Consider unit length, orthogonal vectors , and a vector such that

Prove the following.

- and are orthogonal and the length of is .
- The vectors form a basis for .
- .
- .

*Proof.*We compute,since and since and are orthogonal by assumption. Thus, and are orthogonal. Next,

since by hypothesis and . Hence, from the vector equation we have

*Proof.*Since and are orthogonal (part (a)), we know the vectors are independent. Thus, they form a basis for since any three independent vectors in are a basis*Proof.*We compute, the vector is given byThen the three coordinates of this cross product are given by

Expanding these out we obtain the coordinates

Since we know and since we know . So, simplifying the expressions above, for each of the coordinates we have

Hence, we indeed have

*Proof.*We compute

For 15(a),you calculated ∣∣A-P∣∣² wrong. ∣∣A-P∣∣² should be as follows:

∣∣A-P∣∣² = ∣∣A∣∣² – 2A⋅P + ∣∣P∣∣².

Then you should get ∣∣A∣∣² – 2A⋅P + ∣∣P∣∣² = ∣∣P∣∣². So, A⋅P = ½.

Now you can go to the original equation PXB = A-P, dot product both sides by P and get P⋅(PXB) = P⋅(A-P). (=>) 0 = P.A – P⋅P. (=>) ½ = P⋅P. And you have P = √2/2