### Related Formula textIf vecX text and vecY text are collinear, then vecX = lambda vecY text for some scalar lambda.$\text{If } \vec{X} \text{ and } \vec{Y} \text{ are collinear, then } \vec{X} = \lambda \vec{Y} \text{ for some scalar } \lambda.$ Linear Independence: If vecu$\vec{u}$ and vecv$\vec{v}$ are non-collinear, then xvecu + yvecv = 0 implies x = 0 text and y = 0$x\vec{u} + y\vec{v} = 0 \implies x = 0 \text{ and } y = 0$. ### Core Logic Based on the problem statement: 1) veca + 5vecb$\vec{a} + 5\vec{b}$ is collinear with vecc$\vec{c}$: * quad veca + 5vecb = lambdavecc quad dots(1)$* \quad \vec{a} + 5\vec{b} = \lambda\vec{c} \quad \dots(1)$ 2) vecb + 6vecc$\vec{b} + 6\vec{c}$ is collinear with veca$\vec{a}$: * quad vecb + 6vecc = muveca quad dots(2)$* \quad \vec{b} + 6\vec{c} = \mu\vec{a} \quad \dots(2)$ ### Step 1: Eliminate Vector a From equation (1), isolate veca$\vec{a}$: veca = lambdavecc - 5vecb$\vec{a} = \lambda\vec{c} - 5\vec{b}$ Substitute this into equation (2): vecb + 6vecc = mu(lambdavecc - 5vecb)$\vec{b} + 6\vec{c} = \mu(\lambda\vec{c} - 5\vec{b})$ vecb + 6vecc = mulambdavecc - 5muvecb$\vec{b} + 6\vec{c} = \mu\lambda\vec{c} - 5\mu\vec{b}$ Rearrange to group coefficients for vecb$\vec{b}$ and vecc$\vec{c}$: (1 + 5mu)vecb + (6 - mulambda)vecc = 0$(1 + 5\mu)\vec{b} + (6 - \mu\lambda)\vec{c} = 0$ ### Step 2: Determine Scalars Since vectors vecb$\vec{b}$ and vecc$\vec{c}$ are given as non-collinear, their linear combination resolving to zero dictates that their scalar coefficients must both individually be zero. 1 + 5mu = 0 Rightarrow mu = -frac15$1 + 5\mu = 0 \Rightarrow \mu = -\frac{1}{5}$ 6 - mulambda = 0 Rightarrow 6 - left(-frac15right)lambda = 0 Rightarrow 6 + fraclambda5 = 0$6 - \mu\lambda = 0 \Rightarrow 6 - \left(-\frac{1}{5}\right)\lambda = 0 \Rightarrow 6 + \frac{\lambda}{5} = 0$ fraclambda5 = -6 Rightarrow lambda = -30$\frac{\lambda}{5} = -6 \Rightarrow \lambda = -30$ ### Step 3: Match the Final Equation Substitute lambda = -30$\lambda = -30$ back into equation (1): veca + 5vecb = -30vecc$\vec{a} + 5\vec{b} = -30\vec{c}$ veca + 5vecb + 30vecc = 0$\vec{a} + 5\vec{b} + 30\vec{c} = 0$ The problem provides the structure veca + alphavecb + betavecc = 0$\vec{a} + \alpha\vec{b} + \beta\vec{c} = 0$. Comparing the two yields: alpha = 5, quad beta = 30$\alpha = 5, \quad \beta = 30$ Thus: alpha + beta = 5 + 30 = 35$\alpha + \beta = 5 + 30 = 35$ ### Pattern Recognition Double-collinearity equations should always be mapped out by eliminating the "third" vector (veca$\vec{a}$ in this case) and funneling everything into the two known non-collinear vectors. The resulting zero-equation perfectly exposes the scalar unknowns. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra