Question

\( \vec{r} \) and \( \vec{s} \) are unit vectors. If \( |\vec{r}+\vec{s}|=\sqrt{2} \), find: 

i) the value of \( (4 \vec{r}-\vec{s}) \cdot(2 \vec{r}+\vec{s}) \).

ii) the angle between \( \vec{r} \) and \( \vec{s} \). Show your steps.

.

Answer 1

      .............

Answer 2

(i) (4\(\vec r-\vec s\)) . (2\(\vec r+\vec s\)) = 8\(\vec r.\vec r+4\vec r.\vec s-2\vec s.\vec r-\vec s.\vec s\)

 = 8|\(\vec r\)|² + 2\(\vec r.\vec s-|\vec s|^2\)

 = 8 + 0 - 1 (∵ |\(\vec r\)|  = |\(\vec s\)| = 1 & \(\vec r.\vec s=0\))

= 7

(ii) \(|\vec r+\vec s|=\sqrt2\) 

⇒ \(|\vec r+\vec s|^2=2\) 

⇒ (\(\vec r+\vec s\)) .(\(\vec r-\vec s\)) = 2 (∵ \(\vec a.\vec a=|\vec a|^2\) )

⇒ \(\vec r.\vec r+2\vec r.\vec s+\vec s.\vec s=2\)

⇒ 2\(\vec r.\vec s= 2-2 = 0\)

⇒ \(\vec r.\vec s=0\) ( ∵ \(\vec r.\vec r=|\vec r|^2=1,\vec s.\vec s=|\vec s|^2=1\))

⇒ \(|\vec r||\vec s|cos\theta = 0\)

⇒ cos θ = 0 = cos 90°

∴ Angle between \(\vec r\) & \(\vec s\) is θ 90°