Question

If \(\vec a,\vec b,\vec c\) are three vectors such that \[ |\vec a|=a,\qquad |\vec b|=b,\qquad |\vec c|=c \] and each one of them is perpendicular to the sum of the other two, then \[ |\vec a+\vec b+\vec c| \] equals

• A. \(a+b+c\)
• B. \(a^2+b^2+c^2\)
• C. \(\sqrt{a^2+b^2+c^2}\)
• D. \(\sqrt{a+b+c}\)

Hint:

If vectors are mutually perpendicular, then \[ |\vec a+\vec b+\vec c|^2 = |\vec a|^2+|\vec b|^2+|\vec c|^2 \] since all cross-product terms vanish.

Answer 1

The Correct Option is C

Step 1: Start from the square of the sum.
We want $|\vec a+\vec b+\vec c|$, so look at its square: $|\vec a+\vec b+\vec c|^2=(\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)$.

Step 2: Expand the dot product.
\[ |\vec a+\vec b+\vec c|^2=|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) \]

Step 3: Use the perpendicular conditions.
Each vector is perpendicular to the sum of the other two. So $\vec a\cdot(\vec b+\vec c)=0$, which means $\vec a\cdot\vec b+\vec a\cdot\vec c=0$. Similarly $\vec b\cdot(\vec c+\vec a)=0$ and $\vec c\cdot(\vec a+\vec b)=0$.

Step 4: Add these three to find the cross terms.
Adding all three equations: $2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0$, so the entire cross-term group is zero.

Step 5: Simplify the square.
The whole middle bracket vanishes, leaving $|\vec a+\vec b+\vec c|^2=|\vec a|^2+|\vec b|^2+|\vec c|^2=a^2+b^2+c^2$.

Step 6: Take the square root.
Therefore $|\vec a+\vec b+\vec c|=\sqrt{a^2+b^2+c^2}$. \[ \boxed{\sqrt{a^2+b^2+c^2}} \]