If \(\mathbf{a}=\alpha \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\) and \(\mathbf{b}=2…

If \(\mathbf{a}=\alpha \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\) and \(\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\beta \hat{\mathbf{k}}\), then the values of \(\alpha, \beta\) so that \(\mathbf{a}\) and \(\mathbf{b}\) may be collinear are
  1. \((5,3)\)
  2. \((6,2)\)
  3. \((2,-6)\)
  4. \((-6,2)\)

Solution

\(\mathbf{a}=\alpha \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\beta \hat{\mathbf{k}}\) For (a) and (b) may be collinear \((\mathbf{a} \times \mathbf{b})=0\) \(\begin{array}{lcc} & \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \alpha & 3 & -6 \\ 2 & -1 & \beta \end{array}\right|=0 \\ \Rightarrow \quad & \hat{\mathbf{i}}(3 \beta-6)-\hat{\mathbf{j}}(\alpha \beta+12)+\hat{\mathbf{k}}(-\alpha-6)=0 \\ \Rightarrow \quad(3 \beta-6)=0,(\alpha \beta+12)=0 \text { and }(\alpha+6)=0 \\ \Rightarrow \quad(\beta=2),(\alpha=-6) \end{array}\)

Asked in: AP EAMCET 2020 (17 Sep Shift 1)

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