Question

Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.

Answer 1

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

\[\left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) = \left( 1 + 3\lambda \right) \hat{i} + \left( 1 - \lambda \right) \hat{j} - \hat{k} \]

\[\left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right) = \left( 4 + 2\mu \right) \hat{i} + 0 \hat{j } + \left( 3\mu - 1 \right) \hat{k} \]

If the lines intersect, then they have a common point. So, for some values of \[\lambda \text{ and } \mu\] ,

we must have

\[\left( 1 + 3\lambda \right) \hat{i} + \left( 1 - \lambda \right) \hat{j} - \hat{k} = \left( 4 + 2\mu \right) \hat{i}+ 0 \hat{j} + \left( 3\mu - 1 \right) \hat{k}\]

Equating the coefficients of \[\hat{i}  , \hat{j}  \text{ and }  \hat{k} \]

we get

\[1 + 3\lambda = 4 + 2\mu . . . (1)\]

\[1 - \lambda = 0 . . . (2)\]

\[3\mu - 1 = - 1 . . . (3)\]

Solving (2) and (3), we get

\[\lambda = 1 \]

\[\mu = 0\]

Substituting the values  \[\lambda = 1 \text{ and }  \mu = 0\]

we get ,

\[LHS = 1 + 3\lambda\]

\[ = 1 + 3\left( 1 \right)\]

\[ = 4\]

\[RHS = 4 + 2\mu\]

\[ = 4 + 2\left( 0 \right)\]

\[ = 4\]

\[ \Rightarrow LHS = RHS\]

\[\text{ Since } \lambda = 1 \text{ and  } \mu = 0 \text{ satisfy (3), the given lines intersect }  .\]  Substituting \[\mu = 0\] in the second line, we get \[\vec{r} = 4 \hat{i} + 0 \hat{j} - \hat{k} \] as the position vector of the point of intersection.
Thus, the coordinates of the point of intersection are (4, 0,-1) .