AP EAMCET · Maths · Vector Algebra
If the position vectors of the vertices \(A, B\) and \(C\) of \(\triangle A B C\) are \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\) respectively, then \(\angle B=\)
- A \(\cos ^{-1}\left(\frac{7}{3 \sqrt{10}}\right)\)
- B \(\cos ^{-1}\left(\frac{8}{\sqrt{105}}\right)\)
- C \(\cos ^{-1}\left(\frac{1}{\sqrt{42}}\right)\)
- D \(\cos ^{-1}\left(-\frac{7}{3 \sqrt{10}}\right)\)
Answer & Solution
Correct Answer
(B) \(\cos ^{-1}\left(\frac{8}{\sqrt{105}}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \because \mathbf{B A}=+3 \hat{\mathbf{i}}-6 \hat{\mathbf{k}} \text { and } \mathbf{B C}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \begin{array}{l} \therefore \cos (\angle B)=\frac{|\mathbf{B A} \cdot \mathbf{B}|}{|\mathbf{B A}| \mid \mathbf{B…
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