Chapter 1: Vectors, Lines and Planes

Section 1.5: Applications of Vector Products

Example 1.5.7

Derive the formula given in Table 1.5.1 for the distance from a point to a line. Solution •  Figure 1.5.7(a) is obtained from Figure 1.5.6(a) by adding the angle $\mathrm{\θ}$ between vectors A and B, and by adding the dotted red line from point P orthogonal to $\overline{\mathrm{QR}}$, the line through Q and R. The length of this dotted line is the distance from P to $\overline{\mathrm{QR}}$.   •  From Figure 1.5.7(a) and simple right-triangle trigonometry, $d\= ∥\mathbf{B}∥ \sin \left(\mathrm{\θ}\right)$. But   $\sin \left(\mathrm{\θ}\right)\=\frac{\parallel \mathbf{A}\times \mathbf{B}\parallel }{∥\mathbf{A}∥ ∥\mathbf{B}∥}$     so $d\=∥\mathbf{B}∥$$\frac{\parallel \mathbf{A}\times \mathbf{B}\parallel }{∥\mathbf{A}∥ ∥\mathbf{B}∥}$ = $\frac{∥\mathbf{A}\times \mathbf{B}∥}{\∥\mathbf{A}\∥}$  Figure 1.5.7(a)   Distance from P to the line $\overline{\mathrm{QR}}$

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