Skip to main contentSkip to solution

If ∣a⃗∣=4,∣b⃗∣=3\lvert\vec{a}\rvert = 4, \lvert\vec{b}\rvert = 3 and ∣a⃗+b⃗∣=6\lvert\vec{a}+\vec{b}\rvert = 6, then ∣a⃗−b⃗∣=\lvert\vec{a}-\vec{b}\rvert =

Solution

✅ Correct Option: 1

Use ∣a⃗+b⃗∣2+∣a⃗−b⃗∣2=2(∣a⃗∣2+∣b⃗∣2)|\vec a+\vec b|^2+|\vec a-\vec b|^2 = 2(|\vec a|^2+|\vec b|^2).

So 36+∣a⃗−b⃗∣2=2(16+9)=5036+|\vec a-\vec b|^2 = 2(16+9)=50, giving ∣a⃗−b⃗∣2=14|\vec a-\vec b|^2=14 and ∣a⃗−b⃗∣=14|\vec a-\vec b|=\sqrt{14}.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question