Question

If the angle between the unit vectors a and b is 60 then |a-b|

Answer 1

Answer:

The answer is 1.

Step-by-step explanation:

Since, we know that,

The angle between two vectors a and b is,

$cos \theta = \frac{a.b}{|a||b|}$

Given,

a and b are unit vectors,

⇒ |a| = 1 and |b| = 1,

⇒ a² = 1 and b² = 1

$\implies a.b = \frac{cos \theta}{|a||b|}=cos \theta$

Now,

$(a-b)^2=a^2 - 2 a.b + b^2$

$= 1 - 2(cos \theta)+1$

$= 2 - 2\times cos 60^{\circ}$     ( Given $\theta$ = 60° )

$= 2 - 2\times \frac{1}{2}= 2 - 1 = 1$

Hence, |a-b| = 1

Answer 2

Given :

The angle between the unit vectors a and b is 60

To Find:

|a-b|

Solution:

a and b are unit vectors,

So, |a| = 1 and |b| = 1,

$a^2 = 1$ and $b^2= 1$

$a.b = \frac{cos \theta}{|a||b|}=cos \theta$

Formula :$(a-b)^2=a^2 - 2 a.b + b^2$

Using formula :

$= 1 - 2(cos \theta)+1\\= 2 - 2\times cos 60^{\circ} ( Given \theta = 60^{\circ} )\\= 2 - 2\times \frac{1}{2}= 2 - 1 = 1$

Hence, |a-b| = 1