Vector Algebra Question 6
Question 6 - 2024 (29 Jan Shift 1)
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 $\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{b}}+6 \overrightarrow{\mathrm{c}}$ is collinear with $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{a}}+\alpha \overrightarrow{\mathrm{b}}+\beta \overrightarrow{\mathrm{c}}=\overrightarrow{0}$, then $\alpha+\beta$ is equal to
(1) 35
(2) 30
(3) -30
(4) -25
Show Answer
Answer (1)
Solution
$\vec{a}+5 \vec{b}=\lambda \vec{c}$
$\vec{b}+6 \vec{c}=\mu \vec{a}$
Eliminating $\vec{a}$
$\lambda \overrightarrow{\mathrm{c}}-5 \overrightarrow{\mathrm{b}}=\frac{6}{\mu} \overrightarrow{\mathrm{c}}+\frac{1}{\mu} \overrightarrow{\mathrm{b}}$
$\therefore \mu=\frac{-1}{5}, \lambda=-30$
$\alpha=5, \beta=30$
