Question

Three vectors \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \) are such that \( |\vec{a}|=1 \), \( |\vec{b}|=2 \) and \( |\vec{c}|=4 \) along with \( (\vec{a} + \vec{b} + \vec{c}) = 0 \). Then, the value of \( 4\vec{a}\cdot\vec{b} + 3\vec{b}\cdot\vec{c} + 3\vec{c}\cdot\vec{a} \) will be

• A. 27
• B. -26
• C. -68
• D. -34

Hint:

Whenever you have an asymmetric-looking vector sum expression like \( A\vec{a}\cdot\vec{b} + B\vec{b}\cdot\vec{c} + B\vec{c}\cdot\vec{a} \), grouping the symmetric parts together as \( (A-B)\vec{a}\cdot\vec{b} + B(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \) simplifies the arithmetic and reduces the number of individual dot products you need to find.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given three vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) whose sum is zero, and their respective magnitudes. We need to calculate the value of the scalar expression \( 4\vec{a}\cdot\vec{b} + 3\vec{b}\cdot\vec{c} + 3\vec{c}\cdot\vec{a} \).

Step 2: Key Formula or Approach:
1. Squaring the sum vector identity:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \]
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \]
2. To find the specific term \( \vec{a}\cdot\vec{b} \), we rearrange \( \vec{a} + \vec{b} = -\vec{c} \) and square both sides.

Step 3: Detailed Explanation:
First, calculate the sum of the pairwise dot products:
Given \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), and \( |\vec{c}| = 4 \):
\[ 1^2 + 2^2 + 4^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \]
\[ 1 + 4 + 16 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \]
\[ 21 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \]
\[ \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{21}{2} \]
Next, isolate \( \vec{a}\cdot\vec{b} \) using \( \vec{a} + \vec{b} = -\vec{c} \):
Squaring both sides:
\[ |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 \]
\[ |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = |\vec{c}|^2 \]
Substitute the magnitudes:
\[ 1 + 4 + 2\vec{a}\cdot\vec{b} = 16 \]
\[ 5 + 2\vec{a}\cdot\vec{b} = 16 \implies \vec{a}\cdot\vec{b} = \frac{11}{2} \]
Now, rewrite the required expression to utilize our calculated terms:
\[ 4\vec{a}\cdot\vec{b} + 3\vec{b}\cdot\vec{c} + 3\vec{c}\cdot\vec{a} = \vec{a}\cdot\vec{b} + 3(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \]
Substitute the values:
\[ X = \frac{11}{2} + 3\left(-\frac{21}{2}\right) \]
\[ X = \frac{11 - 63}{2} = \frac{-52}{2} = -26 \]

Step 4: Final Answer:
The value of the expression is -26.