If \(\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+4…

If \(\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\), \(\mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) then, \(\left[\begin{array}{lll}\mathbf{a} \times \mathbf{b} & \mathbf{b} \times \mathbf{c} \quad \mathbf{c} \times \mathbf{a}\end{array}\right]=\)
  1. 4900
  2. 6400
  3. 8100
  4. 12100

Solution

Given vector \(\begin{aligned} & \mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & \mathbf{b}=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \text { and } \mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned}\) and as we know that \(\left[\begin{array}{lll}\mathbf{a} \times \mathbf{b} & \mathbf{b} \times \mathbf{c} & \mathbf{c} \times \mathbf{a}\end{array}\right]=\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]^2\) and \(\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]=\left|\begin{array}{ccc}2 & -3 & -4 \\ 1 & 4 & -2 \\ 3 & -1 & 4\end{array}\right|\) \(\begin{aligned} & =2(16-2)+3(4+6)-4(-1-12) \\ & =28+30+52=110 \end{aligned}\) \(\therefore\left[\begin{array}{lll}\mathbf{a} \times \mathbf{b} & \mathbf{b} \times \mathbf{c} & \mathbf{c} \times \mathbf{a}\end{array}\right]=\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]^2=(110)^2=12100\) Hence, option (d) is correct.

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

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