WBJEE · Physics · Mathematics in Physics
In a triangle \(A B C\), the sides \(A B\) and \(A C\) are represented by the vectors \(3 \mathrm{i}+\mathrm{j}+\mathrm{k}\) and \(\mathrm{i}+2 \mathrm{j}+\mathrm{k}\), respectively. Calculate the angle \(\angle A B C\)
- A \(\cos ^{-1} \sqrt{\frac{5}{11}}\)
- B \(\cos ^{-1} \sqrt{\frac{6}{11}}\)
- C \(\left(90^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)\)
- D \(\left(180^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)\)
Answer & Solution
Correct Answer
(A) \(\cos ^{-1} \sqrt{\frac{5}{11}}\)
Step-by-step Solution
Detailed explanation
\(A B=3 i+j+\hat{k}\) and \(A C=i+2 j+i\) \(8 A=-(3 i+j+k)\) \(\angle A B C\) is angle between \(B A\) and \(B C\) \(B C=A C+B A\) \(=A C-A B\) \(=\hat{i}+2 \hat{j}+\hat{k}-(3 \hat{i}+\hat{j}+\hat{k})=-2 \hat{i}+\hat{j}\) \(B A \cdot B C=|B A \| B C| \cos \theta\)…
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