Question

Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j}  + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.

Answer 1

The position vectors of two arbitrary points on the given lines are 

\[3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( i + 2 \hat{j} + 2 \hat{k} \right) = \left( 3 + \lambda \right) \hat{i} + \left( 2 + 2\lambda \right) \hat{j} + \left( 2\lambda - 4 \right) \hat{k}\]

\[5 \hat{i} - 2 \hat{j} + \mu\left( 3 \hat{i} + 2 \hat{j}+ 6 \hat{k} \right) = \left( 5 + 3\mu \right) \hat{i} + \left( - 2 + 2\mu \right) \hat{j} + 6\mu \hat{k}\]

If the lines intersect, then they have a common point. So, for some values of 

\[\lambda \text{ and } \mu\] we must have

\[\left( 3 + \lambda \right) \hat{i} + \left( 2 + 2\lambda \right) \hat{j} + \left( 2\lambda - 4 \right) \hat{k} = \left( 5 + 3\mu \right) \hat{i} + \left( - 2 + 2\mu \right) j^^ + 6\mu \hat{k} \]  Equating the coefficients of \[\hat{i} , \hat{j} \text{ and }  \hat{ k} \]

\[3 + \lambda = 5 + 3\mu . . . (1)\]

\[2 + 2\lambda = - 2 + 2\mu . . . (2) \]

\[2\lambda - 4 = 6\mu . . . (3)\]

Solving (1) and (2), we get 

\[\lambda = - 4\]

\[\mu = - 2\]

Substituting the values

\[\lambda = - 4 \text{ and }  \mu = - 2\]  in (3), we get

\[LHS = 2\lambda - 4\]

\[ = 2\left( - 4 \right) - 4\]

\[ = - 12\]

\[RHS = 6\mu\]

\[ = 6\left( - 2 \right)\]

\[ = - 12\]

\[ \Rightarrow LHS = RHS\]

\[\text{ Since } \lambda = - 4 \text{ and \mu = - 2 satisfy (3), the lines intersect }  .\]

\[\mu = - 2\]  in the second line, we get

\[\vec{r} = 5 \hat{i} - 2 \hat{j} - 6 \hat{i} - 4 \hat{j} - 12 \hat{k} = - \hat{i}- 6 \hat{j} - 12 \hat{k}\] as the position vector of the point of intersection.

Thus, the coordinates of the point of intersection are (-1,-6,-12).