### Related Formula cos theta = fracveca cdot vecb|veca||vecb|$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ For an acute angle, cos theta > 0$\cos \theta > 0$, which strictly means veca cdot vecb > 0$\vec{a} \cdot \vec{b} > 0$. ### Core Logic Given vectors vecA = alphahati - 2hatj + 2hatk$\vec{A} = \alpha\hat{i} - 2\hat{j} + 2\hat{k}$ and vecB = alphahati + 2alphahatj - 2hatk$\vec{B} = \alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$. For the angle to be acute, their dot product must be strictly positive: vecA cdot vecB > 0$\vec{A} \cdot \vec{B} > 0$ (alpha)(alpha) + (-2)(2alpha) + (2)(-2) > 0$(\alpha)(\alpha) + (-2)(2\alpha) + (2)(-2) > 0$ alpha^2 - 4alpha - 4 > 0$\alpha^2 - 4\alpha - 4 > 0$ ### Step 1: Solving the Inequality Complete the square to find critical points: alpha^2 - 4alpha + 4 > 8$\alpha^2 - 4\alpha + 4 > 8$ (alpha - 2)^2 > 8$(\alpha - 2)^2 > 8$ Extracting roots gives: alpha - 2 > 2sqrt2 quad textor quad alpha - 2 < -2sqrt2$\alpha - 2 > 2\sqrt{2} \quad \text{or} \quad \alpha - 2 < -2\sqrt{2}$ alpha > 2 + 2sqrt2 quad textor quad alpha < 2 - 2sqrt2$\alpha > 2 + 2\sqrt{2} \quad \text{or} \quad \alpha < 2 - 2\sqrt{2}$ ### Step 2: Finding Least Positive Integer Approximate the boundary values. Since sqrt2 approx 1.414$\sqrt{2} \approx 1.414$: 2 + 2(1.414) = 4.828$2 + 2(1.414) = 4.828$ The ranges are alpha in (-infty, -0.828) cup (4.828, infty)$\alpha \in (-\infty, -0.828) \cup (4.828, \infty)$. We need the *least positive integral value* of alpha$\alpha$. Looking at the interval (4.828, infty)$(4.828, \infty)$, the smallest integer present is 5$5$. ### Pattern Recognition Acute angle translates directly to a positive dot product. Set up the quadratic inequality, compute numerical bounds of irrational roots, and select the immediate next integer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra Class 11 Maths: Linear Inequalities