Let $A=(\alpha, 1,2 \alpha), B=(3,1,2)$ and $\mathbf{C}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$. If $\mathbf{A B} imes \mathbf{C}=6 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$, then $\alpha^2+\alpha+5=$

Question

Let $A=(\alpha, 1,2 \alpha), B=(3,1,2)$ and $\mathbf{C}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$. If $\mathbf{A B} 	imes \mathbf{C}=6 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$, then $\alpha^2+\alpha+5=$

Aicharya · Accepted Answer

Given, $ \begin{aligned} & A=(\alpha, 1,2 \alpha) \ & B=(3,1,2) \ & \mathbf{C}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $ and Now, $\quad \mathbf{A B}=(3-\alpha) \hat{\mathbf{i}}+(2-2 \alpha) \hat{\mathbf{k}}$ $ \begin{aligned} 	herefore & \mathbf{A B} 	imes \mathbf{C}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 3-\alpha & 0 & 2-2 \alpha \ 4 & -1 & 3 \end{array}ight| \ & =\hat{\mathbf{i}}(2-2 \alpha)-\hat{\mathbf{j}}(9-3 \alpha-8+8 \alpha)+\hat{\mathbf{k}}(-3+\alpha) \ & =(2-2 \alpha) \hat{\mathbf{i}}+(5 \alpha-1) \hat{\mathbf{j}}+(\alpha-3) \hat{\mathbf{k}} \end{aligned} $ Comparing it with $6 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$, then $ \begin{aligned} & 2-2 \alpha=6,5 \alpha-1=9 	ext { and } \alpha-3=-5 \ & 2 \alpha=-4 \ & \Rightarrow \quad \alpha=-2 \ & 	herefore \quad \alpha^2+\alpha+5=(-2)^2-2+5 \ & =4-2+5=7 \ & \end{aligned} $

Let $A=(\alpha, 1,2 \alpha), B=(3,1,2)$ and $\mathbf{C}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3…

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