# Magnitude, Unit vectors,unknown variables

• concon
In summary: So, in summary, we have found that:1. The magnitude of 3a-3b is 3√101.2. The unit vector u in the direction of 3a-3b is (1/√101, 10/√101, 0).3. The vector of length 8 in the direction of a is (8/√66, 56/√66, -32/√66).
concon

## Homework Statement

a and b are vectors in R^3 s.t. a=(1,7,-4) and b= -3j-4k

1. Find ||3a-3b|| (magnitude of 3a-3b)
2. Find unit vector u in direction of 3a-3b, write answer in form (u1,u2,u3)
3. Find vector of length 8 in direction of a (write answer in form "-")

## Homework Equations

||3a-3b|| = 3||a-b|| = 3*sqrt((a1-b1)^2 +...(a3-b3)^2)
3 is positve scalar and can be factored out I believe

U = X/ ||X|| , X ≠ 0

Length = sqrt(x^2 + y^2) (might be wrong, I am confused on finding length in same dir)

## The Attempt at a Solution

Starting solving 1.
3||a-b|| = 3 sqrt((1+3j+4k)^2 + (7+3j+4k)^2 + (-4+3j+4k)^2)
I don't know what j and k represent

2. I would need to know ||3a-3b|| first right?
then do X/||X||?

3. a= (1,7,-4) and L=8
L is in same direction of a so
8 = sqrt((u1-1)^2 + (u2-7)^2 + (u3 +4)^2) -> or maybe the a vals and u's are switched i don't know

concon said:
3||a-b|| = 3 sqrt((1+3j+4k)^2 + (7+3j+4k)^2 + (-4+3j+4k)^2)

Not sure where this comes from.

The formula is that if we are given ##a = (a_1,a_2,a_3)## and ##b = (b_1, b_2, b_3)##, then

$$\|a-b\|= \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2}$$

Here you are given ##a= (1,7,-4)## and ##b=(0,-3,-4)##.

1 person
micromass said:
Not sure where this comes from.

The formula is that if we are given ##a = (a_1,a_2,a_3)## and ##b = (b_1, b_2, b_3)##, then

$$\|a-b\|= \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2}$$

Here you are given ##a= (1,7,-4)## and ##b=(0,-3,-4)##.
Wow wait a minute, how do you know the values for b as (0,-3, -4)?
What do j and k represent?

Also, do you have any input regarding the other two questions I posted w/in this post?

concon said:
Wow wait a minute, how do you know the values for b as (0,-3, -4)?
What do j and k represent?

The notation ##-3j -4k## is just another notation for ##(0,-3,-4)##.

The vector ##i## actually equals ##(1,0,0)##. The vector ##j=(0,1,0)## and ##k=(0,0,1)##. So

$$-3-4k = -3(0,1,0) -4 (0,0,1) = (0,-3,0) + (0,0,-4) = (0,-3,-4)$$

Also, do you have any input regarding the other two questions I posted w/in this post?

Sure, but let's do (a) first.

micromass said:
The notation ##-3j -4k## is just another notation for ##(0,-3,-4)##.

The vector ##i## actually equals ##(1,0,0)##. The vector ##j=(0,1,0)## and ##k=(0,0,1)##. So

$$-3-4k = -3(0,1,0) -4 (0,0,1) = (0,-3,0) + (0,0,-4) = (0,-3,-4)$$

Sure, but let's do (a) first.
Okay gotcha, thanks! My prof did not cover that this lecture so I was unclear.

Okay so with that in mind ||a-b|| = sqrt(101) right?
and since 3 is a postive number it can be factored out thus,

3*sqrt(101) = ||3a-3b|| right?

concon said:
Okay gotcha, thanks! My prof did not cover that this lecture so I was unclear.

Okay so with that in mind ||a-b|| = sqrt(101) right?
and since 3 is a postive number it can be factored out thus,

3*sqrt(101) = ||3a-3b|| right?

Yes, that's correct.

So, can you use this to solve (b) now?

micromass said:
Yes, that's correct.

So, can you use this to solve (b) now?
I think so but is my equation correct? Which values am I using

micromass said:
Yes, that's correct.

So, can you use this to solve (b) now?
actually scratch that I got the answer. I got:
U = (1/sqrt 101, 10/sqrt 101, 0)

How do I do the last part about length?

Given your vector ##a##, then if ##\lambda>0## is a real number, then ##\lambda a## lies in the same direction of ##a##. So your answer will be ##\lambda a##. Your only job is to figure out what ##\lambda## is exactly.

You want ##\|\lambda a\| = 8##. Can you figure out what ##\lambda## is now?

yep got the answer as (8/sqrt 66 , 56/sqrt 66 , -32/sqrt 66)

Right!

## What is magnitude?

Magnitude refers to the size or quantity of a vector or physical quantity. It is represented by a numerical value and a unit, such as meters or kilograms.

## What are unit vectors?

Unit vectors are vectors with a magnitude of 1 and are used to indicate direction. They are typically represented by a lowercase letter with a caret (^) above it, such as ĵ for the unit vector in the y direction.

## Why do we use unit vectors?

Unit vectors are useful because they allow us to decompose a vector into its directional components. They also simplify calculations, as the magnitude of a vector can be multiplied by the unit vector to determine the component in that direction.

## How do we solve for unknown variables in equations involving magnitude and unit vectors?

To solve for unknown variables, we can use algebraic manipulation to rearrange the equation and isolate the unknown variable. We can also use the properties of vectors, such as the commutative and associative properties, to simplify calculations.

## What are some common physical quantities that have magnitude and unit vectors?

Some common physical quantities that have magnitude and unit vectors include displacement, velocity, acceleration, and force. These quantities are often described in terms of their magnitude and direction, which can be represented by unit vectors.

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