Area of parallelogram =21d1×d2 A=21∣(a+b)×(b+c)∣=221 so, a+b=i^+αj^+2k^b+c=−i^+βj^(a+b)×(b+c)=i^1−1j^αβk^20 $\begin{aligned}
& =\hat{\mathrm{i}}(-2 \beta)-\hat{\mathrm{j}}(2)+\hat{\mathrm{k}}(\beta+\alpha) \
& |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\sqrt{4 \beta^2+4+(\alpha+\beta)^2}=\sqrt{21} \
& 4 \beta^2+4+\alpha^2+\beta^2+2 \alpha \beta=21 \
& \alpha^2+5 \beta^2-12=17 \
& \alpha^2+5 \beta^2=29
\end{aligned}and\alpha \beta=-6andgiven\alpha_{\mathrm{i}} \betaareintegersso,\alpha=-3, \beta=2or\begin{aligned}
& \alpha=3, \beta=-2 \
& \left(\alpha_1, \beta_1\right)=(-3,2) \
& \left(\alpha_2, \beta_2\right)=(3,-2) \
& \alpha_1^2+\beta_1^2-\alpha_2 \beta_2=9+4+6=19
\end{aligned}$
Let a=2 i+α j+ k, b=- i+ k, =β j- k, where α and β are integers and α β=-6. Let the values of the ordered pair (α, β), for which the area of the…
JEE Main 2024 — Mathematics Vectors & 3D Geometry
Let a=2i^+αj^+k^,b=−i^+k^,c=βj^−k^, where α and β are integers and αβ=−6. Let the values of the ordered pair (α,β), for which the area of the parallelogram of diagonals a+b and b+c is 221, be (α1,β1) and (α2,β2). Then α12+β12−α2β2 is equal to
Held on 9 Apr 2024 · Verified 6 Jul 2026.
Options
- A
19
- B
17
- C
24
- D
21
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