Question 8. Let \vec{a}, \vec{b} and \vec{c} be three non-zero vectors such that \vec{b} and \vec{c} are non-collinear. Given that \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
(1) 35
(2) 30
(3) -30
(4) -25
Answer (1)
Explanation:
Since \vec{a}+5\vec{b} is collinear with \vec{c},
\vec{a}+5\vec{b}=\lambda\vec{c}Since \vec{b}+6\vec{c} is collinear with \vec{a},
\vec{b}+6\vec{c}=\mu\vec{a}From the first equation,
\vec{a}=\lambda\vec{c}-5\vec{b}Substitute this in the second equation,
\vec{b}+6\vec{c}=\mu(\lambda\vec{c}-5\vec{b})Rearranging,
\lambda\vec{c}-5\vec{b}=\frac{6}{\mu}\vec{c}+\frac{1}{\mu}\vec{b}Comparing coefficients of \vec{b} and \vec{c},
\mu=-\frac{1}{5} and \lambda=-30
Now, from \vec{a}+5\vec{b}=\lambda\vec{c},
\vec{a}+5\vec{b}-30\vec{c}=\vec{0}Comparing with \vec{a}+\alpha\vec{b}+\beta\vec{c}=\vec{0},
\alpha=5 and \beta=30
Hence, \alpha+\beta=35