Let the lines L1 : r = i + 2j + 3k + λ(2i + 3j + 4k), λ ∈ R and L2 : r = (4i + j) + μ(5i + 2j + k),

Q: Wait, why do we reject the negative root for $\lambda_P$?

The problem explicitly states that point $P$ lies in the first octant. In the first octant, all coordinates $(x, y, z)$ must be strictly positive. When we solved the quadratic equation for $\lambda_P$, we obtained two potential values: $0$ and $-2$. Testing $\lambda_P = -2$ leads to negative coordinates, which violates our constraint. Always check your constraints against the physical reality of the problem!

Q: Can we use vector projection instead of the distance formula?

You absolutely could, but it would be like using a sledgehammer to crack a nut. The distance formula is direct and robust here. Since we already have the parametric coordinates, the distance formula is the most efficient path to the solution. Keep your methods simple and elegant whenever possible.

Q: Is there a faster way to find the intersection point $R$?

Equating the parametric coordinates is the standard, foolproof method. While you could use the cross product of direction vectors to find the shortest distance (which is zero here), equating the components is the most reliable way to find the exact coordinates of $R$. It ensures you don't miss any subtle details.

Question

Let the lines $L_1 : \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}), \lambda \in \mathbb{R}$ and $L_2 : \vec{r} = (4\hat{i} + \hat{j}) + \mu(5\hat{i} + 2\hat{j} + \hat{k}), \mu \in \mathbb{R}$, intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|\vec{PR}| = \sqrt{29}$ and $|\vec{PQ}| = \sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to

Options:

(A) 348

(B) 340

(C) 320

(D) 360

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the lines $L_1 : \vec{r} = (1, 2, 3) + \lambda(2, 3, 4)$ and $L_2 : \vec{r} = (4, 1, 0) + \mu(5, 2, 1)$. To find the intersection point $R$, we equate the components of the two lines:

$$1 + 2\lambda = 4 + 5\mu$$
$$2 + 3\lambda = 1 + 2\mu$$
$$3 + 4\lambda = \mu$$

Substituting $\mu = 3 + 4\lambda$ into the first two equations, we solve the system to find $\lambda = -1$ and $\mu = -1$. Substituting $\lambda = -1$ back into the equation for $L_1$, we determine the anchor point:

$$R = (-1, -1, -1)$$

### The Constraint of the First Octant

We define a point $P$ on line $L_1$ such that the distance $|PR| = \sqrt{29}$. Using the distance formula, we set up the following equation:

$$|PR|^2 = (1 + 2\lambda_P - (-1))^2 + (2 + 3\lambda_P - (-1))^2 + (3 + 4\lambda_P - (-1))^2 = 29$$

This simplifies to $29(1 + \lambda_P)^2 = 29$, which implies $(1 + \lambda_P)^2 = 1$. This yields two potential values: $\lambda_P = 0$ or $\lambda_P = -2$.

The "First Octant" constraint requires all coordinates to be positive. If $\lambda_P = -2$, we obtain $P = (-3, -4, -5)$, which lies outside the first octant. Choosing $\lambda_P = 0$ gives us the valid point:

$$P = (1, 2, 3)$$

### The Dance of $Q$ and the Quadratic Surprise

With $P = (1, 2, 3)$ fixed, we locate point $Q$ on line $L_2$ given the distance $|PQ| = \sqrt{\frac{47}{3}}$. Applying the distance formula:

$$|PQ|^2 = (4 + 5\mu_Q - 1)^2 + (1 + 2\mu_Q - 2)^2 + (\mu_Q - 3)^2 = \frac{47}{3}$$

Expanding the terms leads to:

$$(3 + 5\mu_Q)^2 + (2\mu_Q - 1)^2 + (\mu_Q - 3)^2 = \frac{47}{3}$$

$$(25\mu_Q^2 + 30\mu_Q + 9) + (4\mu_Q^2 - 4\mu_Q + 1) + (\mu_Q^2 - 6\mu_Q + 9) = \frac{47}{3}$$

Combining like terms results in $30\mu_Q^2 + 20\mu_Q + 19 = \frac{47}{3}$. Multiplying by $3$ and simplifying, we obtain the perfect square:

$$9\mu_Q^2 + 6\mu_Q + 1 = (3\mu_Q + 1)^2 = 0$$

This confirms that $\mu_Q = -\frac{1}{3}$.

### Final Calculation

Using $\mu_Q = -\frac{1}{3}$, we find the coordinates of $Q$:

$$Q = \left(4 + 5\left(-\frac{1}{3}ight), 1 + 2\left(-\frac{1}{3}ight), -\frac{1}{3}ight) = \left(\frac{7}{3}, \frac{1}{3}, -\frac{1}{3}ight)$$

Next, we calculate the squared distance $QR^2$ between $Q\left(\frac{7}{3}, \frac{1}{3}, -\frac{1}{3}ight)$ and $R(-1, -1, -1)$:

$$QR^2 = \left(\frac{7}{3} + 1ight)^2 + \left(\frac{1}{3} + 1ight)^2 + \left(-\frac{1}{3} + 1ight)^2$$

$$QR^2 = \left(\frac{10}{3}ight)^2 + \left(\frac{4}{3}ight)^2 + \left(\frac{2}{3}ight)^2 = \frac{100 + 16 + 4}{9} = \frac{120}{9} = \frac{40}{3}$$

The final required value is $27(QR)^2$:

$$27 	imes \frac{40}{3} = 9 	imes 40 = \mathbf{360}$$

Select Answer:

Visualized Solution

Given Lines $L_{1}$ and $L_{2}$

Intersection Point $R$

Solving for Parameters

Coordinates of $R$

Point $P$ on $L_{1}$

Distance Formula for $P R$

First Octant Condition for $P$

Point $Q$ on $L_{2}$

Distance Formula for $P Q$

Simplifying the Equation

Solving the Quadratic for $μ_{Q}$

Coordinates of $Q$

Distance $Q R^{2}$

Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Constraint of the First Octant

The Dance of $Q$ and the Quadratic Surprise

Final Calculation

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Visualized Solution

Given Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Solving for Parameters

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

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

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PR

First Octant Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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​

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ