Question

If the equation of a line through a point a and parallel to vector b is $\mathbf{r} = \mathbf{a} + t\mathbf{b},$ where t is a parameter, then its perpendicular distance from the point c is

• A. $|(\mathbf{c} - \mathbf{b}) \times \mathbf{a}| \div |\mathbf{a}|$
• B. $|(\mathbf{c} - \mathbf{a}) \times \mathbf{b}| \div |\mathbf{b}|$
• C. $|(\mathbf{a} - \mathbf{b}) \times \mathbf{c}| \div |\mathbf{c}|$
• D. $|(\mathbf{a} - \mathbf{b}) \times \mathbf{c}| \div |\mathbf{a} + \mathbf{c}|$

Answer 1

Answer: (B)

$|(\mathbf{c} - \mathbf{a}) \times \mathbf{b}| \div |\mathbf{b}|$

Answer 2

For point $P$ on the line $r = \mathbf{a} + t\mathbf{b}$

$\therefore\overset{\rightarrow}{PC} = (\mathbf{c} - \mathbf{a}) - t\mathbf{b}$, $\because\overset{\rightarrow}{PC}\bot\mathbf{b}$

$\therefore|(\mathbf{c} - \mathbf{a}) - t\mathbf{b}|.\mathbf{b} = 0$ or $t = \frac{(\mathbf{c} - \mathbf{a}).\mathbf{b}}{\mathbf{b}^{2}}$ …..(i)

Distance of $\mathbf{c}$ from line $|\overset{\rightarrow}{PC}|\mspace{6mu} = d = |\mathbf{c} - \mathbf{a} - t\mathbf{b}|$

$d = \left| \mathbf{c} - \mathbf{a} - \frac{(\mathbf{c} - \mathbf{a}).\mathbf{bb}}{\mathbf{b}^{2}} \right| = \left| \frac{(\mathbf{c} - \mathbf{a})\mathbf{b}.\mathbf{b} - (\mathbf{c} - \mathbf{a}).\mathbf{bb}}{\mathbf{b}^{2}} \right|$

$d = \left| \frac{\mathbf{b} \times (\mathbf{c} - \mathbf{a}) \times \mathbf{b}}{\mathbf{b}^{2}} \right| = \frac{|\mathbf{b}||(\mathbf{c} - \mathbf{a}) \times \mathbf{b}|\sin 90{^\circ}}{|\mathbf{b}|^{2}}$,

$(\because\mathbf{b}\bot(\mathbf{c} - \mathbf{a}) \times \mathbf{b})$

$d = \frac{|(\mathbf{c} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|}$.