Verified Solution • 2025–26 Session moderate

QJEEM24Ma866b: JEE Main 2024 Physics - Class 11 JEEM24Ma866b Solution: Let , where and are integers and . Let the values of the…

Let a⃗=2i^+αj^+k^,b⃗=−i^+k^,c⃗=βj^−k^\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k} , where α\alpha and β\beta are integers and αβ=−6\alpha \beta=-6 . Let the values of the ordered pair (α,β)(\alpha, \beta) , for which the area of the parallelogram of diagonals a⃗+b⃗\vec{a}+\vec{b} and b⃗+c⃗\vec{b}+\vec{c} is 212\frac{\sqrt{21}}{2} , be (α1,β1)\left(\alpha_1, \beta_1\right) and (α2,β2)\left(\alpha_2, \beta_2\right) . Then α12+β12−α2β2\alpha_1^2+\beta_1^2-\alpha_2 \beta_2 is equal to

Multiple Choice Options

  • A 21
  • B 24
  • C 19
  • D 17

Step-by-Step Verified Solution

Area of parallelogram whose diagonals are a⃗+b⃗\vec{a}+\vec{b} and b⃗+c⃗\vec{b}+\vec{c} is

amp;=12∣(a⃗+b⃗)×(b⃗+c⃗)∣amp;=12∣a⃗×b⃗+a⃗×c⃗+b⃗×c⃗∣amp;=12∣−2βi^−2j^+(α+β)k^∣\begin{aligned}& =\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\& =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}| \\& =\frac{1}{2}|-2 \beta \hat{i}-2 \hat{j}+(\alpha+\beta) \hat{k}|\end{aligned}

=124β2+4+(α+β)2=\frac{1}{2} \sqrt{4 \beta^2+4+(\alpha+\beta)^2}

Which is given 212\frac{\sqrt{21}}{2}

∴amp;4β2+4+(α+β)2=21⇒amp;(α+β)2+4β2=17⇒amp;α2+5β2+2αβ=17⇒amp;α2+5β2=29∴amp;(α,β)∈{(3,2),(−3,−2),(−3,2),(3,−2)}∵amp;αβ=−6∴amp;(α,β)∈{(−3,2),(3,−2)}∴amp;α12+β12−α2β2=amp;9+4−(−6)=19\begin{array}{ll}\therefore & 4 \beta^2+4+(\alpha+\beta)^2=21 \\\Rightarrow & (\alpha+\beta)^2+4 \beta^2=17 \\\Rightarrow & \alpha^2+5 \beta^2+2 \alpha \beta=17 \\\Rightarrow & \alpha^2+5 \beta^2=29 \\\therefore & (\alpha, \beta) \in\{(3,2),(-3,-2),(-3,2),(3,-2)\} \\\because & \alpha \beta=-6 \\\therefore & (\alpha, \beta) \in\{(-3,2),(3,-2)\} \\\therefore & \alpha_1^2+\beta_1^2-\alpha_2 \beta_2 \\= & 9+4-(-6)=19\end{array}

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