A) 20
B) 10
C) 30
D) 40
Correct Answer: A
Solution :
Since, the coordinates of A, B and C are (1, 2, -1), (2, 0, 3) and (3, -1, 2), then \[\overrightarrow{AB}=(2-1)\,\hat{i}+(0-2)\hat{j}+(3+1)\hat{k}\] \[=\hat{i}-2\hat{j}+4\hat{k}\] and \[\overrightarrow{AC}=(3-1)\hat{i}+(-1-2)\hat{j}+(2+1)\hat{k}\] \[=2\hat{i}-3\hat{j}+3\hat{k}\] \[\cos \theta =\frac{(\hat{i}-2\hat{j}+4\hat{k})\cdot (2\hat{i}-3\hat{j}+3\hat{k})}{\sqrt{1+4+16}\,\sqrt{4+9+9}}\] \[=\frac{2+6+12}{\sqrt{21}\sqrt{22}}=\frac{20}{\sqrt{462}}\] \[\Rightarrow \] \[\sqrt{462}\cos \theta =20.\]
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