For the vector v find its magnitude and direction: <15,-8>?

Answer 1

For a vector the magnitude and direction are defined as:

magnitude = sqrt(a^2+b^2)
angle = arctan(b/a)

where the angle starts from the x-axis and rotates counter-clockwise. Therefore:

magnitude = sqrt(15^2+(-8)^2) = sqrt(225+64) = sqrt(289) = 17
angle = arctan(-8/15) = -28.07 degrees

Rotate the vector 28.07 degrees clockwise because of the negative. If you graph this vector, you see that it lies within Quadrant IV.

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Hope this helps

Answer 2

<15,-8> means a vector which goes right 15 units, and then down 8 units. So we essentially have a right-angled triangle with perpendicular sides of length 8 and 15.
By Pythagoras's theorem, the magnitude of the vector (which corresponds to the hypotenuse of the triangle) is the square root of 8² + 15² = 289 = 17², so the magnitude is 17.

For the direction, trigonometry must be used. The angle the vector makes below the horizontal is equal to arctan(8/15) = 28.1º to 3 significant figures (by basic right-angled trig).

Answer 3

For the vector v find its magnitude and direction: <15,-8>

|<15,-8>| = √(15² + (-8)²) (= √(v.v))
= 17

Direction = arctan (-8/15) (direction into 4th quadrant as x > 0 and y < 0)
≈ -0.4900 rad (~-28.07°)

Answer 4

Assuming you mean 15 units in the x direction and -8 in the y direction:

for magnitude use Pythagoras

[15^2+(-8)^2]^(1/2)

for angle use :

tan(x) = -8/15

x = arc tan (-8/15)

Answer 5

magnitude is: Square root of (15^2 + 8^2) = 17
direction is arctan (-8/15) = -28.07 degrees or 331.93 degrees

Answer 6

15*15=225
-8*-8=64
225+64=289
square root(289)=17

|v|=17

draw an arrow from origin to the point for the direction
x=15
y=-8

Answer 7

keep in concepts the dot product: length is sqrt(v*v) = sqrt(2^2+a million^2) = sqrt(5) The dot product with a unit vector alongside the x axis (i) = (a million,0) supplies you 2, and keep in mind that the dot made from 2 vectors a and b is ||a||*||b||cos(t), there t is the perspective between a and b. So the cosine of the perspective between v and the x axis is two/sqrt(5), and the perspective it makes with the x-axis is arccos(2/sqrt(5)) = 26.6deg