Let \veca=2 \hati+\alpha \hatj+\hatk, \vecb=-\hati+\hatk, \vecc=\beta \hatj-\hatk , 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 \veca+\vecb and \vecb+\vecc is \frac\sqrt212 , 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 PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q50 with Solution

View written solutionFree

Correct answer: C

$\vec a=(2,\alpha,1),\quad \vec b=(-1,0,1),\quad \vec c=(0,\beta,-1)$

We are given:

$\alpha\beta=-6$

The diagonals are:

$\vec d_1=\vec a+\vec b=(2-1,\alpha+0,1+1)=(1,\alpha,2)$

$\vec d_2=\vec b+\vec c=(-1+0,0+\beta,1-1)=(-1,\beta,0)$

If a parallelogram has diagonals $\vec d_1$ and $\vec d_2$ , then its area is

$\frac{1}{2}\,|\vec d_1\times \vec d_2|$

Given area is $\dfrac{\sqrt{21}}{2}$ , so

$|\vec d_1\times \vec d_2|=\sqrt{21}$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & \alpha & 2\\ -1 & \beta & 0 \end{vmatrix}$$ $$=\hat i(\alpha\cdot 0-2\beta)-\hat j(1\cdot 0-2(-1))+\hat k(1\cdot \beta-\alpha(-1))$$ $$=(-2\beta)\hat i-2\hat j+(\alpha+\beta)\hat k$$ Hence, $$|\vec d_1\times \vec d_2|^2=4\beta^2+4+(\alpha+\beta)^2$$ Since $|\vec d_1\times \vec d_2|=\sqrt{21}$, $$4\beta^2+4+(\alpha+\beta)^2=21$$ $$4\beta^2+(\alpha+\beta)^2=17$$ 5. **Use the condition $\alpha\beta=-6$** Possible integer pairs $(\alpha,\beta)$ are: $$(1,-6),\ (-1,6),\ (2,-3),\ (-2,3),\ (3,-2),\ (-3,2),\ (6,-1),\ (-6,1)$$ Now test in $$4\beta^2+(\alpha+\beta)^2=17$$ - For $(2,-3)$: $$4(9)+(2-3)^2=36+1=37 \neq 17$$ - For $(-2,3)$: $$4(9)+(-2+3)^2=36+1=37 \neq 17$$ - For $(3,-2)$: $$4(4)+(3-2)^2=16+1=17$$ ✓ - For $(-3,2)$: $$4(4)+(-3+2)^2=16+1=17$$ ✓ - For $(6,-1)$: $$4(1)+(6-1)^2=4+25=29 \neq 17$$ - For $(-6,1)$: $$4(1)+(-6+1)^2=4+25=29 \neq 17$$ - For $(1,-6)$ and $(-1,6)$ give much larger values. Thus the two ordered pairs are: $$ (\alpha_1,\beta_1)=(3,-2),\quad (\alpha_2,\beta_2)=(-3,2) $$ 6. **Compute the required expression** $$\alpha_1^2+\beta_1^2-\alpha_2\beta_2=3^2+(-2)^2-((-3)(2))$$ $$=9+4-(-6)=19$$ 7. **Final answer** $$\boxed{19}$$ So the correct option is **C**.