# Displacement problem with unit vectors

• Ursa
In summary: So in summary, the dot product of two vectors is a single number obtained by multiplying the corresponding components and adding them together. The component of one vector along another is found by dividing the dot product of the two vectors by the magnitude of the second vector. In order to find the component of one vector along another, the dot product must be calculated correctly first. The magnitude of a vector is the square root of the dot product of the vector with itself.
Ursa
Homework Statement
In a meeting of mimes, mime 1 goes through a displacement d1 = (7.07 m)i + (3.94 m)j and and mime 2 goes through a displacement d2 = (-7.03 m)i + (4.1 m)j. What are (a)|d1 × d2|, (b)d1 · d2, (c)(d1 + d2) · d2, and (d) the component of d1 along the direction of d2?
Relevant Equations
a · b = ab *cos ϕ
a · b = ax*bx+ ay*by+az*bz
a = sqrt(ax^2+ay^2)
rx=ax+bx
rx=ax+bx
(a) I did (7.07*4.1)-(-7.03*3.94)=56.7 with this method I got this answer correct in my first attempt.

(b) This where I seem to have gone wrong. I used a · b = (axbx +ayby) then I used a = sqrt(ax2+ay2) to get a single number for the answer. Filling in the numbers 7.07*-7.03 + 4.94*4.1 = -49.7i+16.2j. Then sqrt(-49.2+16.22)=52.3

(c) I first solved for (d1 + d2) with ax+bx 7.07+-7.03 + 4.94+4.1 = -0.04i+8.04j. Then used those unit vectors in the same way as in part (b) to get 32.97

(d) This one I didn't understand at all, I gave it a shot by trying to solve d1 *cos ϕ. first trying to find ϕ by a · b = ab *cos ϕ
cos-1((a · b)/ab). Finding d1 and d2 by a = sqrt(ax2+ay2
d1=8.09
d2= 8.14
cos-1(52.3/8.09 * 8.14)=37.4

Then 8.14 cos 37.4= 6.43 mI just fundamentally don't understand vectors and really don't get how to do math with them. Any analyses of what I'm doing wrong, explanation and/or drawings would be appreciated.

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Ursa said:
(b) This where I seem to have gone wrong. I used a · b = (axbx +ayby) then I used a = sqrt(ax2+ay2) to get a single number for the answer. Filling in the numbers 7.07*-7.03 + 4.94*4.1 = -49.7i+16.2j. Then sqrt(-49.2+16.22)=52.3
The dot product is just a number so: ##\vec d_1 \cdot \vec d_2 = 7.07*-7.03 + 3.94*4.1 = -49.7+16.2##

Note that should be ##3.94## in there.

Ursa said:
(d) This one I didn't understand at all,
The component of one vector along another is what you get if you imagine taking the second vector as one of your basis vectors. Specifically, the component of ##\vec d_1## along ##\vec d_2## is defined as:
$$\frac{\vec d_1 \cdot \vec d_2}{|\vec d_2|}$$

PeroK said:
The dot product is just a number so: ##\vec d_1 \cdot \vec d_2 = 7.07*-7.03 + 3.94*4.1 = -49.7+16.2##
So in order to get the answer I should just add ##-49.7+16.2## to get ##-33.5##?

PeroK said:
The component of one vector along another is what you get if you imagine taking the second vector as one of your basis vectors. Specifically, the component of ##\vec d_1## along ##\vec d_2## is defined as:
$$\frac{\vec d_1 \cdot \vec d_2}{|\vec d_2|}$$
Once I grasp how to get ##\vec d_1 \cdot \vec d_2## correct I have the top part, but could you expand on what exactly the denominator is? Will is be the ##d_2## = 8.14 which I calculated in my attempt?

Ursa said:
So in order to get the answer I should just add ##-49.7+16.2## to get ##-33.5##?Once I grasp how to get ##\vec d_1 \cdot \vec d_2## correct I have the top part, but could you expand on what exactly the denominator is? Will is be the ##d_2## = 8.14 which I calculated in my attempt?
Yes, and yes. If ##\vec d## is a vector, then ##|\vec d| \equiv d \equiv \sqrt{\vec d \cdot \vec d}## where ##|\vec d|## and ##d## are both common notations for the magnitude of a vector.

PeroK said:
Yes, and yes. If ##\vec d## is a vector, then ##|\vec d| \equiv d \equiv \sqrt{\vec d \cdot \vec d}## where ##|\vec d|## and ##d## are both common notations for the magnitude of a vector.
Thank you, now I understand it a bit better.

PeroK

## 1. What is displacement in physics?

Displacement in physics is a vector quantity that measures the distance and direction between an object's initial position and its final position.

## 2. How is displacement different from distance?

Displacement is a vector quantity that takes into account the direction of an object's motion, while distance is a scalar quantity that only measures the total length traveled by an object.

## 3. What are unit vectors?

Unit vectors are vectors with a magnitude of 1 that point in a specific direction. They are used to represent the direction of motion in displacement problems.

## 4. How do you solve a displacement problem with unit vectors?

To solve a displacement problem with unit vectors, you first need to determine the initial and final positions of the object. Then, calculate the difference between the two positions to find the displacement vector. Finally, use unit vectors to represent the direction of the displacement.

## 5. What is the importance of using unit vectors in displacement problems?

Unit vectors are important in displacement problems because they help to accurately represent the direction of motion. They also allow for easier calculations and provide a standard way to express direction in physics.