1. Key Concepts and Formulas

• Vector Representation of a Line: A line in 3D space passing through a point $A(x_1, y_1, z_1)$ and parallel to a vector $\vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}$ can be represented as $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\lambda$ is a scalar parameter.
• Distance from a Point to a Line (Geometric Method): The perpendicular distance ($d$) from a point $P$ to a line passing through point $A$ with direction vector $\vec{b}$ can be found using the Pythagorean theorem. If $N$ is the foot of the perpendicular from $P$ to the line, then in the right-angled triangle $\triangle APN$: $d = PN = \sqrt{|\vec{AP}|^2 - AN^2}$ where $\vec{AP}$ is the vector from point $A$ to point $P$, and $AN$ is the scalar projection of $\vec{AP}$ onto the direction vector $\vec{b}$. The length of the scalar projection is given by: $AN = \frac{|\vec{AP} \cdot \vec{b}|}{|\vec{b}|}$
• Alternative (Cross-Product) Formula: The perpendicular distance $d$ can also be calculated directly using the cross product: $d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$

2. Step-by-Step Solution

We are asked to find the distance of point $P(4, 6, -2)$ from a line passing through $A(-3, 2, 3)$ and parallel to a line with direction ratios $(3, 3, -1)$.

Step 1: Identify the Given Point and Define the Line Parameters

• The given point is $P(4, 6, -2)$.
• A point on the line is $A(-3, 2, 3)$.
• The direction ratios of the line are $(3, 3, -1)$, so the direction vector of the line is $\vec{b} = 3\hat{i} + 3\hat{j} - \hat{k}$.

Step 2: Form the Vector $\vec{AP}$ and Calculate its Magnitude

We need to form a vector from a point on the line ($A$) to the given point ($P$). This vector will be one side of our right-angled triangle.

• Calculate $\vec{AP}$: Subtract the coordinates of $A$ from $P$. $\vec{AP} = (4 - (-3))\hat{i} + (6 - 2)\hat{j} + (-2 - 3)\hat{k}$ $\vec{AP} = 7\hat{i} + 4\hat{j} - 5\hat{k}$
• Calculate $|\vec{AP}|$ (Magnitude of $\vec{AP}$): $|\vec{AP}| = \sqrt{7^2 + 4^2 + (-5)^2} = \sqrt{49 + 16 + 25} = \sqrt{90}$ For the Pythagorean theorem, we will use $|\vec{AP}|^2 = 90$.

Step 3: Identify the Direction Vector $\vec{b}$ and Calculate its Magnitude

The direction vector of the line is given by its direction ratios. Its magnitude is needed for the scalar projection.

• Direction Vector $\vec{b}$: $\vec{b} = 3\hat{i} + 3\hat{j} - \hat{k}$
• Calculate $|\vec{b}|$ (Magnitude of $\vec{b}$): $|\vec{b}| = \sqrt{3^2 + 3^2 + (-1)^2} = \sqrt{9 + 9 + 1} = \sqrt{19}$ For calculations, we will use $|\vec{b}|^2 = 19$.

Step 4: Calculate the Length of the Scalar Projection ($AN$)

The scalar projection $AN$ is the length of the segment along the line from point $A$ to the foot of the perpendicular $N$ from point $P$.

• Calculate the dot product $\vec{AP} \cdot \vec{b}$: $\vec{AP} \cdot \vec{b} = (7)(3) + (4)(3) + (-5)(-1)$ $\vec{AP} \cdot \vec{b} = 21 + 12 + 5 = 38$
• Calculate $AN$: $AN = \frac{|\vec{AP} \cdot \vec{b}|}{|\vec{b}|} = \frac{38}{\sqrt{19}}$
• Calculate $AN^2$: This is often more convenient for the Pythagorean theorem. $AN^2 = \left(\frac{38}{\sqrt{19}}\right)^2 = \frac{38^2}{19} = \frac{1444}{19}$ Since $38 = 2 \times 19$, we can simplify: $AN^2 = \frac{(2 \times 19)^2}{19} = \frac{4 \times 19^2}{19} = 4 \times 19 = 76$

Step 5: Apply the Pythagorean Theorem to Find the Perpendicular Distance ($PN$)

In the right-angled triangle $\triangle APN$, $AP$ is the hypotenuse, $AN$ is one leg, and $PN$ (the perpendicular distance, $d$) is the other leg.

• Formula: $d^2 = PN^2 = |\vec{AP}|^2 - AN^2$
• Substitute the calculated values: $d^2 = 90 - 76$ $d^2 = 14$
• Calculate the distance $d$: $d = \sqrt{14}$

3. Common Mistakes & Tips

• Sign Errors: Be very careful with negative signs when calculating vector components, dot products, and magnitudes.
• Magnitude vs. Squared Magnitude: Remember to square magnitudes when using the Pythagorean theorem, and take the square root at the end for the final distance.
• Vector Subtraction Order: $\vec{AP} = P - A$, not $A - P$. The order matters for the components, though $|\vec{AP}| = |\vec{PA}|$.
• Cross Product Method: While the geometric method provides intuition, the cross-product formula ($d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$) is often faster for competitive exams if you are proficient with cross-product calculations. Practice both methods!

4. Summary

To find the distance from a point $P$ to a line, we first identify a point $A$ on the line and its direction vector $\vec{b}$. We then form the vector $\vec{AP}$. The perpendicular distance is found by projecting $\vec{AP}$ onto $\vec{b}$ (to find the length $AN$) and then using the Pythagorean theorem with $|\vec{AP}|$ and $AN$. Following this method with the given coordinates and direction ratios, the perpendicular distance is calculated to be $\sqrt{14}$.

5. Final Answer

The final answer is $\boxed{\sqrt{14}}$, which corresponds to option (B).