Related Formula
u × v = 0 u and v are parallel ($ u = λ v $)$$\vec{u} \times \vec{v} = 0 \implies \vec{u} \text{ and } \vec{v} \text{ are parallel ($ \vec{u} = \lambda\vec{v} $)}$$
| x + y|² = | x|² + | y|² + 2 x· y$$|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2\vec{x}\cdot\vec{y}$$
Core Logic
a × b - 2( a × c) = 0$$\vec{a} \times \vec{b} - 2(\vec{a} \times \vec{c}) = 0$$
a × ( b - 2 c) = 0$$\vec{a} \times (\vec{b} - 2\vec{c}) = 0$$
This implies that ( b - 2 c)$(\vec{b} - 2\vec{c})$ is collinear with a$\vec{a}$.
So, b - 2 c = λ a$\vec{b} - 2\vec{c} = \lambda \vec{a}$ for some scalar λ$\lambda$.
Step 1: Finding Lambda
Square both sides of the relation:
|λ a|² = | b - 2 c|²$$|\lambda \vec{a}|^2 = |\vec{b} - 2\vec{c}|^2$$
λ² | a|² = | b|² + 4| c|² - 4( b· c)$$\lambda^2 |\vec{a}|^2 = |\vec{b}|^2 + 4|\vec{c}|^2 - 4(\vec{b}\cdot\vec{c})$$
Given | a| = 1, | b| = 4, | c| = 2$|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$ and the angle between b$\vec{b}$ and c$\vec{c}$ is 60^°$60^\circ$:
b· c = | b|| c| 60^° = (4)(2)((1)/(2)) = 4$$\vec{b}\cdot\vec{c} = |\vec{b}||\vec{c}|\cos 60^\circ = (4)(2)\left(\frac{1}{2}\right) = 4$$
Substitute the values:
λ²(1)² = 4² + 4(2)² - 4(4)$$\lambda^2(1)^2 = 4^2 + 4(2)^2 - 4(4)$$
λ² = 16 + 16 - 16 = 16$$\lambda^2 = 16 + 16 - 16 = 16$$
λ = ± 4$\lambda = \pm 4$
Step 2: Calculating Dot Product
We have b - 2 c = ± 4 a$\vec{b} - 2\vec{c} = \pm 4\vec{a}$.
Take the dot product with c$\vec{c}$ on both sides:
( b - 2 c) · c = (± 4 a) · c$$(\vec{b} - 2\vec{c}) \cdot \vec{c} = (\pm 4\vec{a}) \cdot \vec{c}$$
b· c - 2| c|² = ± 4( a· c)$$\vec{b}\cdot\vec{c} - 2|\vec{c}|^2 = \pm 4(\vec{a}\cdot\vec{c})$$
4 - 2(2)² = ± 4( a· c)$$4 - 2(2)^2 = \pm 4(\vec{a}\cdot\vec{c})$$
4 - 8 = ± 4( a· c)$$4 - 8 = \pm 4(\vec{a}\cdot\vec{c})$$
-4 = ± 4( a· c)$$-4 = \pm 4(\vec{a}\cdot\vec{c})$$
| a· c| = 1$$|\vec{a}\cdot\vec{c}| = 1$$
Pattern Recognition
A cross-product equation structured as a×( X)=0$\vec{a}\times(\vec{X})=0$ immediately gives X = λ a$\vec{X} = \lambda\vec{a}$. Expanding the magnitude squared is the standard method to expose the dot products and solve for λ$\lambda$.
Chapter Mix
Class 12 Maths: Vector Algebra
