Vector Algebra
Parallelogram Diagonal Area — Finding $\alpha_1^2+\beta_1^2-\alpha_2\beta_2$
nta_pyq_2024_apr
Grade 12
Question:
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 $\vec{b}+\vec{c}$ is $\dfrac{\sqrt{21}}{2}$, be $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$. Then $\alpha_1^2+\beta_1^2-\alpha_2\beta_2$ is equal to
Step-by-Step Solution
Key Concept: Area $=\frac{1}{2}|(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})|=\dfrac{\sqrt{21}}{2}\Rightarrow|(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})|=\sqrt{21}$. $\vec{a}+\vec{b}=\hat{i}+\alpha\hat{j}+2\hat{k}$, $\vec{b}+\vec{c}=-\hat{i}+\beta\hat{j}+0\hat{k}$.
$(\alpha_1,\beta_1)=(-3,2)$, $(\alpha_2,\beta_2)=(3,-2)$. $\alpha_1^2+\beta_1^2-\alpha_2\beta_2=19$.
Correct Answer: 1