If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non coplanar vectors, then the point of intersection of the line…

1. $\mathbf{a}+\mathbf{b}+\mathbf{c}$
2. $\mathbf{a}+2 \mathbf{b}$
3. $\mathbf{a}+\mathbf{c}$
4. $\frac{\mathbf{a}+2 \mathbf{b}+\mathbf{c}}{2}$

Solution

Let, $\mathbf{O A}=2 \mathbf{a}+3 \mathbf{b}-\mathbf{c}, \mathbf{O B}=3 \mathbf{a}+4 \mathbf{b}-2 \mathbf{c}$ $ \mathbf{O C}=\mathbf{a}-2 \mathbf{b}+3 \mathbf{c} \text {, and } \mathbf{O D}=\mathbf{a}-6 \mathbf{b}+6 \mathbf{c} $ The vector equation of line joining the points $A$ and $B$ is $ \begin{aligned} \mathbf{r} & =\mathbf{O A}+t(\mathbf{O B}-\mathbf{O A}), t \in R \\ & =(2 \mathbf{a}+3 \mathbf{b}-\mathbf{c})+t(3 \mathbf{a}+4 \mathbf{b}-2 \mathbf{c})-(2 \mathbf{a}+3 \mathbf{b}-\mathbf{c})] \end{aligned} $
Vector equation of the line joining the points $C$ and $D$ is $ \begin{aligned} \mathbf{r} & =\mathbf{O C}+s(\mathbf{O D}-\mathbf{O C}) \\ & =\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}+s(\mathbf{a}-6 \mathbf{b}+6 \mathbf{c})-(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) \end{aligned} $
Comparing coefficient of a in Eq. (ii) and (iii) we get $2+t=1 \Rightarrow t=-1$ Put in Eq. (i) $ \begin{aligned} \mathbf{r} & =2 \mathbf{a}+3 \mathbf{b}-\mathbf{c}+(-1)[\mathbf{a}+\mathbf{b}-\mathbf{c}] \\ & =2 \mathbf{a}+3 \mathbf{b}-\mathbf{c}-\mathbf{a}-\mathbf{b}+\mathbf{c} \\ \mathbf{r} & =\mathbf{a}+2 \mathbf{b} \end{aligned} $ So, the point of in intersection is $\mathbf{a}+2 \mathbf{b}$

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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