JEE Main 2024 — Mathematics Vectors & 3D Geometry

$\begin{aligned} & \text { Area of parallelogram } \left.=\frac{1}{2} \right\rvert\, \overrightarrow{\mathrm{d}}_1 \times \overrightarrow{\mathrm{d}}_2 \\ & \mathrm{~A}=\frac{1}{2}|(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\frac{\sqrt{21}}{2} \\ & \text { so, } \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+\beta \hat{\mathrm{j}} \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & \alpha & 2 \\ -1 & \beta & 0\end{array}\right|\end{aligned}$​ Area of parallelogram =21​​d1​×d2​ A=21​∣(a+b)×(b+c)∣=221​​ so, a+b=i^+αj^​+2k^b+c=−i^+βj^​(a+b)×(b+c)=​i^1−1​j^​αβ​k^20​​​ $\begin{aligned}
& =\hat{\mathrm{i}}(-2 \beta)-\hat{\mathrm{j}}(2)+\hat{\mathrm{k}}(\beta+\alpha) \
& |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\sqrt{4 \beta^2+4+(\alpha+\beta)^2}=\sqrt{21} \
& 4 \beta^2+4+\alpha^2+\beta^2+2 \alpha \beta=21 \
& \alpha^2+5 \beta^2-12=17 \
& \alpha^2+5 \beta^2=29
\end{aligned}$and$and\alpha \beta=-6$and given$andgiven\alpha_{\mathrm{i}} \beta$are integers so,$areintegersso,\alpha=-3, \beta=2$or$or\begin{aligned}
& \alpha=3, \beta=-2 \
& \left(\alpha_1, \beta_1\right)=(-3,2) \
& \left(\alpha_2, \beta_2\right)=(3,-2) \
& \alpha_1^2+\beta_1^2-\alpha_2 \beta_2=9+4+6=19
\end{aligned}$