Answer to Question #266067 in Analytic Geometry for arbee

Question

1). Given the following vectors: u=(1, 2, -4) and v=(2, 3, 5)

Find:

a) u.v

b) u.u

c) (u.v) v

d) angle between u and v

e) unit vector

2. )Suppose that B =(2, 2, 1) and makes 30° with A and A•B = 6. What is the magnitude of A?

3. )What is the angle between A and B if A• B = 0

4.) Find (u+ v)•(2u – v) given that u•u= 4 u•v =-5 and v•v=10

5. )Determine if u and u = (cos 0, sin 0,–1); v=(sin 0,-cos 0,0) V are orthogonal

Answer 1

1)

a)

$u\cdot v=2+2\cdot 3-4\cdot 5=-12$

b)

$u\cdot u=1+2\cdot 2+4\cdot 4=21$

c)

$(u\cdot v) v=-12(2, 3, 5)=(-24,-36,-60)$

d)

$u\cdot v=|u||v|cos\alpha$

$|u|=\sqrt{1+2^2+4^2}=\sqrt{21}$

$|v|=\sqrt{2^2+3^2+5^2}=\sqrt{38}$

$cos\alpha=-\frac{12}{\sqrt{21\cdot38}}=-0.4248$

$\alpha=arccos(-0.4248)=115\degree$

e)

$\tilde{u}=u/|u|=(1/\sqrt{21}, 2/\sqrt{21}, -4/\sqrt{21})$

$\tilde{v}=v/|v|=(2/\sqrt{38}, 3/\sqrt{38}, 5/\sqrt{38})$

2)

$A•B =|A||B|cos30\degree= 6$

$|B|=\sqrt{2^2+2^2+1}=3$

$|A|=\frac{A\cdot B}{|B|cos30\degree}=\frac{6}{3\sqrt 3/2}=4/\sqrt 3$

3)

$A•B =|A||B|cos\alpha= 0$

$\alpha=90\degree$

4)

$(u+ v)•(2u – v)=2 u•u+u\cdot v- v•v=2\cdot4-5-10=-7$

5)

$u\cdot v = (cos 0, sin 0,–1)\cdot(sin 0,-cos 0,0)=0+0+0=0$

u and v are are orthogonal