The square of the distance of the point (15/7, 32/7, 7) from the line (x+1)/3 = (y+3)/5 = (z+5)/7 in

Q: Why are we creating a new line $L_1$ instead of just using the perpendicular distance formula?

Great question! The perpendicular distance formula is only for the *shortest* distance. Here, the problem explicitly restricts the direction of travel to the vector $\hat{i}+4\hat{j}+7\hat{k}$. We aren't looking for the shortest path; we are looking for the path defined by that specific vector.

Q: Does the value of $\lambda$ matter for the final answer?

Absolutely! $\lambda$ represents the scalar multiple of the direction vector $\vec{d}$ that connects $P$ to $Q$. Since $\vec{PQ} = \lambda \vec{d}$, the square of the distance is simply $\lambda^2 |\vec{d}|^2$. Finding $\lambda$ is the key to unlocking the magnitude of the displacement.

Q: What if the lines were skew and never intersected?

In this specific problem, the geometry is constructed such that an intersection exists. If they were skew, you would not be able to reach $L_2$ by moving strictly along $\vec{d}$ from $P$. Always check if the direction vector allows for an intersection!

Q: Is there a faster way to find $t$ without equating all three parts?

You only need two independent equations to solve for one variable $t$. By equating the first and third parts of the parametric expressions, we simplify the algebra significantly. You don't need to use all three parts simultaneously unless you want to verify your result.

Q: Why did we square the components at the end?

The distance formula in 3D space is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. Since the question asks for the *square* of the distance, we simply calculate the sum of the squares of the components of the vector $\vec{PQ}$.

Question

The square of the distance of the point $(\frac{15}{7},\frac{32}{7},7)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4\hat{j}+7\hat{k}$ is:

Options:

(A) 54

(B) 44

(C) 41

(D) 66

Sigma Percentile · Accepted Answer

### Analyzing the Setup

You are standing at a specific coordinate in 3D space, point $P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$. Before you lies a line $L_2$, defined by the symmetric equations:

$$\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$$

You are tasked to travel from $P$ in the exact direction of the vector $\vec{d} = \hat{i} + 4\hat{j} + 7\hat{k}$ until you strike the line $L_2$. This is a problem of intersection, representing a dance between vectors and lines.

### Defining the Path

To solve this, we first define your path. Since you start at $P$ and move along $\vec{d}$, any point on your path can be described by the parametric equation of a line $L_1$:

$$\frac{x - \frac{15}{7}}{1} = \frac{y - \frac{32}{7}}{4} = \frac{z - 7}{7} = \lambda$$

Here, $\lambda$ is your scalar parameter. It represents how far you have traveled along the direction $\vec{d}$. If $\lambda$ is positive, you move in the direction of the vector; if negative, you move backward.

### The Intersection Point $Q$

Any point $Q$ on the line $L_2$ can be represented by a parameter $t$:

$$\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} = t$$

This yields the coordinates $Q(3t-1, 5t-3, 7t-5)$. Because $Q$ must also lie on your path $L_1$, these coordinates must satisfy the equation of $L_1$.

By substituting these into the equation for $L_1$, we create a bridge between the two lines:

$$\frac{(3t-1) - \frac{15}{7}}{1} = \frac{(5t-3) - \frac{32}{7}}{4} = \frac{(7t-5) - 7}{7} = \lambda$$

### Solving the Algebraic Puzzle

We simplify the numerators to reveal the underlying structure:

$$\frac{3t - \frac{22}{7}}{1} = \frac{5t - \frac{53}{7}}{4} = \frac{7t - 12}{7} = \lambda$$

To find $t$, we equate the first and third expressions. This avoids the middle term and leads to a clean linear equation:

$$3t - \frac{22}{7} = \frac{7t - 12}{7}$$

Multiplying the entire equation by $7$, we obtain $21t - 22 = 7t - 12$. A quick rearrangement gives $14t = 10$, or $t = \frac{5}{7}$.

With $t$ in hand, we find $\lambda$ by plugging $t$ back into our first expression:

$$\lambda = 3\left(\frac{5}{7}\right) - \frac{22}{7} = \frac{15-22}{7} = -1$$

### The Final Destination

We have discovered that $\lambda = -1$. This means the vector $\vec{PQ}$ is simply $-1$ times our direction vector $\vec{d}$:

$$\vec{PQ} = -1(\hat{i} + 4\hat{j} + 7\hat{k}) = -\hat{i} - 4\hat{j} - 7\hat{k}$$

The square of the distance is the square of the magnitude of this vector:

$$PQ^2 = (-1)^2 + (-4)^2 + (-7)^2 = 1 + 16 + 49 = 66$$

Through the systematic application of parametric lines and vector algebra, we have navigated from a point to a line along a specific trajectory. The final result is **66**.

Animated Solution for Mathematics - Three Dimensional Geometry: The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

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Visualized Solution

Visualizing the Setup

Defining Line $L_{1}$

General Point $Q$ on $L_{2}$

Substituting $Q$ into $L_{1}$

Simplifying the Equations

Solving for Parameter $t$

Finding Parameter $λ$

Finding the Vector $P Q$

Final Calculation: $P Q^{2}$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Defining the Path

The Intersection Point $Q$

Solving the Algebraic Puzzle

The Final Destination

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Select Answer:

Visualized Solution

Visualizing the Setup

Defining Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Simplifying the Equations

Solving for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Finding Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ2