Find the position vector of a point A in space, so that is inclined at 60⁰ to
and 45⁰ to
and
= 10 units.
Given, is inclined at 600 to and at
450 to
= 10 units.
We want to find the position vector of point A in space, which is nothing but
We know, there are three axes in space: X, Y, and Z.
Let OA be inclined with OZ at an angle α.
We know, directions cosines are associated by the relation:
l² + m² + n² = 1 ….(i)
In this question, direction cosines are the cosines of the angles inclined by on ,
,
and
So,
Substituting the values of l, m, and n in equation (i),
We know the values of and
, i.e. 1/2 and 1/√2 respectively.
Therefore, we get
So is given as
..........(ii)
We have,
Inserting these values of l, m and n in equation (ii),
Also ,Put
Thus, the position vector of A in space