Let's determine the vector
knowing that its magnitude is
3√2 and it makes angles of
and
with the
y and
z-axes respectively.
First, we use the relationship between the direction cosines and the angles they make with the coordinate axes. Let the direction cosines be
l,m, and
n corresponding to the
x,y, and
z-axes respectively. These are given by:
l=cosαm=cosβn=cosγ where
α,β,γ are the angles the vector makes with the
x,y, and
z-axes.
From the given information:
β=γ= We'll find the direction cosines:
cosβ=cos=cosγ=cos=0 We utilize the fact that the sum of the squares of the direction cosines must be 1 :
l2+m2+n2=1 Plugging in the known values:
l2+()2+02=1l2+=1 Solving for
l2 :
l2=1−=l=± Now, the vector
can be expressed using the direction cosines and the magnitude:
=r⋅(l+m+n) With the magnitude
r=3√2,l=±,m=,n=0 :
The vector component values:
=3√2(±++0) Simplifying:
=±3+3Therefore, the correct option is:
Option D
±3+3