JEE Mains · Maths · STD 12 - 10. vector algebra
Let for a triangle \(ABC\), \(\overline{A B}=-2 \hat{i}+\hat{j}+3 \hat{k}\) \(\overline{C B}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\) \(\overline{C A}=4 \hat{i}+3 \hat{j}+\delta \hat{k}\) If \(\delta > 0\) and the area of the triangle \(ABC\) is \(5 \sqrt{6}\), then \(\overline{C B} \cdot \overline{C A}\) is equal to
- A \(60\)
- B \(120\)
- C \(108\)
- D \(54\)
Answer & Solution
Correct Answer
(A) \(60\)
Step-by-step Solution
Detailed explanation
Sol. \(\overline{ AB }+\overline{ BC }+\overline{ CA }=\overrightarrow{0}\) \(\alpha= 2 , \beta= 4 , \gamma-\delta= 3\) \(\frac{1}{2}|\overline{ AB } \times \overline{ AC }|=5 \sqrt{6}\) \((\delta-9)^2+(2 \delta+12)^2+100=600\) \(\Rightarrow \delta=5, \gamma=8\) Hence…
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