If the foot of perpendicular drawn from the point (1, 0, 3) on a line passing through (α, 7, 1) is (

Q: Wait, why are we using the dot product of $\vec{PQ}$ and $\vec{AQ}$? Why not $\vec{PQ}$ and the line's direction vector?

That is a brilliant observation! Actually, they are the same thing. Since $A$ and $Q$ both lie on the line $L$, the vector $\vec{AQ}$ is essentially the direction vector of the line. Using $\vec{AQ}$ is just a convenient way to define the line's orientation using the points we already have.

Q: Do I really need to keep the fractions? Can't I just multiply everything by 3 early on?

You absolutely can! Multiplying by 3 to clear denominators is a great way to avoid arithmetic errors. Just remember to be consistent—if you scale a vector, ensure you are scaling the components correctly so the orthogonality condition remains valid.

Q: What if I made a mistake in calculating the direction ratios? How can I check my work?

A quick sanity check is to verify if the vector $\vec{PQ}$ is indeed orthogonal to the line. If you have your final $\alpha$, plug it back into the direction ratios of the line and perform the dot product. If the result is zero, your geometry is perfect!

Q: Is there a way to solve this using the projection formula instead of dot products?

Yes, you could express the line parametrically as $L(t) = A + t\vec{v}$ and find the value of $t$ that corresponds to point $Q$. However, the dot product method is often faster for JEE problems because it bypasses the need to find the full parametric form of the line.

Q: Why does the dot product equal zero? I always forget the physical meaning.

Think of the dot product as a measure of 'alignment'. When two vectors are perpendicular, they have zero alignment with each other—they are essentially 'blind' to each other's direction. Mathematically, $\vec{A} \cdot \vec{B} = |A||B| \cos(\theta)$. If $\theta = 90^\circ$, then $\cos(90^\circ) = 0$, forcing the entire product to zero!

Question

If the foot of perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(\alpha, 7, 1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$, then $\alpha$ is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are floating in a three-dimensional coordinate system. You have a fixed point $P(1, 0, 3)$ suspended in space, and a line $L$ stretching out infinitely, passing through a mysterious point $A(\alpha, 7, 1)$.

We are told that if you were to drop a plumb line from $P$ to this line, it would land perfectly at the point $Q(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$. Our mission is to uncover the identity of $\alpha$.

### Defining the Vectors

To solve this, we must first understand the relationship between the line $L$ and the segment $PQ$. The segment $PQ$ represents the shortest distance from the point to the line, which, by definition, must be perpendicular to the line itself.

First, let us find the direction ratios of the segment $PQ$. By subtracting the coordinates of $P$ from $Q$, we get:

$$\vec{PQ} = \left( \frac{5}{3} - 1, \frac{7}{3} - 0, \frac{17}{3} - 3 \right) = \left( \frac{2}{3}, \frac{7}{3}, \frac{8}{3} \right)$$

This vector $\vec{PQ}$ is the 'plumb line' we dropped. Now, consider the line $L$. We know it passes through $A(\alpha, 7, 1)$ and $Q(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$.

Therefore, the vector $\vec{AQ}$ must lie along the line $L$. Its direction ratios are:

$$\vec{AQ} = \left( \frac{5}{3} - \alpha, \frac{7}{3} - 7, \frac{17}{3} - 1 \right) = \left( \frac{5}{3} - \alpha, -\frac{14}{3}, \frac{14}{3} \right)$$

### The Power of the Dot Product

Because $PQ$ is perpendicular to the line $L$, the vector $\vec{PQ}$ must be orthogonal to the vector $\vec{AQ}$. In the language of linear algebra, their dot product must vanish into nothingness:

$$\vec{PQ} \cdot \vec{AQ} = 0$$

Substituting our components, we get:

$$\left( \frac{2}{3} \right) \left( \frac{5}{3} - \alpha \right) + \left( \frac{7}{3} \right) \left( -\frac{14}{3} \right) + \left( \frac{8}{3} \right) \left( \frac{14}{3} \right) = 0$$

### The Elegant Simplification

We can multiply the entire equation by $9$ to clear the denominators, turning this into a clean, manageable linear equation:

$$2(5 - 3\alpha) + 7(-14) + 8(14) = 0$$

Notice the beauty of the constants: $7(-14) + 8(14)$ is simply $14(8 - 7)$, which is just $14$. Our equation simplifies to:

$$2(5 - 3\alpha) + 14 = 0$$

Expanding this, we get $10 - 6\alpha + 14 = 0$, which leads us to $24 - 6\alpha = 0$. Solving for $\alpha$, we find:

$$6\alpha = 24 \implies \alpha = 4$$

### Final Conclusion

By respecting the geometric requirement of perpendicularity, we have unmasked $\alpha$. The value of the mysterious coordinate is **$\alpha = 4$**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the foot of perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$ , then $α$ is equal to ________.

Enter Numerical Value:

Visualized Solution

Visualize the Given Points

The Foot of the Perpendicular

The Core Logic: Perpendicular Vectors

Direction Ratios of $P Q$

Simplify DRs of $P Q$

Direction Ratios of Line $L$

Simplify DRs of Line $L$

Apply the Dot Product Condition

Clear the Denominators

Simplify the Constants

Expand and Solve

Final Value of $α$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Defining the Vectors

The Power of the Dot Product

The Elegant Simplification

Final Conclusion

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