A) An equilateral triangle
B) Isosceles triangle
C) A right angled triangle
D) No triangle
Correct Answer: C
Solution :
\[\vec{A}=3\hat{i}-2\hat{j}+\hat{k}\], \[\vec{B}=\hat{i}-3\hat{j}+5\hat{k}\], \[\vec{C}=2\hat{i}-\hat{j}+4\hat{k}\] \[|\vec{A}|=\sqrt{{{3}^{2}}+{{(-2)}^{2}}+{{1}^{2}}}=\sqrt{9+4+1}=\sqrt{14}\] \[|\vec{B}|=\sqrt{{{1}^{2}}+{{(-3)}^{2}}+{{5}^{2}}}=\sqrt{1+9+25}=\sqrt{35}\] \[|\vec{A}|=\sqrt{{{2}^{2}}+{{1}^{2}}+{{(-4)}^{2}}}=\sqrt{4+1+16}=\sqrt{21}\] As \[B=\sqrt{{{A}^{2}}+{{C}^{2}}}\]therefore ABC will be right angled triangle.
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