Q.Find the distance of a point $(2, 4, -1)$ from the line $\dfrac{x+5}{1} = \dfrac{y+3}{4} = \dfrac{z-6}{-9}$.

Concept: Distance from a point to a line in 3D — we use the perpendicular distance formula involving the cross product of the direction vector and the vector from a point on the line to the given point.

Step 1: Identify a point on the line. From the symmetric form, let $P = (-5, -3, 6)$. The direction vector of the line is $\vec{d} = (1, 4, -9)$.

Step 2: Find the vector from $P$ to the given point $Q = (2, 4, -1)$:

Step 3: Compute the cross product $\vec{PQ} \times \vec{d}$:

$= \mathbf{i}(-63 + 28) - \mathbf{j}(-63 + 7) + \mathbf{k}(28 - 7) = (-35, 56, 21).$ …

Concept and Intuition

The distance from a point to a line in 3D is the length of the perpendicular segment from the point to the line. Unlike in 2D, we can't just use a formula with coordinates — we need vector geometry.

Think of it this way: pick any point $A$ on the line. Draw the vector $\overrightarrow{AP}$ from $A$ to the given point $P$. This vector has two components relative to the line: one parallel to the line (along its direction) and one perpendicular to it. The perpendicular component is what we want — its length is the distance.

The trick: the parallel component is just the projection of $\overrightarrow{AP}$ onto the direction vector $\vec{d}$ of the line. Once we subtract that projection from $\overrightarrow{AP}$, what remains is perpendicular to the line. The magnitude of that remainder is our answer.

This cross-product formula is the cleanest way — it directly gives the perpendicular component's magnitude without separately computing the projection.

Step-by-step Solution

1. Identify the given line and point.

The line is $\dfrac{x+5}{1} = \dfrac{y+3}{4} = \dfrac{z-6}{-9}$.

From the symmetric form, we read:

• A point on the line: $A(-5, -3, 6)$ (set each numerator to zero)
• Direction vector: $\vec{d} = (1, 4, -9)$

The given point is $P(2, 4, -1)$.

2. Form the vector from the point on the line to the given point.

3. Compute the cross product $\overrightarrow{AP} \times \vec{d}$.

We need:

Expand:

• $\mathbf{i}$-component: $(7)(-9) - (-7)(4) = -63 + 28 = -35$
• $\mathbf{j}$-component: $-[ (7)(-9) - (-7)(1) ] = -[ -63 + 7 ] = -(-56) = 56$ (Careful: the $\mathbf{j}$ term has a minus sign in the determinant expansion)
• $\mathbf{k}$-component: $(7)(4) - (7)(1) = 28 - 7 = 21$

So:

$\overrightarrow{AP} \times \vec{d} = (-35, 56, 21)$ …

Method: Shortest distance from a point to a line using the cross product

Use this when you only need the perpendicular distance from a point to a line, not the foot.

Steps

Step 1: Extract a point and the direction from the line.

From $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$, read the on-line point $A(x_0,y_0,z_0)$ and direction $\vec d=(a,b,c)$.

Step 2: Form the join.

Compute $\overrightarrow{AP}=\vec p-\vec a$ from the on-line point to the given point $P$.

Step 3: Apply the formula.

$d=\frac{|\overrightarrow{AP}\times\vec d|}{|\vec d|}$ …

Common Mistakes

Mistake 1: Forgetting to divide by $|\vec d|$.

Why it's wrong: $|\overrightarrow{AP}\times\vec d|$ is an area that scales with the length of $\vec d$; only after dividing by $|\vec d|$ is it a genuine distance. Correct approach: use $d=\dfrac{|\overrightarrow{AP}\times\vec d|}{|\vec d|}$ — here $\dfrac{49\sqrt2}{7\sqrt2}=7$.

Mistake 2: Sign error on the middle term of the cross product. …