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: Set up using a direct substitution trick:
Let $\vec s = \vec a+\vec b+\vec c$ be the vector whose magnitude we want.
Notice that $\vec a\cdot(\vec b+\vec c) = \vec a\cdot(\vec s-\vec a) = \vec a\cdot\vec s - |\vec a|^2$.

Step 2: Apply the given condition to each vector:
Since $\vec a$ is perpendicular to $\vec b+\vec c$, we have $\vec a\cdot(\vec s-\vec a)=0$, giving $\vec a\cdot\vec s=|\vec a|^2=9$.
Similarly for $\vec b$: $\vec b\cdot(\vec s-\vec b)=0$ gives $\vec b\cdot\vec s=|\vec b|^2=16$.
And for $\vec c$: $\vec c\cdot(\vec s-\vec c)=0$ gives $\vec c\cdot\vec s=|\vec c|^2=25$.

Step 3: Add the three results together:
Adding $\vec a\cdot\vec s+\vec b\cdot\vec s+\vec c\cdot\vec s$ gives $(\vec a+\vec b+\vec c)\cdot\vec s = \vec s\cdot\vec s = |\vec s|^2$.
\[ |\vec s|^2 = 9+16+25 = 50 \]
This is a quicker route than expanding the square directly, since it reuses the perpendicularity condition on each vector individually.

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

Final Answer:
The same magnitude is obtained through this substitution method. \[ \boxed{|\vec a+\vec b+\vec c| = 5\sqrt{2}} \]