If A=<2,1> and B=<-1,1>, find the magnitude and direction angle for
-2A +3B
Find all specified roots: Cube roots of 8i
A parallelogram has sides of length 32.7cm and 20.2cm. If the longer diagonal has a length of 35.1cm, what is the angle opposite this diagonal?
I assume <2,1> and <-1,1) are vectors
then -2A + 3B
= <-4,-2) + <-3,3) = <-7,1>
magnitude = √(49+1) = √50 or 5√2
direction: let the angle be Ø
tanØ = -1/7
Ø = appr 171.9° with the x-axis
For the 2nd: are you familiar with D' Moivres Theorem?
let me know, and I will go through with this one
3rd:
make a sketch and use the cosine law
35.1^2 = 32.7^2 + 20.2^2 - 2(20.2)(32.7)cosØ
( see if you get Ø = 100.7° )
Not really, can you go through it with me?
2nd:
let z = 0 + 8i
= 8( 0 + i)
we need 0+i in the form cosØ + isinØ
we need Ø, so that cosØ = 0 and sinØ = 1
mmmmhhh, Ø = 90 comes to mind
so z = 8(cos90° + isin90°)
so the cube root, and using De Moivre's
z^(1/3) = 8^(1/3) (cos 30 + isin30)
= 2( √3/2 + (1/2)i )
= √3 + i
our angle could also have been Ø = 450°
then z^(1/3) = 2(cos 150 + isin150)
= 2( -√3/2 + 1/2 i )
= -√3 + i
or Ø could have been 810
then z^(1/3) = 2(cos 270 + isin270)
= 2(0 + i(-1))
= -2i ----> a surprise ? no ?
or Ø could have been -270
z^(1/3) = 2(cos -90 + isin -90)
= 2(cos90 - isin90)
= 2(0 - i) = -2i , mmmhhhh?
check: is (-2i)^ = 8i ???
LS = (-2i)^3
= -8i^3
= -8(i^2)(i)
= -8(-1)i = 8i , YES
so going to left by subtracting multiples of 360 will always yield -2i
to the right, adding 360 to previous angles with always yield either √3 + i or -√3+i
so cube root of 8i = -2i, √3+i , -√3+i
how did you get appr 171.9° in the first question?
Jane, you will have to be able to change a complex number from standard form to trig form
in general a + bi
= (√(a^2 + b^2) [cosØ + isinØ]
where Ø is such that tanØ = b/a
e.g. change 3 + 4i to trig form
= √(3^2+4^2) (cosØ + isinØ), Ø = appr 53.13°
= 5(cos 53.13 + isin53.13)
De Moivre said
of z = r(cosØ + isinØ)
then z^n = r^n( cos (Ø/n) + i sin (Ø/n) )
e.g. let do (3+4i)^2
(since we could easily do this algebraically ...
(3+4i)^2 = 9 + 24i + 16i^2
= 9 + 24i - 16
= -7 + 24i )
using De Moivre:
(3+4i)^2
= 5^2( cos 2(53.13) + isin 2(53.13))
= 25(cos 106.26 + isin 106.26)
= -7 + 24i (I kept all the decimals in my calculator to get the exact answer.
This is what I did in problem #2, you might want to print out this explanation and go over it slowly and try a few of your own.
It is a great theorem
How did I get 171.9 ?
from tanØ = -1/7
I know that the tangent is negative in II and IV by the CAST rule
but <-7,1> ends up in quadrant II
so I took tan-inverse of 1/7 to get 8.13° , had it been in quad I
in quad II we have 180-8.13 = appr 171.9
In all that lengthy typing, I just noticed an error.
When I stated De Moivre's Theorem, it should of course have been:
then z^n = r^n( cos (nØ) + i sin (nØ) )
I did use it correctly in my work.
got it. thank you so much!
Let's start by finding the magnitude and direction angle for -2A + 3B.
To find the magnitude of a vector, we use the formula:
Magnitude = sqrt(x^2 + y^2)
In this case, we have -2A + 3B, where A = <2, 1> and B = <-1, 1>.
So, -2A + 3B = -2<2, 1> + 3<-1, 1>
= <-4, -2> + <-3, 3>
= <-7, 1>
To find the magnitude of -2A + 3B, we can substitute the values into the formula:
Magnitude = sqrt((-7)^2 + 1^2)
= sqrt(49 + 1)
= sqrt(50)
≈ 7.07
Next, let's find the direction angle. The direction angle, θ, can be found using the formula:
θ = tan^(-1) (y / x)
In this case, the coordinates of -2A + 3B are (-7, 1). So:
θ = tan^(-1) (1 / -7)
≈ -0.143 radians (in radians) or -8.19 degrees (in degrees)
Now, let's move on to finding the cube roots of 8i.
To find the cube roots of a complex number, we can use the formula:
Cube Root = ∛(r) * ∛(cis(θ + 2nπ)), where n = 0, 1, 2
In this case, we have 8i, which can be represented as 0 + 8i.
First, let's find the magnitude, r:
r = sqrt(0^2 + 8^2)
= 8
Next, let's find the angle, θ:
θ = tan^(-1)(8 / 0)
= π / 2 (radians) or 90 degrees (degrees)
Now, let's find the cube roots:
∛(r) = ∛(8) = 2
For n = 0:
Cube Root = 2 * ∛(cis(π / 2 + 2(0)π))
= 2 * ∛(cis(π / 2))
= 2 * ∛(0 + 1i)
= 2 * (1 * (cos(π / 6) + i * sin(π / 6)))
= 2(cos(π / 6) + i * sin(π / 6))
For n = 1:
Cube Root = 2 * ∛(cis(π / 2 + 2(1)π))
= 2 * ∛(cis(π / 2 + 2π))
= 2 * ∛(cis(5π / 6))
= 2 * (1 * (cos(5π / 6) + i * sin(5π / 6)))
= 2(cos(5π / 6) + i * sin(5π / 6))
For n = 2:
Cube Root = 2 * ∛(cis(π / 2 + 2(2)π))
= 2 * ∛(cis(π / 2 + 4π))
= 2 * ∛(cis(9π / 6))
= 2 * (1 * (cos(9π / 6) + i * sin(9π / 6)))
= 2(cos(9π / 6) + i * sin(9π / 6))
Lastly, let's find the angle opposite the longer diagonal in the parallelogram.
In a parallelogram, opposite angles are congruent. Therefore, the angle opposite the longer diagonal is also the angle opposite the longer side.
We are given that the shorter side lengths of the parallelogram are 32.7 cm and 20.2 cm, and the longer diagonal has a length of 35.1 cm.
Using the Law of Cosines, we can find the angle opposite the longer diagonal:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, a = 32.7 cm, b = 32.7 cm, and c = 35.1 cm. We want to find angle C.
35.1^2 = 32.7^2 + 32.7^2 - 2 * 32.7 * 32.7 * cos(C)
1227.01 = 2139.29 - 2 * 1070.29 * cos(C)
-912.28 = -2140.58 * cos(C)
cos(C) = -912.28 / -2140.58 ≈ 0.4267
C = cos^(-1)(0.4267) ≈ 64.32 degrees
Therefore, the angle opposite the longer diagonal in the parallelogram is approximately 64.32 degrees.