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

Q: Why is the dot product of $\vec{QP}$ and $\vec{QA}$ equal to zero?

The dot product of two vectors $\vec{u}$ and $\vec{v}$ is defined as $|\vec{u}| |\vec{v}| \cos(\theta)$, where $\theta$ is the angle between them. Since $Q$ is the foot of the perpendicular from $P$ to the line $L$, the segment $PQ$ is perpendicular to the line $L$. Therefore, the angle between $\vec{QP}$ and any vector lying on the line $L$ (like $\vec{QA}$) is $90^\circ$. Since $\cos(90^\circ) = 0$, the dot product must be zero.

Q: Can I use the projection formula to solve this?

Yes, you absolutely can! The foot of the perpendicular $Q$ is the projection of point $P$ onto the line $L$. However, using the dot product condition $\vec{QP} \cdot \vec{QA} = 0$ is often more direct when you already know the coordinates of the foot of the perpendicular. It avoids the need to define the line's direction vector explicitly in terms of a parameter.

Q: Is there a simpler way to handle these fractions?

Fractions can look intimidating, but they are just numbers! The best strategy is to keep them as fractions until the very end. Notice how in this problem, the denominators were all $3$ or $9$. By keeping them as fractions, you can often see patterns or cancellations that would be hidden if you converted to decimals. Stay patient, and the algebra will simplify itself.

Question

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

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### Analyzing the Setup

Imagine you are standing in a vast, empty room. You have a point $P$ floating in the air at coordinates $(1, 0, 3)$. Across the room, there is a line $L$ passing through a point $A(\alpha, 7, 1)$.

You drop a plumb line from $P$ straight down to the line $L$, and it hits the line at a specific point $Q(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$. This point $Q$ is the **foot of the perpendicular**.

Our goal is to find the value of $\alpha$ that makes this geometric setup possible. This is not just a calculation; it is a dance of vectors in three-dimensional space.

### The Perpendicularity Condition

To solve this, we need to translate our geometric intuition into the language of algebra. We know that the line segment $PQ$ is perpendicular to the line $L$.

Since the point $A$ lies on the line $L$, the vector $\vec{QA}$ must also lie along the line $L$. Therefore, the vector $\vec{QP}$ must be perpendicular to the vector $\vec{QA}$.

In the realm of vectors, perpendicularity is synonymous with a dot product of zero. So, our master equation is:

$$\vec{QP} \cdot \vec{QA} = 0$$

This is the key that unlocks the entire problem. It connects the spatial relationship to a solvable algebraic equation.

### Calculating the Vectors

First, let us find the vector $\vec{QP}$. This is simply the position vector of $P$ minus the position vector of $Q$:

$$\vec{QP} = P - Q = \left(1 - \frac{5}{3}, 0 - \frac{7}{3}, 3 - \frac{17}{3}ight) = \left(-\frac{2}{3}, -\frac{7}{3}, -\frac{8}{3}ight)$$

Next, we find the vector $\vec{QA}$ by subtracting $Q$ from $A$:

$$\vec{QA} = A - Q = \left(\alpha - \frac{5}{3}, 7 - \frac{7}{3}, 1 - \frac{17}{3}ight) = \left(\alpha - \frac{5}{3}, \frac{14}{3}, -\frac{14}{3}ight)$$

Take a deep breath. I know these fractions look intimidating, but they are just placeholders. We are about to see them simplify in a very satisfying way.

### The Algebraic Dance

Now, we apply our dot product condition. We multiply the corresponding components of $\vec{QP}$ and $\vec{QA}$ and sum them up:

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

Let us expand the constant terms. The second term is $-\frac{98}{9}$, and the third term is $+\frac{112}{9}$. Combining these is straightforward because they share the same denominator:

$$\left(-\frac{2}{3}ight)\left(\alpha - \frac{5}{3}ight) - \frac{98}{9} + \frac{112}{9} = 0$$

$$\left(-\frac{2}{3}ight)\left(\alpha - \frac{5}{3}ight) + \frac{14}{9} = 0$$

See how the fractions are already behaving? We are left with a much cleaner equation. Now, we isolate the term with $\alpha$:

$$\left(-\frac{2}{3}ight)\left(\alpha - \frac{5}{3}ight) = -\frac{14}{9}$$

To solve for $\alpha - \frac{5}{3}$, we multiply both sides by the reciprocal of $-\frac{2}{3}$, which is $-\frac{3}{2}$:

$$\alpha - \frac{5}{3} = \left(-\frac{14}{9}ight) 	imes \left(-\frac{3}{2}ight)$$

$$\alpha - \frac{5}{3} = \frac{7}{3}$$

### Final Calculation

We are almost there. The final step is to add $\frac{5}{3}$ to both sides:

$$\alpha = \frac{7}{3} + \frac{5}{3} = \frac{12}{3} = 4$$

And there it is! After navigating through the sea of fractions, we arrive at a clean, beautiful integer: **$\alpha = 4$**.

It is a reminder that even in complex 3D geometry, the underlying structure is often elegant and simple. You have successfully navigated the vectors and solved the problem. Well done!

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

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

The Perpendicularity Condition

Calculating Vector $QP$

Calculating Vector $Q A$

Applying the Dot Product

Expanding the Equation

Combining Constant Terms

Isolating the Unknown

Final Value of $α$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Perpendicularity Condition

Calculating the Vectors

The Algebraic Dance

Final Calculation

Similar Questions

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 ________.

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x + 2 y + z = 0$ and $3 y - z = 3$ is $(α, β, γ)$ , then $α + β + γ$ is equal to

If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

Let $A$ and $B$ be two distinct points on the line $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Both $A$ and $B$ are at a distance $217$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$ . If $O$ is the origin, then $O A \cdot O B$ is equal to:

If the length of the perpendicular drawn from the point $P (a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $26$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point $P$ in this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to :

The foot of perpendicular of the point $(2, 0, 5)$ on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1}$ is $(α, β, γ)$ . Then Which of the following is NOT correct?

Let $P$ be the point $(10, - 2, - 1)$ and $Q$ be the foot of the perpendicular drawn from the point $R (1, 7, 6)$ on the line passing through the points $(2, - 5, 11)$ and $(- 6, 7, - 5)$ . Then the length of the line segment $P Q$ is equal to ________

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

The Perpendicularity Condition

Calculating Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QP​

Calculating Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QA​

Applying the Dot Product

Expanding the Equation

Combining Constant Terms

Isolating the Unknown

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

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Perpendicularity Condition

Calculating the Vectors

The Algebraic Dance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the foot of perpendicular drawn from the point (1,0,3) on a line passing through (α,7,1) is (35​,37​,317​), then α is equal to ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the length of the perpendicular from the point (β,0,β)(β=0) to the line 1x−1​=0y−0​=−1z+1​ is 3/2​, then β is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The foot of perpendicular of the point (2,0,5) on the line 2x+1​=5y−1​=−1z+1​ is (α,β,γ). Then Which of the following is NOT correct?

Calculating Vector $QP$

Calculating Vector $Q A$

Final Value of $α$

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 ________.

If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

The foot of perpendicular of the point $(2, 0, 5)$ on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1}$ is $(α, β, γ)$ . Then Which of the following is NOT correct?