Unit vector along the direction of vector

→

a

,

→

a

=

→ a

| → a |

Let

→

a

=

^

i

−2

^

j

+2

^

k

|

→

a

| = √(1)2+(−2)2+(2)2 = ± 3
i.e.

^

a

=

1

3

(

^

i

−2

^

j

+2

^

k

)
So, the vector whose magnitude is 15 and has direction along the vector

^

i

−2

^

j

+2

^

k

is given by,
15 × (

1

3

)(

^

i

−2

^

j

+2

^

k

)
= 5 (

^

i

−2

^

j

+2

^

k

)
= (5

^

i

−10

^

j

+10

^

k

)
So the required vector is (5

^

i

−10

^

j

+10

^

k

)