Dot Product and Angle Between?

Question

Find the dot product v times w and the angle between v and w.

(1) v = i + j, w = - i + j

(2) v = i + sqrt{3}(j), w = i - j

What exactly is the angle between them anyway?

Answer 1

the angle between two vectors is given by:
cos(theta) = u dot w/ (|v|*|w|), where |v| is the magnitude of the vector
for number one
v = (1,1) and w = (-1, 1)
in the dot product all you do is multipliy across from row to row and add like this:
(1*-1) +(1*1) , and in this case the dot product is 0
and the angle is found like this:
cos(theta) = 0/(sqrt(2)*sqrt(2) whic is
cos(theta) = 0
solve for theta, take the arccosine of each side:
theta = cors^-1(0)
thehta is 90 degrees.

Note that the magnitude is simply found by
taking the square root of (i^2 + j^2).

2)
v = (1, sqrt(3)), w = (1,-1)
dot product is
(1*1)+(sqrt(3) * -1))
the dot product is -0.732 or 1-sqrt(3)

the angle in between is
cos(theta)= 1-sqrt(3)/(sqrt(2) * sqrt(4)
theta = cos^-1(1-sqrt(3)/(sqrt(2) * sqrt(4))
theta= 105

Answer 2

1. dot product = (1)(-1) + (1)(1) = 0
dot product = |v||w|cos(Angle) = 0
so the angle is pi/2 or 90 degrees

2. dot product = (1)(1) + SQRT(3)(-1) = 1 - SQRT(3)
|v| = SQRT(1*1 + SQRT(3)*SQRT(3)) = SQRT(4) = 2
|w| = SQRT(1*1 + (-1)*(-1)) = SQRT(2)

|v||w| = 2*SQRT(2)cos(Angle) = 1-SQRT(3)
cos(Angle) = (1-SQRT(3))/(2*SQRT(2))
cos(Angle) = -0.258819
Angle = 105 degrees

The "angle between" is the angle formed at the point where the two vectors join. In these cases draw them using an x-y coordinate system with each vector starting at (0,0).

Answer 3

for (1), the dot product is 1*1 + (-1)(1) = 0.

To find the angle between them, use the fact that the dot product is |v||w|cos(angle).

In this case, you have 0 = Sqrt(2)Sqrt(2)Cos(angle), or

Cos(angle) = 0, or angle = Pi/2.

The vectors are orthogonal.

For (2), do the same as above.

Answer 4

(1) We know, by definition, that:

v.w = | v |*| w |cos(v, w)............(A),

where | v | denotes magnitude of vector v, cos(v, w) indicates the angle between vectors v and w.

Now | v | = √(1² + 1²) = √2, | w | = √[(-1)² + 1²] = √2 and v.w = (1)(-1) + (1)(1) = 0. So that (A) becomes:

0 = √2√2*cos(v, w) or cos(v, w) = 0 ---> Angle between v and w is 90° or they are exactly at rt. angles to each other.

(2) Do it exactly in the same manner as above.

Answer 5

the thank you to stipulate the dot and perspective between must be spelled out extraordinarily of course on your e book. have you ever atempted to paintings those out following the examples on your math e book? a million) v = i + j, w = -i + j - ok v = < a million, a million, 0> w = < -a million, a million, -a million> those are basically the coefficients The dot product is (a million)(-a million) + (a million)(a million) + (0)(-a million) = 0 (those vectors are orthogonal. the perspective between them is ninety tiers (pi/2) Do you recognize what the dot product tells you? It tells you pertaining to to the perspective between the vectors. If the dot product is 0, the the two vectors are perpendicular to a minimum of one yet another (orthogonal, usual are additionally words used). If the dot product is constructive then the perspective between the vectors is acute (under ninety tiers), and if the dot product is damaging, then the perspective between them is obtuse (better than ninety tiers) This makes losts of expertise in case you would be able to work out the vectors on a cartesian coordinate equipment. See in case you could start this one: (2) v = 3i -4j + ok, w = 6i -8j + 2k

Answer 6

a.b=a*b*cos(theta)
i,j,k are unit vectors perpendicular to each other
1)v.w=-1cos0+1cos90-cos90+1cos0
=-1+1=0
cos (theta)=v.w/|v||w|
=0
theta=pi/2 degree

2)v.w=1-sqrt(3)
|v|=sqrt(1+3)=2
|w|=sqrt(1+1)=sqrt(2)
cos(theta)=(1-sqrt(3))/(2*sqrt(2))
theta=cos inverse of((1-sqrt(3))/2*sqrt(2))