Given, \(\triangle A B C\) such that \(A\) is \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, B\) is…

Given, \(\triangle A B C\) such that \(A\) is \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, B\) is \(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\) and \(C\) is \(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\), then \(\triangle A B C\) is
  1. An equilateral triangle
  2. A right-angled triangle
  3. An isosceles triangle
  4. A scalene triangle

Solution

Vertices of \(\triangle A B C\) are given as \(A\) is \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) \(B\) is \(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\) and \(C\) is \(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\) \(\therefore \quad \mathbf{A B}=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}, \mathbf{B C}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\quad \mathbf{C A}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\) \(\begin{gathered} \because \quad|\mathbf{A B}|=\sqrt{1+4+36}=\sqrt{41} \\ |\mathbf{B C}|=\sqrt{4+1+1}=\sqrt{6} \end{gathered}\) and \(|\mathbf{C A}|=\sqrt{1+9+25}=\sqrt{35}\) \(\because \quad|\mathbf{A B}|^2=|\mathbf{B C}|^2+|\mathbf{C A}|^2\) \(\therefore \triangle A B C\) is an right-angled triangle. Hence, option (b) is correct.

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

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