A line passing through the point A with position vector vector(a) = (4i + 2j + 2k) is parallel to the vector

Question

A line passing through the point A with position vector \(\vec{a}=4\hat{i}+2\hat{j}+2\hat{k}\) is parallel to the vector \(\vec{b}=2\hat{i}+3\hat{j}+6\hat{k}.\) Find the length of the perpendicular drawn on this line from a point P with position vector \(\vec{r_1}=\hat{i}+2\hat{j}+3\hat{k}.\)

Answer 1

The equation of line passing through the point A and parallel to \(\vec{b}\) is given in cartesian form as

Let Q \((\alpha, \beta, \gamma)\) be foot of perpendicular drawn from point P to the line (i).

Co-ordinate or point P \(\equiv\) (1, 2, 3)

Since, Q lie on line (i)

\(\frac{x-4}{2}=\frac{y-2}{3}=\frac{z-2}{6}\)

Putting the value of \(\alpha, \beta, \gamma;\) we get

Hence, the co-ordinate of Q \(\equiv\) (4, 2, 2)

\(\therefore\) Length of perpendicular  Q \(\equiv\) (4, 2, 2)

\(=\sqrt{9+0+1}=\sqrt{10}\) units.