MHT CET · Maths · Vector Algebra
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 \(t_{1} t_{2}=\)
- A \(-16\)
- B 16
- C 160
- D \(-160\)
Answer & Solution
Correct Answer
(D) \(-160\)
Step-by-step Solution
Detailed explanation
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 \(t_{1}=-16\) \(\therefore \mathrm{t}_{1} \mathrm{t}_{2}=-160\)
Solving, we get \(t_{2}=10\) and \(t_{1}=-16\) \(\therefore \mathrm{t}_{1} \mathrm{t}_{2}=-160\)
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