1. ## vectors help

Hi, is this categorized as number theory?....the only text book my college libary has for this kind of stuff is like, 30 years old and not the easiest read!

This is an example in my workbook...if someone could spare the time to go through it that would be awesome.
Any help much appreciated!

Given that
a = 2i + 3j 4k, b = i 2j and c = 5i + j + k. Find:

(a) the vector
a + b c

(b) the magnitude of
a + b c

(c) a unit vector parallel to
a + b

2. Originally Posted by bobchiba
Hi, is this categorized as number theory?....the only text book my college libary has for this kind of stuff is like, 30 years old and not the easiest read!

This is an example in my workbook...if someone could spare the time to go through it that would be awesome.
Any help much appreciated!

Given that
a = 2i + 3j 4k, b = i 2j and c = 5i + j + k. Find:

(a) the vector
a + b c

(b) the magnitude of
a + b c

(c) a unit vector parallel to
a + b

This is vector geometry

a) a + b - c = 2i + 3j − 4k + i − 2j - (5i + j + k)

= 2i + 3j − 4k + i − 2j - 5i - j - k

= -2i - 5k

b) Magnitude is square root of (-2)^2 + (-5)^2 = square root of 29

c) (-2 - 5k)/(square root of 29)

3. Originally Posted by bobchiba
Hi, is this categorized as number theory?....the only text book my college libary has for this kind of stuff is like, 30 years old and not the easiest read!

This is an example in my workbook...if someone could spare the time to go through it that would be awesome.
Any help much appreciated!

Given that a = 2i + 3j 4k, b = i 2j and c = 5i + j + k. Find:

(a) the vector a + b c

(b) the magnitude of a + b c

(c) a unit vector parallel to a + b c
The terms i, j, and k are unit vectors (i is a vector in the x-direction, j is a vector in the y-direction, and k is a vector in the z-direction).

(a) the vector a + b c

When you add vectors (a, b, and c are vectors), you add i's with i's, j's with j's, and k's with k's:

a + b c = (2i + 3j 4k) + (i 2j) - (5i + j + k)
. . . - . - .= (2i + i - 5i) + (3j - 2j - j) + (-4k - k)
. . . - . - .= -2i + 0j - 5k

4. Hi again,...ok I understand part a and b now, thats great...but could someone shed i little more light on part c.
Thanks.

5. Originally Posted by Glaysher
c) (-2 - 5k)/(square root of 29)
As Glaysher demonstrated, the unit vector parallel to -2i - 5k is the vector -2i - 5k divided by its magnitude, sqrt(29).

Semi-quick explanation: All vectors have a direction and a magnitude. When a vector is written in the form: ai + bj + ck, it's magnitude can be found as the sqrt(a^2 + b^2 + c^2), and its direction is determined by the magnitude of a, b, and c with respect to each other (if a is very big while b and c are small, then the vector will look almost horizontal along the x-axis, if a, b, and c are about the same in value, we will have a vector that shoots off from the origin at almost 45% from each axis).

Any vector with the same direction as our initial vector: -2i - 5k, but different magnitude is just multiplied by some scalar. For example, -10i - 25k is parallel to -2i - 5k, but its magnitude is 5 times as great.

Since a "unit vector" has a magnitude of 1, if we want a unit vector in the direction of -2i - 5k, then by dividing by the magnitude of -2i - 5k, which is sqrt(29), we will have a unit vector in the direction of -2i - 5k that is 1 unit in length.