Cosine of the acute angle between the lines x-3 2 = y-2 1 = z-5 2 and x-1 6 = y-3 -3 = z+6 2 is

The angle between two lines is the angle between their direction vectors.

For the first line, the direction vector is $\vec{v}_1 = 2\hat{i} + 1\hat{j} + 2\hat{k}$v1​=2i^+1j^​+2k^, and for the second line, it is $\vec{v}_2 = 6\hat{i} - 3\hat{j} + 2\hat{k}$v2​=6i^−3j^​+2k^.

The cosine of the angle $\theta$θ is given by the formula $\cos \theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}$cosθ=∣v1​∣∣v2​∣∣v1​⋅v2​∣​.

Calculating the dot product gives $(2)(6) + (1)(-3) + (2)(2) = 12 - 3 + 4 = 13$(2)(6)+(1)(−3)+(2)(2)=12−3+4=13.

The magnitude of the first vector is $\sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3$22+12+22​=9​=3.

The magnitude of the second vector is $\sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$62+(−3)2+22​=36+9+4​=49​=7.

Substituting these into the formula, we get $\cos \theta = \frac{13}{3 \times 7} = \frac{13}{21}$cosθ=3×713​=2113​