Question

Suppose that \(\vec a\), \(\vec b\) and \(\vec c\) are three vectors such that \(|\vec a|=3\), \(|\vec b|=4\), \(|\vec c|=5\), and each of them is perpendicular to the sum of the other two vectors. Find \(|\vec a+\vec b+\vec c|\).

Hint:

Add the three perpendicularity conditions to get \(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=0\), then expand the square.

Answer 1

Step 1: Understanding the Concept:
Each vector being perpendicular to the sum of the other two means their dot products with that sum are zero.
This gives us three separate equations which we can add together to find the sum of all pairwise dot products.

Step 2: Key Formula or Approach:
The condition translates to \(\vec a\cdot(\vec b+\vec c)=0\), \(\vec b\cdot(\vec c+\vec a)=0\), \(\vec c\cdot(\vec a+\vec b)=0\).
We use \(|\vec a+\vec b+\vec c|^2=|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)\).

Step 3: Detailed Explanation:
Write the three perpendicularity conditions as equations:
\[ \vec a\cdot\vec b+\vec a\cdot\vec c=0, \quad \vec b\cdot\vec c+\vec b\cdot\vec a=0, \quad \vec c\cdot\vec a+\vec c\cdot\vec b=0 \]
Adding all three equations, each dot product term appears exactly twice:
\[ 2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0 \implies \vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=0 \]
Now expand the square of the magnitude of the sum vector using this result:
\[ |\vec a+\vec b+\vec c|^2 = |\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) \]
\[ = 3^2+4^2+5^2+2(0) = 9+16+25 = 50 \]

Step 4: Take the square root:
\[ |\vec a+\vec b+\vec c| = \sqrt{50} = 5\sqrt{2} \]

Final Answer:
The magnitude of the resultant vector sum is found using the vanishing pairwise dot product sum. \[ \boxed{|\vec a+\vec b+\vec c| = 5\sqrt{2}} \]