If \( |\vec{a}|=|\vec{b}|=|\vec{a}+\vec{b}|=1 \), then
Match List-I with List-II
| LIST-I | LIST-II | ||
| A. | \( |\vec{a}-\vec{b}| \) | I. | \( \displaystyle\frac{2\pi}{3} \) |
| B. | Angle between \( \vec{a} \) and \( \vec{b} \) | II. | \( -\displaystyle\frac{1}{2} \) |
| C. | Angle between \( \vec{a} \) and \( \vec{a}+\vec{b} \) | III. | \( \sqrt{3} \) |
| D. | \( \vec{a}.\vec{b} \) | IV. | \( \displaystyle\frac{\pi}{3} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-I, C-IV, D-II
Squaring the given magnitude of the sum gives \( \vec{a}\cdot\vec{b} \):
\( \displaystyle |\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle 1=1+1+2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle \vec{a}\cdot\vec{b}=-\frac{1}{2} \)
(A) \( |\vec{a}-\vec{b}| \)
This matches (III) \( \sqrt{3} \).
\( \displaystyle |\vec{a}-\vec{b}|^2=1+1-2\left(-\frac{1}{2}\right)=3 \)
\( \displaystyle \Rightarrow\ |\vec{a}-\vec{b}|=\sqrt{3} \)
(B) Angle between \( \vec{a} \) and \( \vec{b} \)
This matches (I) \( \frac{2\pi}{3} \).
\( \displaystyle \cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}=-\frac{1}{2} \)
\( \displaystyle \Rightarrow\ \theta=\frac{2\pi}{3} \)
(C) Angle between \( \vec{a} \) and \( \vec{a}+\vec{b} \)
This matches (IV) \( \frac{\pi}{3} \).
\( \displaystyle \vec{a}\cdot(\vec{a}+\vec{b})=|\vec{a}|^2+\vec{a}\cdot\vec{b}=1-\frac{1}{2}=\frac{1}{2} \)
\( \displaystyle \cos\phi=\frac{1/2}{|\vec{a}|\,|\vec{a}+\vec{b}|}=\frac{1}{2} \)
\( \displaystyle \Rightarrow\ \phi=\frac{\pi}{3} \)
(D) \( \vec{a}\cdot\vec{b} \)
This matches (II) \( -\frac{1}{2} \), since \( 1=1+1+2\,\vec{a}\cdot\vec{b} \) gives \( \vec{a}\cdot\vec{b}=-\frac{1}{2} \).
So the matching is A-III, B-I, C-IV, D-II.