Page 426 - 35Linear Algebra

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However, the basis is not orthonormal so we know nothing about the lengths of
the basis vectors (save that they cannot vanish).
To complete the hint, lets use the dot product to compute a formula for c 1
in terms of the basis vectors and v. Consider
2
1
1
n
2
v 1 v = c v 1 v 1 + c v 1 v + · · · + c v 1 v n = c v 1 v 1 .
1
Solving for c (remembering that v 1 v 1 6= 0) gives
v 1 v
1
c = .
v 1 v 1
This should get you started on this problem.
Hint for Review Problem 3

Lets work part by part:
⊥
(a) Is the vector v = v − u·v u in the plane P?
u·u
Remember that the dot product gives you a scalar not a vector, so if you
think about this formula u·v is a scalar, so this is a linear combination
u·u
of v and u. Do you think it is in the span?
(b) What is the angle between v ⊥ and u?
This part will make more sense if you think back to the dot product for-
mulas you probably first saw in multivariable calculus. Remember that

u · v = kukkvk cos(θ),
π
and in particular if they are perpendicular θ = π and cos( ) = 0 you will
2 2
get u · v = 0.
⊥
Now try to compute the dot product of u and v ⊥ to find kukkv k cos(θ)

⊥ u · v
u · v = u · v − u
u · u
u · v

= u · v − u · u
u · u
u · v

= u · v − u · u
u · u
Now you finish simplifying and see if you can figure out what θ has to be.
(c) Given your solution to the above, how can you find a third vector perpen-
⊥
dicular to both u and v ?
Remember what other things you learned in multivariable calculus? This
might be a good time to remind your self what the cross product does.

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