Question

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three non-zero vectors such that $\overrightarrow{b}$ and $\overrightarrow{c}$ are non-collinear. If $\overrightarrow{a}+5\overrightarrow{b}$ is collinear with $\overrightarrow{c},\overrightarrow{b}+6\overrightarrow{c}$ is collinear with $\overrightarrow{a}$ and $\overrightarrow{a}+ α\overrightarrow{b} + β\overrightarrow{c} = 0$, then α + β is equal to

• A. 35
• B. 30
• C. -30
• D. -25

Answer 1

Answer: (A)

35

Answer 2

Set Up Collinearity Conditions: - Since $\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$, we can write:$\vec{a} + 5 \vec{b} = \lambda \vec{c}$

Similarly, since $\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$, we write:$\vec{b} + 6 \vec{c} = \mu \vec{a}$

Eliminate $\vec{a}$ and Find Relations: - Eliminating $\vec{a}$ from these equations, we get:

$\lambda \vec{c} - 5 \vec{b} = \frac{6}{\mu} \vec{c} + \frac{1}{\mu} \vec{b}$
Solving for $\mu$ and $\lambda$, we find:

$\mu = -\frac{1}{5}, \quad \lambda = -30$
Determine $\alpha$ and $\beta$: - With $\alpha = 5$ and $\beta = 30$, we find:
$\alpha + \beta = 5 + 30 = 35$
So, the correct option is: $\mathbf{35}$