Let $\vec{b}_1$ and $\vec{b}_2$ be the vector parallel to the pair to lines, $\frac{-x+2}{-2}$ = $\frac{y-1}{7}$ = $\frac{z+3}{-3}$ and $\frac{x+2}{-1}$ = $\frac{2y-8}{4}$ = $\frac{z-5}{4}$ , respectively. Now, $\frac{-x+2}{-2}$ = $\frac{y-1}{7}$ = $\frac{z+3}{-3}$ ⇒ $\frac{x-2}{2}$ = $\frac{y-1}{7}$ = $\frac{z+3}{-3}$ $\frac{x+2}{-1}$ = $\frac{2y-8}{4}$ = $\frac{z-5}{4}$ ⇒ $\frac{x+2}{-1}$ = $\frac{y-4}{2}$ = $\frac{z-5}{4}$ ∴ $\vec{b}_1$ = $2\hat{i}+7\hat{j}-3\hat{k}$ and $\vec{b}_2$ = $-\hat{i}+2\hat{j}+4\hat{k}$ $\left|\vec{b}_1\right|$ = $\sqrt{(2)^2+(7)^2+(-3)^2}$ = $\sqrt{62}$ $\left|\vec{b}_2\right|$ = $\sqrt{(-1)^2+(2)^2+(4)^2}$ = $\sqrt{21}$ $\vec{b}_1 \cdot \vec{b}_2$ = = 2 (- 1) + 7 × 2 + (- 3) . 4 = - 2 + 14 - 12 = 0 The angle θ between the given pair of lines is given by the relation, cos θ = $\left|\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}\right|$ ⇒ cos θ = $\frac{0}{\sqrt{62} \times \sqrt{21}}$ = 0 ⇒ θ = $\cos^{-1}$ (0) = $\frac{\pi}{2}$ Thus, the given lines are perpendicular to each other and the angle between them is 90°.