### Related Formula vecacdotvecb = |veca||vecb| cos theta$\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}| \cos \theta$ |veca times vecb|^2 = |veca|^2 |vecb|^2 sin^2 theta$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$ |veca|^2 = a_x^2 + a_y^2 + a_z^2$|\vec{a}|^2 = a_x^2 + a_y^2 + a_z^2$ ### Core Logic Given |vecb|^2 = 6$|\vec{b}|^2 = 6$, angle theta = fracpi4$\theta = \frac{\pi}{4}$, and vecacdotvecb = 3sqrt2$\vec{a}\cdot\vec{b} = 3\sqrt{2}$. |veca||vecb| cos theta = 3sqrt2$|\vec{a}||\vec{b}| \cos \theta = 3\sqrt{2}$ Squaring both sides: |veca|^2 |vecb|^2 cos^2 theta = 18$|\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 18$ |veca|^2 (6) left(frac1sqrt2right)^2 = 18$|\vec{a}|^2 (6) \left(\frac{1}{\sqrt{2}}\right)^2 = 18$ |veca|^2 (6) left(frac12right) = 18 Rightarrow 3|veca|^2 = 18 Rightarrow |veca|^2 = 6$|\vec{a}|^2 (6) \left(\frac{1}{2}\right) = 18 \Rightarrow 3|\vec{a}|^2 = 18 \Rightarrow |\vec{a}|^2 = 6$ ### Step 1: Finding alpha and beta relation We have veca = hati + alpha hatj + beta hatk$\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$. |veca|^2 = 1^2 + alpha^2 + beta^2$|\vec{a}|^2 = 1^2 + \alpha^2 + \beta^2$ 1 + alpha^2 + beta^2 = 6 Rightarrow alpha^2 + beta^2 = 5$1 + \alpha^2 + \beta^2 = 6 \Rightarrow \alpha^2 + \beta^2 = 5$ ### Step 2: Evaluating the Target Expression We need to find the value of (alpha^2 + beta^2)|veca times vecb|^2$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$. |veca times vecb|^2 = |veca|^2 |vecb|^2 sin^2 theta$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$ |veca times vecb|^2 = (6) (6) sin^2left(fracpi4right) = 36 left(frac12right) = 18$|\vec{a} \times \vec{b}|^2 = (6) (6) \sin^2\left(\frac{\pi}{4}\right) = 36 \left(\frac{1}{2}\right) = 18$ Thus, the required value is: (alpha^2 + beta^2)|veca times vecb|^2 = (5)(18) = 90$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = (5)(18) = 90$ ### Pattern Recognition Vector magnitude and dot product give direct length values. |vecatimesvecb|^2 + (vecacdotvecb)^2 = |veca|^2|vecb|^2$|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$ (Lagrange's Identity) skips sin^2theta$\sin^2\theta$ calculations completely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra