Brysondennard904idn
Answered

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Vector v is shown in the graph.
vector v with initial point at 0 comma 0 and terminal point at negative 8 comma 6
Which are the component form and magnitude of v?

v = ❬–8, 6❭; ||v|| = –10
v = ❬8, 6❭; ||v|| = –10
v = ❬8, 6❭; ||v|| = 10
v = ❬–8, 6❭; ||v|| = 10

Question 5
Vector v is shown in the graph.
vector v with initial point at 2 comma 5 and terminal point at negative 3 comma negative 2

Which are the magnitude and direction of v? Round the answers to the thousandths place.

||v|| = 8.602; θ = 54.462°
||v|| = 8.602; θ = 234.462°
||v|| = 9.220; θ = 54.462°
||v|| = 9.220; θ = 234.462°

Sagot :

Question 4

Answer: Choice D

Explanation:

If the initial point is the origin, the coordinates of the terminal point form the vector itself in component form. We go from (-8,6) to <-8,6>. The notation change is from ordered pair to vector format.

We have a right triangle with legs of 8 and 6 units. The pythagorean theorem will help us determine the hypotenuse is 10. Therefore, the vector length is 10 units and we would say ||v|| = 10. We have a 6-8-10 right triangle.

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Question 5

Answer: Choice B

Explanation:

Vector v starts at (2,5) and ends at (-3,-2).

The x component of the vector is x2-x1 = -3-2 = -5 meaning we move 5 units to the left when going from the start point to the endpoint.

At the same time we move 7 units down because y2-y1 = -2-5 = -7 which is the y component of the vector.

The component form of vector v is

v = <-5, -7>

it says "move 5 units left, 7 units down".

Apply the pythagorean theorem to find the length of the vector.

a^2+b^2 = c^2

c = sqrt(a^2 + b^2)

||v|| = sqrt( (-5)^2 + (-7)^2 )

||v|| = 8.602 which is approximate.

Now let's use the arctangent function to find the angle

theta = arctan(b/a)

theta = arctan(-7/(-5))

theta = 54.462 which is also approximate.

There's a problem however. This angle is in Q1 but the vector <-5,-7> is in Q3. An easy fix is to add on 180 to rotate to the proper quadrant.

54.462+180 = 234.462

which is the proper approximate angle for theta.

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