What is the value of ||\vec{x}||How to Approach a Vector Projection Problem?

[msimmons]

Problem:
Let \vec{x} and \vec{y} be vectors in Rⁿ and define

p = \frac{x^Ty}{y^Ty}y
and
z = x - p

(a) Show that \vec{p}\bot\vec{z}. Thus \vec{p} is the vector projection of x onto y; that is \vec{x} = \vec{p} + \vec{z}, where \vec{p} and \vec{z} are orthogonal components of \vec{x}, and \vec{p} is a scalar multiple of \vec{y}

(b) If ||\vec{p}|| = 6 and ||\vec{z}|| = 8, determine the value of ||\vec{x}||

My problem:
I understand the question, but have no idea how to approach it. Hints?

 

[trambolin]

Draw any arbitrary two vectors, draw the projection of one onto the other and stare at it until you realize that \sqrt{6^2 + 8^2} is obvious.

 

[HallsofIvy]

And then think "3-4-5 right triangle" twice!

Your "x^T y is a fancy way of writing the dot product. In my simpler mind, what you really want to prove is that
\frac{\vec{x}\cdot\vec{y}}{||\vec{x}||} \vec{x}
is perpendicular to \vec{p}- \vec{x}.

Okay, go ahead and take the dot product:
\frac{\vec{x}\cdot\vec{y}}{||\vec{x}||} \vec{x}\cdot (\vec{x}- \frac{\vec{x}\dot\vec{y}}{||\vec{x}||} \vec{x})[/itex]