Let a and b be two unit vectors and a be the angle between them. Then, a + b is a unit vector if: 1. a= 4 2. a= 3 3. a=2 3 4. a= 2

3. a=2 3 Solution: a and b are unit vectors. | a |= | b |=1 (1) Now, a . b =| a | | b | a . b = (2) [using (1)] Given that | a + b |=1 Squaring both sides, we get | a + b |^2=1 | a |^2+ | b |^2+2 a . b =1 1+1+2 =1 [using (1) and (2)] 2+2 =1 2 =-1 =-1 2 =2 3

Let a and b be two unit vectors and a be the angle between them. …

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Question

Let $\vec{\text{a}}$ and $\vec{\text{b}}$ be two unit vectors and a be the angle between them. Then, $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if:

$\text{a}=\frac{\pi}{4}$
$\text{a}=\frac{\pi}{3}$
$\text{a}=\frac{2\pi}{3}$
$\text{a}=\frac{\pi}{2}$

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Answer

$\text{a}=\frac{2\pi}{3}$

Solution:
$\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors.
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1\dots(1)$
Now,
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\text{a}$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=\cos\text{a}\dots(2)$
[using (1)]
Given that
$\Big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=1$
$\Rightarrow1+1+2\cos\text{a}=1$ [using (1) and (2)]
$\Rightarrow2+2\cos\text{a}=1$
$\Rightarrow2\cos\text{a}=-1$
$\Rightarrow\cos\text{a}=\frac{-1}{2}$
$\Rightarrow\text{a}=\frac{2\pi}{3}$

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