If $\bar{a}, \bar{b}, \bar{c}$ are the position vectors of the points $\mathrm{A}(1,3,0), \mathrm{B}(2,5,0),…
If $\bar{a}, \bar{b}, \bar{c}$ are the position vectors of the points $\mathrm{A}(1,3,0), \mathrm{B}(2,5,0), \mathrm{C}(4,2,0)$
respectively and $\bar{c}=t_{1} \bar{a}+t_{2} \bar{b}$, then value of $t_{1} t_{2}=$
$-16$
16
160
$-160$
Solution
From given conditions, wo war $\begin{aligned} 4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}} &=\mathrm{t}_{1}(\hat{\mathrm{i}}+3 \hat{\mathrm{j}})+\mathrm{t}_{2}(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}) \\ \therefore \quad &=\left(\mathrm{t}_{1}+2 \mathrm{t}_{2}\right) \hat{\mathrm{i}}+\left(3 \mathrm{t}_{1}+5 \mathrm{t}_{2}\right) \hat{\mathrm{j}} \end{aligned}$ $\therefore \mathrm{t}_{1}+2 \mathrm{t}_{2}=4 \quad$ and $\quad 3 \mathrm{t}_{1}+5 \mathrm{t}_{2}=2$
Solving, we get $t_{2}=10$ and $t_{1}=-16$ $\therefore \mathrm{t}_{1} \mathrm{t}_{2}=-160$