Key Concepts and Formulas

1. Direction Vector of a Line in 3D: For a line given in symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$, its direction vector is $\vec{d} = \langle a, b, c \rangle$.
2. Cross Product: The cross product of two vectors $\vec{u}$ and $\vec{v}$ results in a vector that is perpendicular to both $\vec{u}$ and $\vec{v}$. If $\vec{u} = \langle u_1, u_2, u_3 \rangle$ and $\vec{v} = \langle v_1, v_2, v_3 \rangle$, then $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$.
3. Equation of a Line in 3D: Given a point $P_0(x_0, y_0, z_0)$ and a direction vector $\vec{d} = \langle a, b, c \rangle$, the parametric form of the line's equation is $x = x_0 + \lambda a$, $y = y_0 + \lambda b$, $z = z_0 + \lambda c$, where $\lambda$ is a scalar parameter.
4. Equation of the $yz$-plane: The $yz$-plane is defined by the equation $x = 0$.
5. Distance Formula in 3D: The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by $\text{d}(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

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Step-by-Step Solution

Step 1: Determine the Direction Vector of Line $L$

• What we are doing: We need to find the direction vector of line $L$. Since line $L$ is perpendicular to two given lines, its direction vector will be perpendicular to the direction vectors of both those lines.
• Why we are doing it: The cross product of two vectors yields a vector perpendicular to both. This is the standard method to find a common perpendicular direction.

Let the two given lines be $L_1$ and $L_2$. The equation of $L_1$ is $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2}$. Its direction vector, $\vec{d_1}$, is obtained from the denominators: $\vec{d_1} = \langle 2, 1, -2 \rangle$

The equation of $L_2$ is $\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}$. Its direction vector, $\vec{d_2}$, is: $\vec{d_2} = \langle 1, 3, 4 \rangle$

Now, we compute the cross product of $\vec{d_1}$ and $\vec{d_2}$ to find the direction vector of line $L$, denoted as $\vec{d_L}$: $\vec{d_L} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}$ Calculating the determinant: $\vec{d_L} = \hat{i}( (1)(4) - (-2)(3) ) - \hat{j}( (2)(4) - (-2)(1) ) + \hat{k}( (2)(3) - (1)(1) )$ $\vec{d_L} = \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1)$ $\vec{d_L} = \hat{i}(10) - \hat{j}(10) + \hat{k}(5)$ $\vec{d_L} = 10\hat{i} - 10\hat{j} + 5\hat{k}$ We can simplify this direction vector by dividing by the common factor of 5, as any scalar multiple represents the same direction: $\vec{d_L} = \langle 2, -2, 1 \rangle$

Step 2: Write the Equation of Line $L$

• What we are doing: We are writing the parametric equations for line $L$.
• Why we are doing it: We know a point $P(2, -1, 3)$ that line $L$ passes through, and we've found its direction vector. The parametric form is convenient for finding intersection points.

Line $L$ passes through $P(2, -1, 3)$ and has the direction vector $\vec{d_L} = \langle 2, -2, 1 \rangle$. Using the parametric form $x = x_0 + \lambda a$, $y = y_0 + \lambda b$, $z = z_0 + \lambda c$: $x = 2 + 2\lambda$ $y = -1 - 2\lambda$ $z = 3 + \lambda$

Step 3: Find the Intersection Point $Q$ with the $yz$-plane

• What we are doing: We are finding the coordinates of point $Q$, which is the intersection of line $L$ with the $yz$-plane.
• Why we are doing it: The $yz$-plane is defined by $x=0$. By setting the $x$-coordinate of a point on line $L$ to zero, we can find the specific value of $\lambda$ that corresponds to point $Q$.

The equation of the $yz$-plane is $x = 0$. Substitute $x=0$ into the parametric equation for $x$ from Step 2: $0 = 2 + 2\lambda$ Solving for $\lambda$: $2\lambda = -2$ $\lambda = -1$ Now, substitute $\lambda = -1$ back into all three parametric equations to find the coordinates of point $Q$: $x_Q = 2 + 2(-1) = 2 - 2 = 0$ $y_Q = -1 - 2(-1) = -1 + 2 = 1$ $z_Q = 3 + (-1) = 2$ So, the intersection point $Q$ is $(0, 1, 2)$.

Step 4: Calculate the Distance Between $P$ and $Q$

• What we are doing: We are calculating the distance between the given point $P$ and the calculated point $Q$.
• Why we are doing it: This is the final objective of the problem, directly applying the 3D distance formula.

We have the coordinates of point $P(2, -1, 3)$ and point $Q(0, 1, 2)$. Using the 3D distance formula: $\text{d}(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ $\text{d}(P, Q) = \sqrt{(0 - 2)^2 + (1 - (-1))^2 + (2 - 3)^2}$ $\text{d}(P, Q) = \sqrt{(-2)^2 + (1 + 1)^2 + (-1)^2}$ $\text{d}(P, Q) = \sqrt{(-2)^2 + (2)^2 + (-1)^2}$ $\text{d}(P, Q) = \sqrt{4 + 4 + 1}$ $\text{d}(P, Q) = \sqrt{9}$ $\text{d}(P, Q) = 3$

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Common Mistakes & Tips

• Cross Product Arithmetic: Be very careful with signs when calculating the cross product, especially with negative components. A single sign error will propagate through the rest of the problem.
• Parametric Form Substitution: Ensure you substitute the correct value of $\lambda$ into all three coordinate equations to find the intersection point.
• Distance Formula Calculation: Double-check the squaring and addition steps. It's easy to make a small arithmetic error here.

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Summary

We first determined the direction vector of line $L$ by taking the cross product of the direction vectors of the two lines perpendicular to $L$. Then, we used the given point $P$ and the calculated direction vector to write the parametric equations of line $L$. Next, we found the intersection point $Q$ of line $L$ with the $yz$-plane by setting the $x$-coordinate to zero and solving for the parameter $\lambda$. Finally, we calculated the distance between points $P$ and $Q$ using the 3D distance formula. The distance was found to be 3.

The final answer is $\boxed{3}$, which corresponds to option (D).