represent an infinite number of points on the line with just one variable.

### Phase 2: The Perpendicularity Dance

Now, we need to find the specific $\lambda$ that places $M$ exactly where the perpendicular from $A$ lands. We construct the vector $\vec{AM}$ by subtracting the coordinates of $A$ from $M$:

$$\vec{AM} = (2\lambda+1-8, 3\lambda-1-5, 5\lambda+2-7) = (2\lambda-7, 3\lambda-6, 5\lambda-5)$$

Here is the magic: the vector $\vec{AM}$ is perpendicular to the line. The direction vector of our line, $\vec{d}$, is simply the denominators of the line equation: $(2, 3, 5)$.

Because $\vec{AM}$ is perpendicular to the line, their dot product must be zero:

$$\vec{AM} \cdot \vec{d} = 0$$

$$(2\lambda-7)(2) + (3\lambda-6)(3) + (5\lambda-5)(5) = 0$$

Solving this, we get:

$$4\lambda - 14 + 9\lambda - 18 + 25\lambda - 25 = 0$$

$$38\lambda - 57 = 0$$

$$\lambda = \frac{57}{38} = \frac{3}{2}$$

### Phase 3: The Midpoint Bridge

With $\lambda = \frac{3}{2}$, we can find the exact coordinates of $M$:

$$x_M = 2\left(\frac{3}{2}\right) + 1 = 4$$

$$y_M = 3\left(\frac{3}{2}\right) - 1 = \frac{7}{2}$$

$$z_M = 5\left(\frac{3}{2}\right) + 2 = \frac{19}{2}$$

So, $M = (4, \frac{7}{2}, \frac{19}{2})$. Now, the final step: the mirror line bisects the segment $AA'$. This means $M$ is the midpoint of $A$ and $A'$.

Using the midpoint formula, $\frac{A + A'}{2} = M$, we solve for $A'(\alpha, \beta, \gamma)$:

For the x-coordinate: $\frac{\alpha + 8}{2} = 4 \Rightarrow \alpha = 0$.

For the y-coordinate: $\frac{\beta + 5}{2} = \frac{7}{2} \Rightarrow \beta = 2$.

For the z-coordinate: $\frac{\gamma + 7}{2} = \frac{19}{2} \Rightarrow \gamma = 12$.

### Conclusion: The Final Sum

We have found the image $A'(0, 2, 12)$. The question asks for the sum $\alpha + \beta + \gamma$.

Adding them up, $0 + 2 + 12 = 14$.

The final result is **14**. It is a simple, elegant conclusion to a journey through 3D space. Remember, geometry is not just about formulas; it is about visualizing the relationship between points, lines, and the space they inhabit.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $(α, β, γ)$ be the image of the point $(8, 5, 7)$ in the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5}$ . Then $α + β + γ$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Setup

Defining a General Point $M$

Constructing Vector $A M$

The Perpendicularity Condition

Setting up the Dot Product

Solving for $λ$

Finding the Coordinates of $M$

The Midpoint Property

Calculating Image Coordinates

Final Summation

The Sigma Insight: Equation of a Line in Space

The Mirror in 3D Space

Phase 1

Phase 2

Phase 3

Conclusion

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