A) Right angled
B) Isosceles
C) Equilateral
D) Right angled isosceles
Correct Answer: D
Solution :
\[\mathbf{a}=2\mathbf{i}+\mathbf{j}+2\mathbf{k}\Rightarrow |a|=\sqrt{4+1+4}=\sqrt{9}\] \[\mathbf{b}=-\mathbf{i}+\mathbf{j}-4\mathbf{k}\Rightarrow |b|=\sqrt{1+1+16}=\sqrt{18}\] \[\mathbf{c}=-\mathbf{i}-2\mathbf{j}+2\mathbf{k}\Rightarrow |c|=\sqrt{1+4+4}=\sqrt{9}\] \[|\mathbf{a}|\,\,=|\mathbf{c}|\] and also, \[{{\mathbf{b}}^{2}}={{\mathbf{a}}^{2}}+{{\mathbf{c}}^{2}}\] Hence it is isosceles and right angled triangle.
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