Related Formula
If X and Y are collinear, then X = λ Y for some scalar λ.$$\text{If } \vec{X} \text{ and } \vec{Y} \text{ are collinear, then } \vec{X} = \lambda \vec{Y} \text{ for some scalar } \lambda.$$
Linear Independence: If u$\vec{u}$ and v$\vec{v}$ are non-collinear, then x u + y v = 0 x = 0 and y = 0$x\vec{u} + y\vec{v} = 0 \implies x = 0 \text{ and } y = 0$.
Core Logic
Based on the problem statement:
- a + 5 b$\vec{a} + 5\vec{b}$ is collinear with c$\vec{c}$:
* a + 5 b = λ c (1)$* \quad \vec{a} + 5\vec{b} = \lambda\vec{c} \quad \dots(1)$
- b + 6 c$\vec{b} + 6\vec{c}$ is collinear with a$\vec{a}$:
* b + 6 c = μ a (2)$* \quad \vec{b} + 6\vec{c} = \mu\vec{a} \quad \dots(2)$
Step 1: Eliminate Vector a
From equation (1), isolate a$\vec{a}$:
a = λ c - 5 b$$\vec{a} = \lambda\vec{c} - 5\vec{b}$$
Substitute this into equation (2):
b + 6 c = μ(λ c - 5 b)$$\vec{b} + 6\vec{c} = \mu(\lambda\vec{c} - 5\vec{b})$$
b + 6 c = μλ c - 5μ b$$\vec{b} + 6\vec{c} = \mu\lambda\vec{c} - 5\mu\vec{b}$$
Rearrange to group coefficients for b$\vec{b}$ and c$\vec{c}$:
(1 + 5μ) b + (6 - μλ) c = 0$$(1 + 5\mu)\vec{b} + (6 - \mu\lambda)\vec{c} = 0$$
Step 2: Determine Scalars
Since vectors b$\vec{b}$ and c$\vec{c}$ are given as non-collinear, their linear combination resolving to zero dictates that their scalar coefficients must both individually be zero.
1 + 5μ = 0 ⇒ μ = -(1)/(5)$$1 + 5\mu = 0 \Rightarrow \mu = -\frac{1}{5}$$
6 - μλ = 0 ⇒ 6 - (-(1)/(5))λ = 0 ⇒ 6 + (λ)/(5) = 0$$6 - \mu\lambda = 0 \Rightarrow 6 - \left(-\frac{1}{5}\right)\lambda = 0 \Rightarrow 6 + \frac{\lambda}{5} = 0$$
(λ)/(5) = -6 ⇒ λ = -30$$\frac{\lambda}{5} = -6 \Rightarrow \lambda = -30$$
Step 3: Match the Final Equation
Substitute λ = -30$\lambda = -30$ back into equation (1):
a + 5 b = -30 c$$\vec{a} + 5\vec{b} = -30\vec{c}$$
a + 5 b + 30 c = 0$$\vec{a} + 5\vec{b} + 30\vec{c} = 0$$
The problem provides the structure a + α b + β c = 0$\vec{a} + \alpha\vec{b} + \beta\vec{c} = 0$. Comparing the two yields:
α = 5, β = 30$$\alpha = 5, \quad \beta = 30$$
Thus:
α + β = 5 + 30 = 35$$\alpha + \beta = 5 + 30 = 35$$
Pattern Recognition
Double-collinearity equations should always be mapped out by eliminating the "third" vector (a$\vec{a}$ in this case) and funneling everything into the two known non-collinear vectors. The resulting zero-equation perfectly exposes the scalar unknowns.
Chapter Mix
Class 12 Mathematics: Vector Algebra
