Question

Given vectors \(\vec{a}, \vec{b}, \vec{c}\) are non-collinear and \((\vec{a}+\vec{b})\) is collinear with \((\vec{b}+\vec{c})\) which is collinear with \(\vec{a}\), and \(|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}\), find \(|\vec{a}+\vec{b}+\vec{c}|\).

Hint:

When multiple vectors are collinear with equal magnitudes, reduce everything in terms of one vector and use magnitude conditions.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
We use the definitions of collinearity (\(\vec{u} = \lambda \vec{v}\)) and properties of dot products to find the required sum.
Step 2: Detailed Explanation:
1. Use collinearity conditions:
\(\vec{a} + \vec{b} = \lambda \vec{c}\) \dots (i)
\(\vec{b} + \vec{c} = \mu \vec{a}\) \dots (ii)
Subtract (ii) from (i): \(\vec{a} - \vec{c} = \lambda \vec{c} - \mu \vec{a} \implies (1 + \mu) \vec{a} = (1 + \lambda) \vec{c}\).
Since \(\vec{a}\) and \(\vec{c}\) are non-collinear, their coefficients must be zero:
\(1 + \mu = 0\) and \(1 + \lambda = 0 \implies \lambda = \mu = -1\).
Substituting back into (i): \(\vec{a} + \vec{b} = -\vec{c} \implies \vec{a} + \vec{b} + \vec{c} = 0\).
2. Calculate the dot product sum:
\[ |\vec{a} + \vec{b} + \vec{c}|^{2} = 0 \]
\[ |\vec{a}|^{2} + |\vec{b}|^{2} + |\vec{c}|^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
Substitute magnitudes \(|\vec{a}| = |\vec{b}| = |\vec{c}| = \sqrt{2}\):
\[ 2 + 2 + 2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
\[ 6 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3 \]
Taking the absolute value: \(|-3| = 3\).
Step 3: Final Answer:
The value is 3.