Let \(\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \dot{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}\), 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 ( \(\alpha_{1}, \beta_{1}\) ) and ( \(\alpha_{2}, \beta_{2}\) ). Then \(\alpha_{1}^{2}+\beta_{1}^{2}-\alpha_{2} \beta_{2}\) is equal to

Vector Algebra — JEE Main · 2024 · 09 Apr 2024 · Shift 2 Question with Solution

Let

a = 2 \hat{i} + α \hat{j} + \hat{k}, \dot{b} = - \hat{i} + \hat{k}, c = β \hat{j} - \hat{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

\frac{21}{2}

, be (

α_{1}, β_{1}

) and (

α_{2}, β_{2}

).

Then

α_{1}^{2} + β_{1}^{2} - α_{2} β_{2}

is equal to

Answer: (D)

a + b = (2 - 1) \hat{i} + α \hat{j} + (1 + 1) \hat{k} = \hat{i} + α \hat{j} + 2 \hat{k}

;

b + c = - \hat{i} + β \hat{j} + 0 \hat{k}

(a + b) \times (b + c) = \hat{i} 1 - 1 \hat{j} α β \hat{k} 20 = - 2 β \hat{i} - 2 \hat{j} + (α + β) \hat{k}

Area

= \frac{1}{2} (a + b) \times (b + c) = \frac{21}{2}

4 β^{2} + 4 + (α + β)^{2} = 21

With

α β = - 6

(α + β)^{2} = α^{2} + β^{2} - 12

, so

4 β^{2} + α^{2} + β^{2} - 12 = 17 ⟹ α^{2} + 5 β^{2} = 29

Checking integer pairs with

α β = - 6

(3, - 2)

gives

9 + 20 = 29

✓;

(- 3, 2)

gives

9 + 20 = 29

✓ (no other factor pair of

6

works).

Taking

(α_{1}, β_{1}) = (3, - 2)

and

(α_{2}, β_{2}) = (- 3, 2)

α_{1}^{2} + β_{1}^{2} - α_{2} β_{2} = 9 + 4 - (- 6) = 19

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option