Question

find the unit vector in the direction of AB where A and B are the points (1,2,3) and(4,5,6) respectively​

Answer 1

Given :

A( 1,2,3)

B ( 4,5,6)

To find:

The unit vector in the direction of AB

Solution:

$\sf \: Position \: Vector \: of \: A = \hat i + 2\hat j + 3 \hat k$

$\sf \: Position \: Vector \: of \: B= 4\hat i + 5\hat j + 6\hat k$

So,

$\rm \vec {AB} = Position \: Vector \: of \: B - Position \: Vector \: of \: A$

$\rm \vec {AB} =(4\hat i + 5\hat j + 6\hat k) - (\hat i + 2\hat j + 3 \hat k)$

$\rm \vec {AB} =3\hat i + 3\hat j + 3\hat k$

$\rm |\vec {AB}| = \sqrt{ {(3)}^{2} +( {3)}^{2} + {(3)}^{2} }$

$\rm |\vec {AB}| = \sqrt{9 +9 + 9}$

$\rm |\vec {AB}| = \sqrt{27}$

$\rm |\vec {AB}| = \sqrt{9 \times 3}$

$\rm |\vec {AB}| = 3 \sqrt{3}$

The unit vector in the direction AB is

$\rm \: \dfrac{1}{ |\vec{AB}| } \vec {AB} = \dfrac{1}{3 \sqrt{3} } (3\hat i + 3\hat j + 3\hat k)$

$\rm = \dfrac{1}{ \cancel3 \sqrt{3} } \cancel3(\hat i + \hat j + \hat k)$

$\rm = \dfrac{1}{ \sqrt{3} } (\hat i + \hat j + \hat k)$