Area of a parallelogram with vectors

ganondorf29
Messages
49
Reaction score
0

Homework Statement


Determine the area of the parallelogram spanned by the vectors
< 0, 9, 6 > and < −10, −6, −4 >


Homework Equations


Area = A X B

The cross product of < 0, 9, 6 > and < −10, −6, −4 > = 0i - 60j + 90k

The Attempt at a Solution



I know the area is the cross product of A X B, but the examples that I have done did not have a k value. ex <3,-3,0> X <2,3,0> What do I do now that I have a k value?
 
Physics news on Phys.org
Area is a real number. Not a vector. It's |AxB|. You need to find the length of the cross product vector.
 
Thank you, I got it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top