Use the formula d = |[a⃗₂-a⃗₁, b⃗₁, b⃗₂]| / |b⃗₁ × b⃗₂|.

Shortest Distance Between Two Skew Lines

Question

Find the shortest distance between the lines:

L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}

L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}

Solution — Step by Step

From $L_1$ : point $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ , direction $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ .

From $L_2$ : point $\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$ , direction $\vec{b_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ .

We need $\vec{b_1} \times \vec{b_2}$ because this gives a vector perpendicular to both lines — the direction of the shortest connector.

\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}

= \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}

|\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1+4+1} = \sqrt{6}

\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}

The shortest distance formula is:

d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}

(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1

d = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}

Shortest distance = $\dfrac{1}{\sqrt{6}}$ units

Why This Works

Two skew lines never meet and aren’t parallel — they live on different planes. The shortest distance between them lies along the unique line segment that is perpendicular to both simultaneously. That perpendicular direction is exactly $\vec{b_1} \times \vec{b_2}$ .

The scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ measures the projection of the vector joining the two known points onto this perpendicular direction. Dividing by $|\vec{b_1} \times \vec{b_2}|$ converts this projection into an actual length.

If the lines were intersecting or parallel, this formula gives zero or $0/0$ respectively — so the non-zero result here confirms the lines are indeed skew.

Alternative Method (Scalar Triple Product in Determinant Form)

We can write the same formula as a single determinant:

d = \frac{\left| \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} \right|}{\sqrt{(m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 + (l_1 m_2 - l_2 m_1)^2}}

Plugging in directly:

\text{Numerator} = \begin{vmatrix} 1 & 2 & 2 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}

= 1(15-16) - 2(10-12) + 2(8-9) = -1 + 4 - 2 = 1

Same numerator, same answer. Many students find this determinant form faster in exams since you skip computing the cross product separately.

In JEE Main, skew lines questions almost always use symmetric form (Cartesian equations). Practice reading off $\vec{a}$ and $\vec{b}$ from the denominators in under 15 seconds — this alone saves a minute per question.

Common Mistake

The most frequent error: taking $\vec{a_2} - \vec{a_1}$ in the wrong order, getting $-\hat{i} - 2\hat{j} - 2\hat{k}$ instead. Since we take the absolute value in the numerator, the sign doesn’t affect the final answer — but students who skip the absolute value sign get a negative distance and then panic. Always write the modulus bars explicitly around the scalar triple product.

A second trap: computing $|\vec{b_1} \times \vec{b_2}|$ incorrectly by forgetting the negative sign on the $\hat{i}$ component. Here $\vec{b_1} \times \vec{b_2} = -\hat{i} + 2\hat{j} - \hat{k}$ , not $\hat{i} + 2\hat{j} + \hat{k}$ . A wrong cross product gives $\sqrt{6}$ in both cases here (by coincidence of this problem’s numbers), but that won’t save you in every question — always expand the determinant carefully.

Question

Solution — Step by Step

Why This Works

Alternative Method (Scalar Triple Product in Determinant Form)

Common Mistake

Try These Next