The vector in the direction of the vector
that has magnitude 9 is
Given is the vector
.
Let this vector be
, such that
![]()
Let us first find the unit vector in the direction of this vector
.
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
Unit vector in the direction of the vector
is given as,
![]()
As, we have
.
Then, ![]()
![]()
[∵ if ![]()
]
![]()
![]()
![]()
Therefore,
![]()
[∵
and
]
We have found unit vector in the direction of the vector
, but we need to find the unit vector in the direction of
but also with the magnitude 9.
We have the formula:
Vector in the direction of
with a magnitude of 9![]()
![]()
And
as just found.
So,
⇒Vector in the direction of
with a magnitude of 9=![]()
![]()
Thus, vector in the direction of vector
and has magnitude 9 is
.