Question

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be vectors of magnitude 2, 3 and 4 respectively. If $\vec{a} \cdot (\vec{b} + \vec{c}) = 0$, $\vec{b} \cdot (\vec{c} + \vec{a}) = 0$ and $\vec{c} \cdot (\vec{a} + \vec{b}) = 0$, then the magnitude of $\vec{a} + \vec{b} + \vec{c}$ is ______.

• A. 29
• B. $\sqrt{28}$
• C. $\sqrt{29}$
• D. 28

Hint:

If every vector is perfectly perpendicular to the sum of the other two vectors, it means all the mixed dot products sum to absolutely zero! The magnitude of their sum is just the square root of the sum of their individual squares ($\sqrt{A^2+B^2+C^2}$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the individual magnitudes of three vectors and three dot-product relationships. We must calculate the magnitude of the sum of the three vectors.

Step 2: Detailed Explanation:
Let's expand the three given dot-product relationships:
1. $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$
2. $\vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0$
3. $\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$
Because the dot product is commutative ($\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x}$), we can add all three equations together to find the sum of all cross-terms:
$(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}) + (\vec{b}\cdot\vec{c} + \vec{a}\cdot\vec{b}) + (\vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c}) = 0 + 0 + 0$
$2\vec{a}\cdot\vec{b} + 2\vec{b}\cdot\vec{c} + 2\vec{c}\cdot\vec{a} = 0$
$2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$ --- (Equation 1)
We need to find the magnitude $|\vec{a} + \vec{b} + \vec{c}|$.
A fundamental vector identity is:
$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})$
We are given the magnitudes:
$|\vec{a}| = 2 \implies |\vec{a}|^2 = 4$
$|\vec{b}| = 3 \implies |\vec{b}|^2 = 9$
$|\vec{c}| = 4 \implies |\vec{c}|^2 = 16$
Substitute these values and Equation 1 into the identity:
$|\vec{a} + \vec{b} + \vec{c}|^2 = 4 + 9 + 16 + 0$
$|\vec{a} + \vec{b} + \vec{c}|^2 = 29$
Take the square root to find the magnitude:
$|\vec{a} + \vec{b} + \vec{c}| = \sqrt{29}$

Step 3: Final Answer:
The magnitude is $\sqrt{29}$, matching option (c).