Mathematics · Ch 6 — Applications of Vector Algebra

Angle Between Two Straight Lines

6.7.4

Vector form. For two lines $\vec r=\vec a+s\vec b$ $r = a + s b$ and $\vec r=\vec c+t\vec d$ $r = c + t d$ , the acute angle $\theta$ $θ$ between them equals the acute angle between their direction vectors $\vec b,\vec d$ $b, d$ :

$\cos\theta=\left|\frac{\vec b\cdot\vec d}{|\vec b||\vec d|}\right|,\qquad\text{i.e.}\qquad \theta=\cos^{-1}\left(\left|\frac{\vec b\cdot\vec d}{|\vec b||\vec d|}\right|\right).$

(The absolute value keeps $\theta$ $θ$ the acute one, since a line has no preferred sense of direction.) Remark.
1. The two lines are parallel $\iff\theta=0\iff|\vec b\cdot\vec d|=|\vec b||\vec d|$ .
2. Equivalently, the lines are parallel $\iff \vec b=\lambda\vec d$ for some scalar $\lambda$ .
3. The lines are perpendicular $\iff\vec b\cdot\vec d=0$ .
Cartesian form. For lines $\frac{x-x_1}{b_1}=\frac{y-y_1}{b_2}=\frac{z-z_1}{b_3}$ and $\frac{x-x_2}{d_1}=\frac{y-y_2}{d_2}=\frac{z-z_2}{d_3}$ , the acute angle is $\theta=\cos^{-1}\left(\frac{|b_1d_1+b_2d_2+b_3d_3|}{\sqrt{b_1^2+b_2^2+b_3^2}\,\sqrt{d_1^2+d_2^2+d_3^2}}\right).$ Remark. …