Vectors in yz and xz plane dot product, cross product, and angle

[Ursa]

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)

$\vec a \cdot \vec b =ab cos \phi$
$1.19= 2.6*1.3 cos \phi$
$\phi = 69.4°$

 

[kuruman]

Check your trig functions. Remember that $\cos\Phi=\dfrac{\text{adjacent}}{\text{hypotenuse}}$.

 

[Ursa]

So if I understand my mistake the right answers should be
(a) 0.97 m^{2}
(b) -1.19 m^{2}
(c) -1.90 m^{2}
(d) 2.34 m^{2}
(e) 73.3°

or have I veered way of again?

 

[PeroK]

So if I understand my mistake the right answers should be
(a) 0.97 m^{2}
(b) -1.19 m^{2}
(c) -1.90 m^{2}
(d) 2.34 m^{2}
(e) 73.3°

or have I veered way of again?

That's not what I get.

 

[kuruman]

Your answer to (a) is incorrect. That means that your vectors $\vec a$ and/or $\vec b$ are incorrect. What did you change? How about posting them as you did before?

 

[PeroK]

... why not just post the components of your two vectors and get those right before going any further.

 

[Ursa]

These are the components I came up with

$a_y =2.60 sin 63.0 = 2.32$ $a_x=0$ $a_z =2.60 cos 63.0 = 1.18$

$b_y=0$ $b_x =1.30 sin 51.0 = 1.01$ $b_z =1.3 cos 51.0 = 0.82$

 

[PeroK]

These are the components I came up with

$a_y =2.60 sin 63.0 = 2.32$ $a_x=0$ $a_z =2.60 cos 63.0 = 1.18$

$b_y=0$ $b_x =1.30 sin 51.0 = 1.01$ $b_z =1.3 cos 51.0 = 0.82$

You have sines and cosines mixed up.

 

[kuruman]

These are the components I came up with

$a_y =2.60 sin 63.0 = 2.32$ $a_x=0$ $a_z =2.60 cos 63.0 = 1.18$

$b_y=0$ $b_x =1.30 sin 51.0 = 1.01$ $b_z =1.3 cos 51.0 = 0.82$

See post #2.