The angle between the pair of lines given by = i + 2 j - 3 k + λ ( i - 2 j + 2 k) and = 5 i + j + k + μ (3 i - 2 j + 6 k) is

The angle $\theta$θ between two lines is the angle between their respective direction vectors. From the given equations, we identify the direction vectors:

• Vector 1 ($\vec{b}_1$b1​): $\hat{i} - 2\hat{j} + 2\hat{k}$i^−2j^​+2k^
• Vector 2 ($\vec{b}_2$b2​): $3\hat{i} - 2\hat{j} + 6\hat{k}$3i^−2j^​+6k^

The cosine of the angle $\theta$θ is given by the formula:

\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$ **Calculate the Dot Product ($\vec{b}_1 \cdot \vec{b}_2$):**\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (-2)(-2) + (2)(6)$$$$\vec{b}_1 \cdot \vec{b}_2 = 3 + 4 + 12 = 19$

Calculate the Magnitudes ($|\vec{b}_1|$∣b1​∣ and $|\vec{b}_2|$∣b2​∣):

• $|\vec{b}_1| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$∣b1​∣=12+(−2)2+22​=1+4+4​=9​=3
• $|\vec{b}_2| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$∣b2​∣=32+(−2)2+62​=9+4+36​=49​=7

Substitute into the Cosine Formula:

$$\cos \theta = \frac{19}{3 \times 7}$$cosθ=3×719​

$$\cos \theta = \frac{19}{21}$$cosθ=2119​

$$\theta = \cos^{-1}\left(\frac{19}{21}\right)$$θ=cos−1(2119​)