If \( |\vec{a}|=4,|\vec{b}|=3 \) and \( |\vec{a}+\vec{b}|=6 \), then \( |\vec{a}-\vec{b}| \) =
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{14} \)
Expanding the square of the magnitude of the sum gives \( \vec{a}\cdot\vec{b} \):
\( \displaystyle |\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle 36=16+9+2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle 2\,\vec{a}\cdot\vec{b}=11 \)
\( \displaystyle |\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle =16+9-11=14 \)
\( \displaystyle |\vec{a}-\vec{b}|=\sqrt{14} \)
Shortcut: Use the parallelogram law \( |\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2=2\left(|\vec{a}|^2+|\vec{b}|^2\right) \):
\( \displaystyle 36+|\vec{a}-\vec{b}|^2=2(16+9)=50 \)
\( \displaystyle |\vec{a}-\vec{b}|=\sqrt{14} \)
Why the other options are incorrect
Option (2) 14 — Incorrect. 14 is \( |\vec{a}-\vec{b}|^2 \); the square root has not been taken.
Option (3) \( 4\sqrt{2} \) — Incorrect. \( 4\sqrt{2}=\sqrt{32} \), and \( |\vec{a}-\vec{b}|^2=32 \) would need \( 2\,\vec{a}\cdot\vec{b}=-7 \), not 11.
Option (4) 1 — Incorrect. 1 is \( |\vec{a}|-|\vec{b}| \); magnitudes of vectors cannot be subtracted like numbers unless the vectors are in the same direction.