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Worked Examples · Example 10

Q.Find the distance between the lines l1l_1 and l2l_2 given by r⃗=i^+2j^−4k^+λ(2i^+3j^+6k^)\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) and r⃗=3i^+3j^−5k^+μ(2i^+3j^+6k^)\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}).

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Appeared in past exams:GUJCET 2025· Set 03· 1mreworded
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Both lines are parallel (same direction vector). The shortest distance between two parallel lines is the length of the projection of the vector joining a point on each line onto a vector perpendicular to the direction. The distance is 2937\boxed{\frac{\sqrt{293}}{7}} units.

Why This Approach Works

When two lines are parallel, the shortest distance between them is simply the perpendicular distance from any point on one line to the other line. This is because the lines never meet and maintain a constant separation — like two parallel railway tracks.

The key insight: instead of finding a complicated common perpendicular, we can take any point AA on l1l_1 and any point BB on l2l_2, then find the component of AB→\overrightarrow{AB} that is perpendicular to the common direction. That perpendicular component is exactly the shortest distance.

Distance between parallel lines =∣AB→×d⃗∣∣d⃗∣= \frac{|\overrightarrow{AB} \times \vec{d}|}{|\vec{d}|}, where d⃗\vec{d} is the common direction vector.

Step-by-Step Solution

1. Identify the direction vectors and points

Both lines have the same direction vector:

d⃗=2i^+3j^+6k^\vec{d} = 2\hat{i} + 3\hat{j} + 6\hat{k}

A point on l1l_1 (taking λ=0\lambda = 0):

A=(1,2,−4)A = (1, 2, -4)

A point on l2l_2 (taking μ=0\mu = 0):

B=(3,3,−5)B = (3, 3, -5)

2. Find the vector joining the two points

AB→=B⃗−A⃗=(3−1)i^+(3−2)j^+(−5+4)k^\overrightarrow{AB} = \vec{B} - \vec{A} = (3-1)\hat{i} + (3-2)\hat{j} + (-5+4)\hat{k}

AB→=2i^+1j^−1k^\overrightarrow{AB} = 2\hat{i} + 1\hat{j} - 1\hat{k}

3. Compute the cross product AB→×d⃗\overrightarrow{AB} \times \vec{d}

AB→×d⃗=∣i^j^k^21−1236∣\overrightarrow{AB} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}

Expanding:

  • i^\hat{i}-component: (1)(6)−(−1)(3)=6+3=9(1)(6) - (-1)(3) = 6 + 3 = 9
  • j^\hat{j}-component: −[(2)(6)−(−1)(2)]=−[12+2]=−14-[(2)(6) - (-1)(2)] = -[12 + 2] = -14
  • k^\hat{k}-component: (2)(3)−(1)(2)=6−2=4(2)(3) - (1)(2) = 6 - 2 = 4

So:

AB→×d⃗=9i^−14j^+4k^\overrightarrow{AB} \times \vec{d} = 9\hat{i} - 14\hat{j} + 4\hat{k}

4. Find the magnitude of this cross product …

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