Scalar Product/Norm proof

**Paperwings** · July 13th 2008, 08:15 PM

Hello, I am having trouble. I would appreciate if someone could help me with part b of this problem, I have solved part a.

Problem:
Let u = (a,b) and v = (c, d) be nonzero points in the plane $R^n$ and let $\theta$ be the radian measure of the angle with vertex 0 formed by 0, u, and v.
**notations: ||u|| is the norm and <u,v> is the scalar/dot product

a. Prove that $\left\| u \right\|^2 \left\| v \right\|^2 - \left( <u,v> \right)^2 = \left( ||u|| ||v|| sin \theta \right) ^2$

(1) LHS = $\left\| u \right\|^2 \left\| v \right\|^2 - \left( ||u|| ||v|| cos \theta \right)^2$ by definition of dot product
(2) LHS = $<u,u><v,v>-<u,u><v,v>cos^2$ by distribution and the property that $<u,u> = ||u||^2$
(3) LHS = $[<u,u><v,v>](1-cos^2 \theta)$ by factoring
(4) LHS = RHS by sine cosine rule that $sin^2 \theta + cos^2 \theta = 1$

b. I am given that $|ad-bc|= \left\| u \right\| \left\| v \right\| |sin \theta |$
I have to use this to verify that |ad-bc|/2 is the area of the triangle with vertices 0, u, and v and that that as a consequence |ad-bc| is the area of the parallelogram with vertices 0, u, u+v, and v.

I know that the area of a triangle is $\frac {1}{2}bh$ , so if the area of the triangle with vertices 0, u, v. Then shouldn't the area be |ac-bd|/2 instead of |ad-bc|/2. Any help is greatly appreciated. Thank you.

**CaptainBlack** · July 13th 2008, 10:54 PM

Originally Posted by Paperwings

Hello, I am having trouble. I would appreciate if someone could help me with part b of this problem, I have solved part a.

Problem:
Let u = (a,b) and v = (c, d) be nonzero points in the plane $R^n$ and let $\theta$ be the radian measure of the angle with vertex 0 formed by 0, u, and v.
**notations: ||u|| is the norm and <u,v> is the scalar/dot product

a. Prove that $\left\| u \right\|^2 \left\| v \right\|^2 - \left( <u,v> \right)^2 = \left( ||u|| ||v|| sin \theta \right) ^2$

(1) LHS = $\left\| u \right\|^2 \left\| v \right\|^2 - \left( ||u|| ||v|| cos \theta \right)^2$ by definition of dot product
(2) LHS = $<u,u><v,v>-<u,u><v,v>cos^2$ by distribution and the property that $<u,u> = ||u||^2$
(3) LHS = $[<u,u><v,v>](1-cos^2 \theta)$ by factoring
(4) LHS = RHS by sine cosine rule that $sin^2 \theta + cos^2 \theta = 1$

b. I am given that $|ad-bc|= \left\| u \right\| \left\| v \right\| |sin \theta |$
I have to use this to verify that |ad-bc|/2 is the area of the triangle with vertices 0, u, and v and that that as a consequence |ad-bc| is the area of the parallelogram with vertices 0, u, u+v, and v.

I know that the area of a triangle is $\frac {1}{2}bh$ , so if the area of the triangle with vertices 0, u, v. Then shouldn't the area be |ac-bd|/2 instead of |ad-bc|/2. Any help is greatly appreciated. Thank you.

See the attachment (then think about the sign of $sin(\theta)$ when its not less than a right angle)

RonL

**Paperwings** · July 14th 2008, 05:25 PM

If $\theta$ is a right angle, then the area would be $\frac{1}{2}bh = ||u||||v||sin(90) = ||u||||v|| = |ad-bc|/2$

If the angle is less than a right angle, then area would be $||u|| ||v||sin \theta = |ad-bc|/2$

Else if the angle is greater than a right angle, then the area would be $\frac {1}{2} ||u|| ||v|| sin(180- \theta) = ||u|| ||v|| sin( \theta) = |ad-bc|/2$ since $sin( \theta) = sin(180 - \theta)$

Thank you for the help.

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