Question

Let $ \bar{a}, \bar{b}, \bar{c} $ be three non-coplanar vectors. Then the point of intersection of the line joining the points $ \mathbf{a} + \bar{b} + \bar{c} $, $ \bar{a} - \bar{b} + 3\bar{c} $ and the line joining the points $ 2\bar{a} - \bar{b} + \bar{c} $, $ \bar{a} - 2\bar{b} + 4\bar{c} $ is

• A. $ 2\bar{a} + 4\bar{c} $
• B. $ 3\bar{a} - 3\bar{b} + 5\bar{c} $
• C. $ \bar{a} - 2\bar{b} + 4\bar{c} $
• D. $ \bar{a} - \bar{b} + 3\bar{c} $

Hint:

Represent the lines in vector form using a parameter. The intersection point occurs when the position vectors of the two lines are equal for some values of their parameters. Use the non-coplanarity of the vectors to equate coefficients.

Answer 1

The Correct Option is C

Step 1: Vector equation of the first line.
$ \mathbf{r}_1 = \bar{a} + (1 - 2\lambda)\bar{b} + (1 + 2\lambda)\bar{c} $ 
Step 2: Vector equation of the second line.
$ \bar{r}_2 = (2 - \mu)\bar{a} + (-1 - \mu)\bar{b} + (1 + 3\mu)\bar{c} $ 
Step 3: Equate $ \bar{r_1 $ and $ \bar{r}_2 $ and equate coefficients.}
$ 1 = 2 - \mu \implies \mu = 1 $ $ 1 - 2\lambda = -1 - \mu \implies 1 - 2\lambda = -2 \implies \lambda = \frac{3}{2} $ $ 1 + 2\lambda = 1 + 3\mu \implies 1 + 3 = 1 + 3 $ (consistent) 
Step 4: Substitute $ \lambda $ or $ \mu $ to find the intersection point.
Using $ \lambda = \frac{3}{2} $ in $ \mathbf{r}_1 $: $ \mathbf{a} + (1 - 3)\bar{b} + (1 + 3)\bar{c} = \bar{a} - 2\bar{b} + 4\bar{c} $ Using $ \mu = 1 $ in $ \bar{r}_2 $: $ (2 - 1)\bar{a} + (-1 - 1)\mathbf{b} + (1 + 3)\bar{c} = \bar{a} - 2\bar{b} + 4\bar{c} $ 
Step 5: Conclusion.
The point of intersection is $ \bar{a} - 2\bar{b} + 4\bar{c} $.