Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $A(x_1, y_1, z_1)$ with a direction vector $\vec{d} = \langle a, b, c \rangle$ can be represented in symmetric form as $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.
• Perpendicular Distance from a Point to a Line: The shortest (perpendicular) distance $D$ from a point $P(x_0, y_0, z_0)$ to a line passing through a point $A$ with a direction vector $\vec{d}$ is given by the formula: $D = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}$ Here, $\vec{AP}$ is the vector from any point $A$ on the line to the given point $P$. This formula leverages the geometric property that the magnitude of the cross product $\|\vec{AP} \times \vec{d}\|$ represents the area of the parallelogram formed by $\vec{AP}$ and $\vec{d}$. Dividing this area by the base length $\|\vec{d}\|$ yields the height, which is the perpendicular distance.
• Cross Product of Two Vectors: For two vectors $\vec{u} = \langle u_x, u_y, u_z \rangle$ and $\vec{v} = \langle v_x, v_y, v_z \rangle$, their cross product is $\vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix} = (u_y v_z - u_z v_y)\mathbf{i} - (u_x v_z - u_z v_x)\mathbf{j} + (u_x v_y - u_y v_x)\mathbf{k}$.
• Magnitude of a Vector: The magnitude of a vector $\vec{v} = \langle v_x, v_y, v_z \rangle$ is given by $\|\vec{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.

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Step-by-Step Solution

Step 1: Identify a Point on the Line and its Direction Vector

• What we are doing: We first extract the necessary information from the given line equation: a specific point $A$ that lies on the line, and the direction vector $\vec{d}$ which indicates the line's orientation in space.
• Why we are doing this: These components are crucial for constructing the vectors needed in the distance formula.

The given line equation is $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$. Comparing this to the standard symmetric form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$:

• A point on the line is $A(x_1, y_1, z_1) = A(1, -2, -3)$. (Note: $y+2 = y-(-2)$ and $z+3 = z-(-3)$)
• The direction vector of the line is $\vec{d} = \langle a, b, c \rangle = \langle 2, -1, 2 \rangle$. The given point from which we need to find the distance is $P(2, -10, 1)$.

Step 2: Form the Vector $\vec{AP}$

• What we are doing: We construct a vector from the point $A$ (on the line) to the given point $P$.
• Why we are doing this: This vector $\vec{AP}$ is one of the two vectors required for the cross product in our distance formula. It establishes the relative position of point $P$ with respect to the line.

Given $A(1, -2, -3)$ and $P(2, -10, 1)$, the vector $\vec{AP}$ is calculated by subtracting the coordinates of $A$ from $P$: $\vec{AP} = \langle 2-1, \; -10-(-2), \; 1-(-3) \rangle$ $\vec{AP} = \langle 1, \; -10+2, \; 1+3 \rangle$ $\vec{AP} = \langle 1, \; -8, \; 4 \rangle$

Step 3: Calculate the Cross Product $\vec{AP} \times \vec{d}$

• What we are doing: We compute the cross product of the vector $\vec{AP}$ and the direction vector $\vec{d}$.
• Why we are doing this: The magnitude of this cross product, $\|\vec{AP} \times \vec{d}\|$, will form the numerator of our distance formula. It is numerically equal to the area of the parallelogram formed by these two vectors.

Using $\vec{AP} = \langle 1, -8, 4 \rangle$ and $\vec{d} = \langle 2, -1, 2 \rangle$: $\vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix}$ Expanding the determinant:

• i-component: $(-8)(2) - (4)(-1) = -16 - (-4) = -16 + 4 = -12$
• j-component: $-[(1)(2) - (4)(2)] = -[2 - 8] = -[-6] = 6$
• k-component: $(1)(-1) - (-8)(2) = -1 - (-16) = -1 + 16 = 15$

So, $\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle$.

Step 4: Calculate the Magnitudes $\|\vec{AP} \times \vec{d}\|$ and $\|\vec{d}\|$

• What we are doing: We calculate the length (magnitude) of the cross product vector and the length of the direction vector.

• Why we are doing this: These magnitudes are the final components needed to substitute into the distance formula.

• Magnitude of $\vec{AP} \times \vec{d}$: $\|\vec{AP} \times \vec{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2}$ $= \sqrt{144 + 36 + 225}$ $= \sqrt{405}$ To simplify $\sqrt{405}$, we find its prime factors: $405 = 5 \times 81 = 5 \times 9^2$. $= \sqrt{81 \times 5} = 9\sqrt{5}$

• Magnitude of $\vec{d}$: $\|\vec{d}\| = \sqrt{2^2 + (-1)^2 + 2^2}$ $= \sqrt{4 + 1 + 4}$ $= \sqrt{9} = 3$

Step 5: Compute the Perpendicular Distance

• What we are doing: We substitute the calculated magnitudes into the perpendicular distance formula.
• Why we are doing this: This is the final step to obtain the required distance.

Using the formula $D = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}$: $D = \frac{9\sqrt{5}}{3}$ $D = 3\sqrt{5}$

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Common Mistakes & Tips

• Sign Errors: Be extremely vigilant with signs, especially during vector subtraction to form $\vec{AP}$ and when calculating the components of the cross product. A single sign error can propagate and lead to an incorrect final answer.
• Cross Product Expansion: Double-check the determinant expansion for the cross product. A common error is forgetting the negative sign for the $\mathbf{j}$ component or mixing up the terms within the parentheses.
• Simplifying Square Roots: Always simplify square roots to their simplest form (e.g., $\sqrt{405}$ to $9\sqrt{5}$). This is crucial for matching your answer with the given options in an MCQ format.
• Alternative Method (Foot of Perpendicular): While the cross-product method is generally more efficient for competitive exams, an alternative involves finding the coordinates of the foot of the perpendicular. This method requires representing a general point on the line using a parameter (e.g., $t$), forming a vector from P to this general point, and then using the condition that this vector is perpendicular to the line's direction vector (their dot product is zero) to solve for $t$. This approach is usually more computationally intensive.

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Summary

This problem required us to find the perpendicular distance from a given point to a given line in 3D space. We utilized the vector formula $D = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}$. First, we extracted a point $A(1, -2, -3)$ and the direction vector $\vec{d} = \langle 2, -1, 2 \rangle$ from the line's equation. Then, we formed the vector $\vec{AP} = \langle 1, -8, 4 \rangle$. We computed the cross product $\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle$ and found its magnitude to be $9\sqrt{5}$. The magnitude of the direction vector $\|\vec{d}\|$ was found to be $3$. Finally, dividing these magnitudes yielded the perpendicular distance $3\sqrt{5}$.

The final answer is $\boxed{3\sqrt{5}}$, which corresponds to option (C).