Answer to Question #214453 in Electricity and Magnetism for thftdrdrd

Question #214453

The vectors field are specified as:


F = −12ax +5ay −10az

and G = 6ax +5ay −2az

Determine:

(a) The unit vector in the direction of −F + 3G

(b) The magnitude of 5ax + G − 2F

(c) |2F| |3G| (F+ G)

(d) The cross product between F and G


1
Expert's answer
2021-07-08T10:12:11-0400

Gives

F=12ax+5ay10azF=-12a_x+5a_y-10a_z

G=6ax+5ay2azG=6a_x+5a_y-2a_z

Part(a)


-F+3G=


(12ax+5ay10az)+3(6ax+5ay2az)-(12a_x+5a_y-10a_z)+3(6a_x+5a_y-2a_z)

F+3G=6ax+0ay+4az-F+3G=6a_x+0a_y+4a_z

P=F+3GP=-F+3G

P^=PP=152(6ax+4az)\hat{P}=\frac{P}{|P|}=\frac{1}{\sqrt52}(6a_x+4a_z)

Part(b)


5ax+G2F=5ax+6ax+5ay2az2(12ax+5ay10az)=35ax4ay+18az5a_x+G-2F=5a_x+6a_x+5a_y-2a_z-2(-12a_x+5a_y-10a_z)=35a_x-4a_y+18a_z

5ax+G2F=352+42+182=1553=39.40|5a_x+G-2F|=\sqrt{35^2+4^2+18^2}=\sqrt{1553}=39.40


Part(c)

2F=24ax+10ay20az2F=-24a_x+10a_y-20a_z

3G=18ax+15ay6az3G=18a_x+15a_y-6a_z

F+G=6ax+10ay12azF+G=-6a_x+10a_y-12a_z


2F3G(F+G)=1076×585×280=13275.8|2F| |3G|(F+G)=\sqrt{1076}\times\sqrt{585}\times\sqrt{280}=13275.8

Part(d)

F×G=[axayaz12510652]F\times G=\begin{bmatrix} a_x&a_y&a_z\\ -12&5&-10\\ 6&5&-2 \end{bmatrix}


F×G=ax(10+50)ay(24+60)+az(6030)F\times G=a_x(-10+50)-a_y(24+60)+a_z(-60-30)

F×G=40ax84ay90azF\times G=-40a_x-84a_y-90a_z


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