Q.Find the angle between the pair of lines given by $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$.

Concept: Angle Between Lines — the angle between two lines equals the angle between their direction vectors.

Step 1: Identify direction vectors.

First line: $\vec{b}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$

Second line: $\vec{b}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}$

Step 2: Use dot product formula:

$\cos\theta = \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1|\,|\vec{b}_2|}$

Compute dot product:

$\vec{b}_1 \cdot \vec{b}_2 = 1(3) + 2(2) + 2(6) = 3 + 4 + 12 = 19$

Step 3: Find magnitudes:

$|\vec{b}_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$

$|\vec{b}_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9+4+36} = \sqrt{49} = 7$

Step 4:

$\cos\theta = \frac{19}{3 \times 7} = \frac{19}{21}$

Thus $\theta = \cos^{-1}\left(\frac{19}{21}\right)$

The key idea: when two lines are given in parametric vector form $\vec{r} = \vec{a} + \lambda \vec{b}$, the direction of each line is completely determined by its direction vector $\vec{b}$. The fixed points $\vec{a}$ only tell us where the lines are located in space — they have no effect on the angle between the lines.

So the problem reduces to finding the angle between the two direction vectors:

• $\vec{b}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$
• $\vec{b}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}$

The angle $\theta$ between any two vectors is given by the dot product formula:

Let's work through it step by step.

1. Compute the dot product $\vec{b}_1 \cdot \vec{b}_2$:

2. Find the magnitude $|\vec{b}_1|$:

3. Find the magnitude $|\vec{b}_2|$:

4. Substitute into the formula:

Since $\frac{19}{21} < 1$, the angle is acute and well-defined.

5. Write the final expression for $\theta$:

Method: Angle Between Two Lines Given in Vector Form

Use this when both lines are written as $\vec{r}=\vec{a}+\lambda\vec{b}$ and you need the angle between them.

Steps

Step 1: Keep only the direction vectors.

The fixed points $\vec{a_1},\vec{a_2}$ locate the lines in space but do not affect the angle — discard them and keep $\vec{b_1},\vec{b_2}$.

Step 2: Apply the dot-product formula.

Compute the dot product for the numerator and each magnitude for the denominator.

Step 3: Take the inverse cosine.

Write $\theta=\cos^{-1}(\cdots)$. The modulus guarantees the acute angle; leave the answer as an exact $\cos^{-1}$ unless a decimal is requested.

Common Mistakes

Mistake 1: Bringing the fixed points $\vec{a_1},\vec{a_2}$ into the angle calculation.

Why it's wrong: the points only position the lines in space; the angle depends solely on the direction vectors $\vec{b_1},\vec{b_2}$. Correct approach: use only $(1,2,2)$ and $(3,2,6)$ in the dot-product formula.

Mistake 2: Forgetting to divide by the magnitudes.

Why it's wrong: the dot product $19$ alone is not $\cos\theta$; you must divide by $|\vec{b_1}||\vec{b_2}|=3\times7$. Correct approach: $\cos\theta=\tfrac{19}{21}$, then $\theta=\cos^{-1}\tfrac{19}{21}$.