Key Concepts and Formulas

• Distance from a Point to a Line in 3D: The shortest distance ($d$) from a point $P$ to a line passing through point $A$ and parallel to vector $\vec{b}$ is given by the formula: $d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$ where $\vec{AP}$ is the vector from point $A$ to point $P$. This formula leverages the property that the magnitude of the cross product $|\vec{AP} \times \vec{b}|$ equals the area of the parallelogram formed by $\vec{AP}$ and $\vec{b}$, and dividing by the base $|\vec{b}|$ gives the height, which is the perpendicular distance.
• Vector Operations: Essential operations include vector subtraction ($\vec{p} - \vec{a}$), dot product ($\vec{u} \cdot \vec{v}$), cross product ($\vec{u} \times \vec{v}$), and magnitude of a vector ($|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$).

Step-by-Step Solution

Step 1: Identify Given Information and Represent in Vector Form. We are given a point $P$ with position vector $\vec{p} = -\widehat{i} + 2\widehat{j} + 6\widehat{k}$. The line passes through a point $A$ with position vector $\vec{a} = 2\widehat{i} + 3\widehat{j} - 4\widehat{k}$ and is parallel to the vector $\vec{b} = 6\widehat{i} + 3\widehat{j} - 4\widehat{k}$. Our goal is to find the shortest distance from point $P$ to this line.

Step 2: Calculate the Vector $\vec{AP}$. This vector connects a known point on the line ($A$) to the given point ($P$). It is a crucial component for the distance formula. $\vec{AP} = \vec{p} - \vec{a}$ $\vec{AP} = (-\widehat{i} + 2\widehat{j} + 6\widehat{k}) - (2\widehat{i} + 3\widehat{j} - 4\widehat{k})$ $\vec{AP} = (-1 - 2)\widehat{i} + (2 - 3)\widehat{j} + (6 - (-4))\widehat{k}$ $\vec{AP} = -3\widehat{i} - \widehat{j} + 10\widehat{k}$

Step 3: Calculate the Magnitude of the Direction Vector $\vec{b}$. The magnitude of the direction vector is needed as the denominator in the distance formula. $|\vec{b}| = \sqrt{(6)^2 + (3)^2 + (-4)^2}$ $|\vec{b}| = \sqrt{36 + 9 + 16}$ $|\vec{b}| = \sqrt{61}$

Step 4: Calculate the Cross Product $\vec{AP} \times \vec{b}$. The cross product is used to find the area of the parallelogram formed by $\vec{AP}$ and $\vec{b}$. $\vec{AP} \times \vec{b} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ -3 & -1 & 10 \\ 6 & 3 & -4 \end{vmatrix}$ $= \widehat{i}((-1)(-4) - (10)(3)) - \widehat{j}((-3)(-4) - (10)(6)) + \widehat{k}((-3)(3) - (-1)(6))$ $= \widehat{i}(4 - 30) - \widehat{j}(12 - 60) + \widehat{k}(-9 + 6)$ $= -26\widehat{i} + 48\widehat{j} - 3\widehat{k}$

Step 5: Calculate the Magnitude of the Cross Product $|\vec{AP} \times \vec{b}|$. This magnitude represents the area of the parallelogram. $|\vec{AP} \times \vec{b}| = \sqrt{(-26)^2 + (48)^2 + (-3)^2}$ $= \sqrt{676 + 2304 + 9}$ $= \sqrt{2989}$

Step 6: Apply the Distance Formula. Now, we use the formula for the distance from a point to a line. $d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$ $d = \frac{\sqrt{2989}}{\sqrt{61}}$ To simplify, we perform the division inside the square root: $d = \sqrt{\frac{2989}{61}}$ $d = \sqrt{49}$ $d = 7$

Common Mistakes & Tips

• Vector Subtraction Order: Always subtract the position vector of the starting point from the position vector of the ending point (e.g., $\vec{p} - \vec{a}$ for $\vec{AP}$).
• Sign Errors in Cross Product: Be extremely careful with the signs when expanding the determinant for the cross product. A single sign error can lead to a completely wrong result.
• Magnitude vs. Vector: Distinguish between a vector and its magnitude. The distance formula requires the magnitude of the cross product and the magnitude of the direction vector.

Summary

The problem requires finding the shortest distance from a given point to a line in 3D space. This is achieved by using the vector formula involving the cross product. We first find the vector connecting a point on the line to the given point ($\vec{AP}$), then compute the cross product of this vector with the direction vector of the line ($\vec{AP} \times \vec{b}$). The magnitude of this cross product, divided by the magnitude of the direction vector, yields the shortest perpendicular distance. Following these steps, the calculated distance is 7 units.

The final answer is $\boxed{7}$, which corresponds to option (B).