source : yahoo.com
Please help with this vectors problem?
A = -3i + 6j
Means that the terminal side of A passes through the point (-3, 6) which is 3 units to the left of y-axis and 6 above x-axis
So if you draw a line going right and connecting the point to y-axis (distance 3) and another line going down to x-axis (distance 6)
Then you will get a right triangle.
You _should_ know your trig functions.
We want angle that the side makes with y-axis.
tan(blah) = 3/6 = 1/2 = 0.5
blah = arctan(0.5) = 26.6°
θA = blah + 90° = 116.6° = 117° to 3 significant figures
B is already is 1st quadrant so no need to add 90 or anything.
Distance 2 from x axis (opposite vertical) and 8 from y-axis (adjacent horizontal)
tan(θB) = 2.00/8.00 = 1/4 = 0.25
θB = arctan(0.25) = 14.0°
C=A +B so calculate A+B you’re not braindead so you should get this right.
−3.00 i + 6.00 j + 8.00 i + 2.00 j = 5.00i + 8.00j = C
Also in Q1
tan(θC) = 8/5
θC = 58.0°
For vectors in 3rd quadrant figure out angle from negative x-axis and then add 180°. And in 4th do it from negative y-axis and add 270°.
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