Consider the vectors, u =(2, – 4), v = (4, 2), and w = v – u.?

Question

(a) Find the magnitude, and the direction of the vectors, u, v, and w.
(b) Find the area, A, of the triangle, T, the three sides of which are made by the vectors, u, v and w. What kind of triangle is T?
(c) Find the volume of the prism the length of which is, l = 6, and the cross-section of which is the triangle made by the vectors, u and v.

Answer 1

magnitude is the resultant force is given by:

R = sqrt (sum Fx)^2 + (sum Fy)^2
Ru = sqrt (2)^2 + (-4)^2
Rv = sqrt (4)^2 + (2)^2
Rw = {sqrt (2)^2 + (-4)^2} - {sqrt (4)^2 + (2)^2}

Direction if found using

theta = tan^-1 (or tan inverse) sum Fy/sum Fx
or the pythagoras theorem which is:

c^2=a^2+b^2

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Answer 2

Consider the vectors, u = <2, -4>, v = <4, 2>, and w = v – u.

(a) Find the magnitude, and the direction of the vectors, u, v, and w.

Magnitude:

|| u || = √[2² + (-4)²] = √(4 + 16) = √20
|| v || = √[4² + 2²] = √(16 + 4) = √20

Similarly for w.

Direction:
For u:
tanα = ∆y/∆x = -4/2 = -2
α = arctan(-2)

Similarly for v and w.

(b) Find the area, A, of the triangle, T, the three sides of which are made by the vectors, u, v and w.

Area = (1/2) || u || || v || sinθ
where θ is the angle between the vectors u and v

|| u X v || = <2, -4, 0> X <4, 2, 0> = 20

The cross product can also be defined as:

|| u X v || = || u || || v || sinθ

So

Area = (1/2)(|| u || || v || sinθ) = (1/2)|| u X v || = (1/2)(20) = 10


What kind of triangle is T?

Use the Law of Cosines to see if triangle T is acute, right, or obtuse.

(c) Find the volume of the prism the length of which is, l = 6, and the cross-section of which is the triangle made by the vectors, u and v.

Volume = (base)(length)
The base is the area of the triangle.