Engineering Solving Problems Involving Complex Vectors

[Martin Harris]

Homework Statement

1) The following vectors are given written as complex numbers:
a = 3-2i
b= -6-4i
c= 4+ 6i
d= -4+3i
The requirements: To calculate the vectors' modulus.

2) The following system of vectors is given
3a-2b=7
-5a+6b=3i
The requirements: To find a and b as vectors, as well as to calculate their modulus.

Relevant Equations

Let the vectors be of the following type: z = a+bi, where a = real part, b = imaginary part

Hi

Here is my attempt at a solution for problems 1) and 2) that can be found within the summary.

Problem 1)
a = 3-2i
b= -6-4i
c= 4+ 6i
d= -4+3i

Now, to calculate each vector modulus, I applied the following formula:

$$\left| Vector modulus \right| = \sqrt {(a^2 + b^2) }$$
where a = real part of the vector, and b = imaginary part of the vector

By this formula, the following vector modulus, were calculated:
$$a = \sqrt {13} = 3.605551275463989$$
$$b = 2* \sqrt {13} = 7.211102550927978$$
$$c = 2*\sqrt {13} = 7.211102550927978$$
$$d = \sqrt {25} = 5$$

End of Problem 1 solution attempt

Problem 2)
The following system of vectors is given
$$3a-2b=7 (Eq1)$$
$$-5a+6b=3i (Eq2)$$

I am requested to find vectors a,b as well as to calculate their modulus

From Eq(1), $$ vector a = \frac {7a+2b} {3} Eq (3) $$
Substituting vector a with the above form in Eq 2 yields:
$$vector b = 4.375+1.125i$$ or $$vector b = \frac {35} {8} + \frac {9i} {8} $$
Now plugging back vector b in Eq (3) yields:
$$vector a = 5.25 +0.75i$$ or $$vector a = \frac {15.75} {3} + \frac {2.25i} {3} $$

Now, to calculate the vector modulus, the same formula that was used in Problem1 will be applied, hence it will yield the following vector modulus:

$$Modulus for vector a = \frac {15*\sqrt {2}} {4} = 5.303300858899107 $$
$$Modulus for vector b = 4.517327749898163 $$

End of Problem 2 solution attempt

I would be more than grateful if someone could peer-review (check my attempt).
Many thanks!

 

[anuttarasammyak]

It seems all right but would be more accurate to write it using root e.g.
a=\sqrt{13}

 

[Martin Harris]

Cool! Thanks for the advice, I'll follow it.