Step-by-step explanation:
All of these vectors are in DE Moviere Form, so let put them in Vector notation
To this this, multiply the modulus by the exact values of trig functions,
So using this, Vector u
[tex]u= 3j + 3 \sqrt{3} i[/tex]
Vector V
[tex]v = 2 \sqrt{2} j - 2 \sqrt{2} i[/tex]
Vector W
[tex]w = - 6 \sqrt{3}j + 6i[/tex]
So know we can compete the parts.
For Part A:
We first multiply the vector U by -7 then multiply that by v.
[tex] - 7u = - 21j - 21 \sqrt{3} j[/tex]
Know we do cross product
[tex] - 7u(v) = ( - 21j - 21 \sqrt{3} i)(2 \sqrt{2} j \: + 2 \sqrt{2} i[/tex]
First, multiply the j componets.
Then, do the I componets and then add the product together.
[tex] - 42 \sqrt{2} j - 42 \sqrt{6} i[/tex]
So the answer to part A is the vector above.
Part B:
Let first find the dot product of u and w.
First, the cross product of u and w is
[tex] - 18 \sqrt{3} j + 18\sqrt{3} i[/tex]
The magnitudes of each respective vector is
6 and -12. So if we use dot product formula
[tex] \cos( \alpha ) = \frac{xy}{ |x| |y| } [/tex]
So we get
[tex] \cos( \alpha ) = \frac{ - 18 \sqrt{3} + 18 \sqrt{3} }{ - 72} [/tex]
[tex] \cos(a) = \frac{0}{ - 12} [/tex]
[tex] \cos(a) = 0[/tex]
[tex]a = \frac{\pi}{2} [/tex]
Since the angle is pi/2, the vectors are orthogonal.
