Question

Show that each of the given three vectors is a unit vector:

`1/7 (2hati + 3hatj + 6hatj), 1/7(3hati - 6hatj + 2hatk), 1/7(6hati + 2hatj - 3hatk)`

Also, show that they are mutually perpendicular to each other.

Answer 1

Let `veca = 1/7 (2hati + 3hatj + 6hatk), vecb = 1/7 (3hati - 6hatj + 2hatk) "and"  vecc = 1/7 (6hati + 2hatj - 3hatk)`

∴ `|veca| = sqrt((2/7)^2 + (3/7)^2 + (6/7)^2)`

`= sqrt(4/49 + 9/49 + 36/49)`

`= sqrt(49/49)`

`= sqrt1`

= 1

`|vecb| = sqrt((3/7)^2 + ((-6)/7)^2 + (2/7)^2)`

`= sqrt(9/49 + 36/49 + 4/49)`

`= sqrt(49/49)`

`= sqrt1`

= 1

`|vecc| = sqrt((6/7)^2 + (2/7)^2 + ((-3)/7)^2)`

`= sqrt(49/49)`

`= sqrt1`

= 1

Hence `vec a, vec b, vecc` are unit vectors.

Now, `veca . vecb = 1/49 [(2) (3) + (3) (-6) + (6) (2)]`

`= 1/49 [6 - 18 + 12]`

= 0

So, `veca` is perpendicular to `vecb`

`vecb . vecc = 1/49 [(3). (6) + (-6) (2) + (2) (-3)]`

`= 1/49 [18 - 12 - 6]`

= 0

So, `vecb` is perpendicular to `vecc.`

`vecc . vec a = 1/49 [(6) (2) + (2) (3) + (-3) (6)]`

`= 1/49 [12 + 6 - 18]`

= 0

So, `vecc` is perpendicular to `vec a`

Hence, `vec a, vecb  "and"  vec c` are three mutually perpendicular unit vectors.