Q71 / 160 · TS EAMCET 2013 · Mathematics

The points whose position vectors are $2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k}, 3 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$ and $4 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$ are the vertices of

  • A
    an isosceles triangle
  • B
    right angled triangle
  • Correct answer
    C
    equilateral triangle
  • D
    right angled isosceles triangle

Solution

$\begin{aligned} & \text {Let } \mathbf{a}=2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k}=\mathrm{OA} \\ & \mathbf{b}=3 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}=\mathrm{OB} \\ & \text {and } \mathbf{c}=4 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k}=\mathrm{OC} \\ & A B=O B-O A=i+j-2 k \\ & B C=O C-O B=\mathbf{i}-2 \mathbf{j}+\mathbf{k} \\ & \text {and } \quad C A=O A-O C=-2 i+j+k \\ & \text {Now, } \quad A B=\sqrt{1+1+4}=\sqrt{6} \\ & \mathrm{BC}=\sqrt{1+4+1}=\sqrt{6} \\ & \text {and } \mathrm{CA}=\sqrt{4+1+1}=\sqrt{6} \\ & \end{aligned}$ Since, the length of all three sides are equal. So, the triangle is an equilateral triangle.

Hence, the correct answer is option: 'C'.

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