Related Formula
a· b = | a|| b| θ$$\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}| \cos \theta$$
| a × b|² = | a|² | b|² ² θ$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$$
| a|² = aₓ² + ay² + az²$$|\vec{a}|^2 = a_x^2 + a_y^2 + a_z^2$$
Core Logic
Given | b|² = 6$|\vec{b}|^2 = 6$, angle θ = (π)/(4)$\theta = \frac{\pi}{4}$, and a· b = 3√(2)$\vec{a}\cdot\vec{b} = 3\sqrt{2}$.
| a|| b| θ = 3√(2)$$|\vec{a}||\vec{b}| \cos \theta = 3\sqrt{2}$$
Squaring both sides:
| a|² | b|² ² θ = 18$$|\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 18$$
| a|² (6) ( 1√(2))² = 18$$|\vec{a}|^2 (6) \left(\frac{1}{\sqrt{2}}\right)^2 = 18$$
| a|² (6) ((1)/(2)) = 18 ⇒ 3| a|² = 18 ⇒ | a|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 a = i + α j + β k$\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$.
| a|² = 1² + α² + β²$$|\vec{a}|^2 = 1^2 + \alpha^2 + \beta^2$$
1 + α² + β² = 6 ⇒ α² + β² = 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 (α² + β²)| a × b|²$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$.
| a × b|² = | a|² | b|² ² θ$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$$
| a × b|² = (6) (6) ²((π)/(4)) = 36 ((1)/(2)) = 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:
(α² + β²)| a × b|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. | a× b|² + ( a· b)² = | a|²| b|²$|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$ (Lagrange's Identity) skips ²θ$\sin^2\theta$ calculations completely.
Chapter Mix
Class 12 Maths: Vector Algebra
