Question

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

Answer 1

Gives

$F=-12a_x+5a_y-10a_z$

$G=6a_x+5a_y-2a_z$

Part(a)

-F+3G=

$$-(12a_x+5a_y-10a_z)+3(6a_x+5a_y-2a_z)$$

$$-F+3G=6a_x+0a_y+4a_z$$

$P=-F+3G$

$\hat{P}=\frac{P}{|P|}=\frac{1}{\sqrt52}(6a_x+4a_z)$

Part(b)

$$5a_x+G-2F=5a_x+6a_x+5a_y-2a_z-2(-12a_x+5a_y-10a_z)=35a_x-4a_y+18a_z$$

$$|5a_x+G-2F|=\sqrt{35^2+4^2+18^2}=\sqrt{1553}=39.40$$

Part(c)

$2F=-24a_x+10a_y-20a_z$

$3G=18a_x+15a_y-6a_z$

$F+G=-6a_x+10a_y-12a_z$

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

Part(d)

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

$$F\times G=a_x(-10+50)-a_y(24+60)+a_z(-60-30)$$

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