The length of the vector v1 is 1 as
∣∣v1∣∣=(0)2+(3/13)2+(2/13)2=1We want to find two vectors v2,v3 such that {v1,v2,v3}is an orthonormal basis for R3.
Let u=⟨x,y,z⟩ be a vector that is perpendicular to v1.
Then we have u⋅v1=0, and hence we have the relation
0(x)+(3/13)(y)+(2/13)(z)=0
3y+2z=0For example, the vector u=⟨1,2,−3⟩ satisfies the relation, anf hence u⋅v1=0.
Let us define the third vector w to be the cross product of u and v1
w=u×v1=∣∣i10j23/13k−32/13∣∣
=i∣∣23/13−32/13∣∣−j∣∣10−32/13∣∣
+k∣∣1023/13∣∣
=(13/13)i−(2/13)j+(3/13)k By the property of the cross product, the vector w is perpendicular to both u,v1.
∣∣u∣∣=(1)2+(2)2+(−3)2=14
∣∣w∣∣=(13/13)2+(−2/13)2+(3/13)2
=14 Therefore, the set
{v1,v2,v3}
={⎣⎡03/132/13⎦⎤,⎣⎡1/142/14−3/14⎦⎤,⎣⎡13/1823/1822/182⎦⎤}
is an orthonormal basis for R3.
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