Q.Find the angle between the following pairs of lines:

Concept: Angle Between Lines — the angle between two lines equals the angle between their direction vectors. For lines given in vector form $\vec{r} = \vec{a} + \lambda \vec{b}$, use $\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1|\,|\vec{b}_2|}$.

(i) Direction vectors: $\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}$, $\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}$.

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

Magnitudes: $|\vec{b}_1| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9+4+36} = \sqrt{49} = 7$; $|\vec{b}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.

$\cos\theta = \frac{19}{7 \times 3} = \frac{19}{21}$. Since $\frac{19}{21} < 1$, $\theta = \cos^{-1}\left(\frac{19}{21}\right)$.

(ii) Direction vectors: $\vec{b}_1 = \hat{i} - \hat{j} - 2\hat{k}$, $\vec{b}_2 = 3\hat{i} - 5\hat{j} - 4\hat{k}$.

Dot product: $\vec{b}_1 \cdot \vec{b}_2 = 1(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16$. …

The key idea is simple: a line in space is defined by a point and a direction. The direction vector tells you which way the line runs. When two lines are given in the form $\vec{r} = \vec{a} + \lambda \vec{b}$, the angle between them is just the angle between their direction vectors $\vec{b}_1$ and $\vec{b}_2$. The position vectors $\vec{a}$ don't matter at all for the angle — they only tell you where the lines are located, not how they're oriented.

Why does this work? Because the direction vector is like an arrow along the line. If you slide both arrows to the same starting point, the angle between them is exactly the angle between the lines. The dot product formula $\vec{b}_1 \cdot \vec{b}_2 = |\vec{b}_1| |\vec{b}_2| \cos \theta$ gives us $\cos \theta$, and then we take the inverse cosine.

Let's work through each part step by step.

Part (i)

1. Identify the direction vectors.

   For the first line, $\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})$, the direction vector is $\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}$.

   For the second line, $\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$, the direction vector is $\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}$.

2. Compute the dot product.

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

3. Find the magnitudes.

   $|\vec{b}_1| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.

   $|\vec{b}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

4. Apply the formula.

   $\cos \theta = \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} = \frac{19}{7 \times 3} = \frac{19}{21}$.

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

Part (ii)

1. Identify the direction vectors. First line: $\vec{r} = 3\hat{i} + \hat{j} - 2\hat{k} + \lambda(\hat{i} - \hat{j} - 2\hat{k})$, so $\vec{b}_1 = \hat{i} - \hat{j} - 2\hat{k}$. …

Method: Angle between two lines from their direction vectors

The angle between two lines is the angle between their directions — where the lines sit is irrelevant. When lines are given as $\vec r=\vec a+\lambda\vec b$, only the $\vec b$'s matter.

Steps

Step 1: Pull out the two direction vectors $\vec b_1,\vec b_2$ — the coefficients of $\lambda$ and $\mu$. Discard the position vectors $\vec a$ entirely; they never enter the angle.

Step 2: Apply the cosine formula.

The absolute value in the numerator forces the acute angle, the convention for the angle between lines. …

Common Mistakes

Mistake 1: Including the position vectors $\vec a$ in the calculation.

Why it's wrong: the angle depends only on the directions $\vec b_1,\vec b_2$ (the coefficients of $\lambda,\mu$); the constant vectors — even a large $-56\hat k$ — are irrelevant. Correct approach: dot only the direction vectors.

Mistake 2: Omitting the modulus in the numerator. …