Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}$,…

Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}$, where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$, for which the area of the parallelogram of diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$, be $\left(\alpha_1, \beta_1\right)$ and $\left(\alpha_2, \beta_2\right)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to
  1. 19
  2. 17
  3. 24
  4. 21

Solution

$\begin{aligned} & \text { Area of parallelogram } \left.=\frac{1}{2} \right\rvert\, \overrightarrow{\mathrm{d}}_1 \times \overrightarrow{\mathrm{d}}_2 \\ & \mathrm{~A}=\frac{1}{2}|(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\frac{\sqrt{21}}{2} \\ & \text { so, } \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+\beta \hat{\mathrm{j}} \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & \alpha & 2 \\ -1 & \beta & 0\end{array}\right|\end{aligned}$ $\begin{aligned} & =\hat{\mathrm{i}}(-2 \beta)-\hat{\mathrm{j}}(2)+\hat{\mathrm{k}}(\beta+\alpha) \\ & |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\sqrt{4 \beta^2+4+(\alpha+\beta)^2}=\sqrt{21} \\ & 4 \beta^2+4+\alpha^2+\beta^2+2 \alpha \beta=21 \\ & \alpha^2+5 \beta^2-12=17 \\ & \alpha^2+5 \beta^2=29 \end{aligned}$ and $\alpha \beta=-6$ and given $\alpha_{\mathrm{i}} \beta$ are integers so, $\alpha=-3, \beta=2$ or $\begin{aligned} & \alpha=3, \beta=-2 \\ & \left(\alpha_1, \beta_1\right)=(-3,2) \\ & \left(\alpha_2, \beta_2\right)=(3,-2) \\ & \alpha_1^2+\beta_1^2-\alpha_2 \beta_2=9+4+6=19 \end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 2)

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