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

The Correct Option is A

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: \( \mathbf{35} \)