- #1

Ursa

- 11

- 2

- Homework Statement
- Vector ##\vec a## lies in the yz plane 63.0° from the positive direction of the y axis, has a positive z component, and has magnitude ##2.60 m##. Vector ##\vec b## lies in the xz plane 51.0° from the positive direction of the x axis, has a positive z component, and has magnitude ##1.30 m##. Find (a)##\vec a \cdot \vec b## , (b) the x-component of ##\vec a X \vec b## , (c) the y-component of ##\vec a X \vec b## , (d) the z-component of ##\vec a X \vec b## , and (e) the angle between ##\vec a## and ##\vec b## .

- Relevant Equations
- ##a_y =a sin \Phi##

##a_x =a cos \Phi##

##\vec a \cdot \vec b =ab cos \phi##

I tried to find the components of the vectors.

##a_y =2.60 sin 63.0 = 2.32## and assuming the z axis would behave the same as an x-axis ##a_z =2.60 cos 63.0 = 1.18##

##b_z =1.30 sin 51.0 = 1.01## making the same assumption ##b_x =1.3 cos 51.0 = 0.82## I now think I should have switched these two around, seeing the in the positive direction on the zx plane the z looks like the x and the x like the y.

but going further in this logic I attempted (a)

\begin{matrix}

0i & 2.32j & 1.18k\\

0.82i & 0j & 1.01k

\end{matrix}

##0*0.82 + 2.32*0 + 1.18*1.01= 1.19##

(b)

##2.32*1.01 - 1.18 *0 = 2.3##

(c)

##1.18 *0 - 0*1.01= 0##

(d)

##0 *0 - 2.31*0.82= 1.90##

(e)

##\vec a \cdot \vec b =ab cos \phi##

##1.19= 2.6*1.3 cos \phi##

##\phi = 69.4°##

##a_y =2.60 sin 63.0 = 2.32## and assuming the z axis would behave the same as an x-axis ##a_z =2.60 cos 63.0 = 1.18##

##b_z =1.30 sin 51.0 = 1.01## making the same assumption ##b_x =1.3 cos 51.0 = 0.82## I now think I should have switched these two around, seeing the in the positive direction on the zx plane the z looks like the x and the x like the y.

but going further in this logic I attempted (a)

\begin{matrix}

0i & 2.32j & 1.18k\\

0.82i & 0j & 1.01k

\end{matrix}

##0*0.82 + 2.32*0 + 1.18*1.01= 1.19##

(b)

##2.32*1.01 - 1.18 *0 = 2.3##

(c)

##1.18 *0 - 0*1.01= 0##

(d)

##0 *0 - 2.31*0.82= 1.90##

(e)

##\vec a \cdot \vec b =ab cos \phi##

##1.19= 2.6*1.3 cos \phi##

##\phi = 69.4°##