Related Formula
| a × b|² + ( a · b)² = | a|² | b|²$$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$
θ = b · c| b| | c|$$\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$$
Core Logic
Given | a| = 1, | b| = 4, a · b = 2$|\vec{a}| = 1, |\vec{b}| = 4, \vec{a} \cdot \vec{b} = 2$.
Let's evaluate | a × b|²$|\vec{a} \times \vec{b}|^2$ using Lagrange's identity:
| a × b|² = | a|² | b|² - ( a · b)²$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$
| a × b|² = (1)(16) - (2)² = 16 - 4 = 12$$|\vec{a} \times \vec{b}|^2 = (1)(16) - (2)^2 = 16 - 4 = 12$$
Step 1: Dot products with c
We are given:
c = 2( a × b) - 3 b$$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$
To find the angle between b$\vec{b}$ and c$\vec{c}$, we need b · c$\vec{b} \cdot \vec{c}$ and | c|$|\vec{c}|$.
Taking the dot product with b$\vec{b}$ on both sides:
b · c = 2( b · ( a × b)) - 3( b · b)$$\vec{b} \cdot \vec{c} = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3(\vec{b} \cdot \vec{b})$$
Since b · ( a × b) = 0$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ (scalar triple product with repeated vector):
b · c = 0 - 3| b|² = -3(16) = -48 (1)$$\vec{b} \cdot \vec{c} = 0 - 3|\vec{b}|^2 = -3(16) = -48 \quad \dots (1)$$
Step 2: Finding magnitude of c
Now, let's find | c|²$|\vec{c}|^2$:
| c|² = c · c = (2( a × b) - 3 b) · (2( a × b) - 3 b)$$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (2(\vec{a} \times \vec{b}) - 3\vec{b}) \cdot (2(\vec{a} \times \vec{b}) - 3\vec{b})$$
Since ( a × b) · b = 0$(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$, the cross terms vanish:
| c|² = 4| a × b|² + 9| b|²$$|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2$$
| c|² = 4(12) + 9(16) = 48 + 144 = 192$$|\vec{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$$
| c| = √(192) = 8√(3)$$|\vec{c}| = \sqrt{192} = 8\sqrt{3}$$
Step 3: Calculating angle
Now apply the angle formula:
θ = b · c| b| | c| = -48(4)(8√(3))$$\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} = \frac{-48}{(4)(8\sqrt{3})}$$
θ = -4832√(3) = -32√(3) = - √(3)2$$\cos \theta = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$$
θ = ⁻¹(- √(3)2)$$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$$
Pattern Recognition
Always remember that the cross product vector ( a × b)$(\vec{a} \times \vec{b})$ is orthogonal to both a$\vec{a}$ and b$\vec{b}$. This immediately eliminates cross terms when finding the magnitude of linear combinations like x( a × b) + y b$x(\vec{a} \times \vec{b}) + y\vec{b}$.
Chapter Mix
Class 12 Maths: Vector Algebra
