Let a =2 i +α j + k , b =- i + k , c =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 parallelogram of diagonals a + b and b + c is 21 2 , be (α_1

Correct answer: A. 19 aligned & Area of parallelogram .=(1)/(2) \, d_1 × d_2 & A=(1)/(2)|( a + b ) ×( b + c )|= 21 2 & so, a + b = i +α j +2 k & b + c =- i +β j & ( a + b ) ×( b + c )=| array ccc i & j & k 1 & α & 2 -1 & β & 0 array | aligned aligned & = i (-2 β)- j (2)+ k (β+α) & |( a + b ) ×( b + c )|= 4 β^2+4+(α+β)^2 = 21 & 4 β^2+4+α^2+β^2+2 α β=21 & α^2+5 β^2-12=17 & α^2+5 β^2=29 aligned and α β=-6 and given α_ i β are integers so, α=-3, β=2 or aligned & α=3, β=-2 & (α_1, β_1)=(-3,2) & (α_2, β_2)=(3,-2) & α_1^2+β_1^2-α_2 β_2=9+4+6=19 aligned

JEE Main 20249 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Vector Algebra

Let a =2 i +α j + k , b =- i + k , c =β j - k , where α and β are...

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}

\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

Options

Correct answer

A. 19

Step-by-step solution

\begin{aligned} & \text{ Area\text{ 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}

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