Vector Algebra
Collinearity Conditions — Scalar Relations
nta_pyq_2024_jan
Grade 12
Question:
Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$, $\vec{b}+6\vec{c}$ is collinear with $\vec{a}$, and $\vec{a}+\alpha\vec{b}+\beta\vec{c}=\vec{0}$, then $\alpha+\beta$ is equal to:
Step-by-Step Solution
Key Concept: Let $\vec{a}+5\vec{b}=\lambda\vec{c}$ and $\vec{b}+6\vec{c}=\mu\vec{a}$. Eliminate $\vec{a}$: substitute $\vec{a}=\frac{1}{\mu}\vec{b}+\frac{6}{\mu}\vec{c}$ (from second) into the first, then compare coefficients.
$\vec{a}+5\vec{b}=\lambda\vec{c}$ ...(i); $\vec{b}+6\vec{c}=\mu\vec{a}$ ...(ii). From (ii): $\vec{a}=\frac{1}{\mu}\vec{b}+\frac{6}{\mu}\vec{c}$. Sub into (i): $(\frac{1}{\mu}+5)\vec{b}+(\frac{6}{\mu}-\lambda)\vec{c}=\vec{0}$. Since $\vec{b},\vec{c}$ non-collinear: $\frac{1}{\mu}+5=0\Rightarrow\mu=-\frac{1}{5}$; $\lambda=\frac{6}{\mu}=-30$. So $\vec{a}+5\vec{b}+30\vec{c}=\vec{0}$, $\alpha+\beta=35$.
Correct Answer: 1