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

To solve this problem, we have three conditions involving the vectors \( \overrightarrow{a} \), \( \overrightarrow{b} \), and \( \overrightarrow{c} \). Let's analyze each condition step by step: 

1. Condition 1: \( \overrightarrow{a} + 5\overrightarrow{b} \) is collinear with \( \overrightarrow{c} \).
   This implies there exists some scalar \( k_1 \) such that: \(\overrightarrow{a} + 5\overrightarrow{b} = k_1 \overrightarrow{c}\). 
   From this, we can express: \(\overrightarrow{a} = k_1 \overrightarrow{c} - 5\overrightarrow{b}\).
2. Condition 2: \( \overrightarrow{b} + 6\overrightarrow{c} \) is collinear with \( \overrightarrow{a} \).
   This means there exists another scalar \( k_2 \) such that: \(\overrightarrow{b} + 6\overrightarrow{c} = k_2 \overrightarrow{a}\). 
   Substituting \( \overrightarrow{a} = k_1 \overrightarrow{c} - 5\overrightarrow{b} \) from the first condition, we have: \(\overrightarrow{b} + 6\overrightarrow{c} = k_2(k_1 \overrightarrow{c} - 5\overrightarrow{b})\).
3. Rearrange and compare coefficients:
   • Coefficient of \( \overrightarrow{b} \): \( 1 = -5k_2 \) ⟹ \( k_2 = -\frac{1}{5} \)
   • Coefficient of \( \overrightarrow{c} \): \( 6 = k_2 k_1 \) ⟹ \( k_1 = \frac{6}{k_2} = -30\)
4. Condition 3: \( \overrightarrow{a} + α\overrightarrow{b} + β\overrightarrow{c} = 0 \).
   Substitute \( \overrightarrow{a} = k_1 \overrightarrow{c} - 5\overrightarrow{b} = -30\overrightarrow{c} - 5\overrightarrow{b} \):
   • Combine: \( (-5 + \alpha)\overrightarrow{b} + (-30 + \beta)\overrightarrow{c} = 0 \)
   • This implies \( -5 + \alpha = 0 \) and \( -30 + \beta = 0 \)
   • Solving these, we get \( \alpha = 5 \) and \( \beta = 30 \)
5. Thus, calculating \( \alpha + \beta = 5 + 30 = 35 \).

Therefore, the correct answer is 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: \( \mathbf{35} \)