Find the position vector of a point A in space such that 
is inclined at 60° to 
and at 45° to 
and 
= 10 units.
We are given that,
is inclined at 60° to 
and at 45° to 
.
We need to find the position vector of point A in space.
That is,
Position vector of A = 
So, we need to find 
.
We know that, space contains three axes, namely, X, Y and Z.
Let OA be inclined at angle α with OZ.
And we know that direction cosines are associated by the relation,
l2 + m2 + n2 = 1 …(i)
And here, direction cosines are cosines of the angles inclined by 
on 
, 
and 
. So,
l = cos 60°
m = cos 45°
n = cos α
Substituting these values of l, m and n in equation (i), we get
(cos 60°)2 + (cos 45°)2 + (cos α)2 = 1
We know the values of cos 60° and cos 45°.
That is,
We get,
So 
is given as,
…(ii)
We have,
Putting these values of l, m and n in equation (ii), we get
Also, put 
.
Thus, position vector of point A in space is 
.