High School Common Core Mathematics

Front 1

Explain the process for adding vectors using their components.

Back 1

First, find the components of each vector. For example, Vector A can be noted as (Ax, Ay). Next, add the x components of the vectors together. This sum is equal to the x component of the resultant. Next, add the y components of the vectors together. This sum is equal to the y component of the resultant. Write the resultant vector in component form. Then, use the x and y components as lengths for the legs of a right triangle where the hypotenuse is vector C. Finally, use the Pythagorean theorem to find the magnitude of vector C.

Front 2

Find the sum of the vectors using their components.

Back 2

A+B=C
A=(Ax,Ay) A=(2,2)
B=(Bx,By) B=(1,2)
Ax+Bx=Cx 2+1=3
Ay+By=Cy 2+2=4
C=(3,4)
Cx²+Cy²=C²
3²+4²=C²
C=5

Front 3

Find the sum of the vectors using their components.

Back 3

A+B=C
A=(Ax,Ay) A=(3,2)
B=(Bx,By) B=(2,3)
Ax+Bx=Cx 3+2=5
Ay+By=Cy 2+3=5
C=(5,5)
Cx²+Cy²=C²
5²+5²=C²
C=7.07

Front 4

Explain the process for adding vectors using the parallelogram rule.

Back 4

First, draw the two vectors with their tails at the same point. Next, add the two sides necessary to make a parallelogram. Last, start with the corner where the vectors' tails meet, and draw a line from this point to the opposite corner. This vector is A+B or vector C.

Front 5

Show how to add these vectors using the parallelogram rule.

Back 5

Front 6

Show how to add these vectors using the parallelogram rule.

Back 6

Front 7

The magnitude of the sum of two vectors typically (is, is not) the sum of the magnitues.

Back 7

The magnitude of two vectors typically is not the sum of the magnitudes.

Front 8

Explain how to find the direction and magnitude of two vectors when they are stated in magnitude and direction form.

Back 8

First, draw the two vectors tail-to-tail. Then, translate so the vectors are drawn head-to-tail and draw the resultant. Next, find the supplementary angle to find an angle of the triangle, then use the law of cosines to find the magnitude of the resultant vector. Finally, use the law of sines to find the direction of the resultant vector.

Front 9

A vector u has a magnitude of 17m and a direction of 0deg. A vector v has a magnitude of 10m and a direction of 70deg. Find the direction and magnitude of u+v to the nearest whole values.

Back 9

m²=u²+v²-2uv cos(110°)
m²=(17)² + (10)² - 2(17)(10)cos(110°)
m²=289+100-340cos(110°)
m=√289+100-340cos(110°)
m=22.8
sin(110°)/22.8 = sinθ/10
θ=24.4°
u+v= 22.8m @ 24.4°

Front 10

Vector subtraction can be shown by v-w. This expression can also be written as ________.

Back 10

Vector subtraction can be shown by v-w. This expression can also be written as v+(-w).

Front 11

When showing vector subtraction as v+(-w), what does (-w) represent?

Back 11

(-w) represents the additive inverse of w. It is a vector with the same magnitude as w, but pointing in the opposite direction.

Front 12

How can vector subtraction be represented graphically?

Back 12

Vector subtraction can be represented graphically using the idea that, since a-b=a+(-b), vector b can be graphed as vector -b, which has the same magnitude as vector b, but in the opposite direction. Once graphed as such, vector addition is completed.

Front 13

Represent subtraction of vector B from vector A graphically. Find the components of the resultant.

Back 13

A-B=(-1,3)

Front 14

Represent subtraction of vector B from vector A graphically. Find the components of the resultant.

Back 14

A-B=(-4,3)

Front 15

Subtract vector b from a using their components. b=(5,7) a=(13,3)

Back 15

a-b=a+(-b)
a+(-b)=(ax+(-bx),(ay+(-by))
a+(-b)=(5-13),(7-3)
a+(-b)=(-8,4)

Front 16

subtract vector b from a using their components. b=(12,2) a=(2,8)

Back 16

a-b=a+(-b)
a+(-b)=(ax+(-bx), ay+(-by))
a+(-b)=(12-2),(2-8)
a+(-b)=(10,-6)

Front 17

Represent the multiplication of vector A by -2 graphically.

Back 17

Front 18

Represent the multiplication of vector B by 3.5

Back 18

Front 19

Multiply vector A by -4 using the components of vector A. A=(-6,14)

Back 19

A=(-6,14)
-4A=(-6*-4, 14*-6)
-4A=(24, -84)

Front 20

Multiply vector A by 1.75 using the components of vector A. A=(8,-4)

Back 20

A=(8,-4)
1.75A=(8*1.75, -4*1.75)
1.75A=(14, -7)

Front 21

In vector notation, what is the meaning of ||cv||? What is another way this can be written?

Back 21

||CV||means the scalar quantity C multiplied by the length (or norm) of vector V. This can also be written as |c|v

Front 22

How does one find the magnitude and direction of the resultant of scalar C times vector V?

Back 22

First, find the length (or norm) of vector V, then multiply this by the scalar quantity C to find the magnitude. The direction of CV is the same as the direction of vector V if C>0 and is the opposite as the direction of vector V if C<0.

Front 23

Find the magnitude and distance of the resultant R of scalar C times vector V. C=-2 V=(3,4) @ 30deg

Back 23

32+42=|V|2 9+16=|V|2 25=|V|2 |V|=5 ||CV||=-2*5 ||CV||=R Magnitude R=10 Direction R=-30deg because c<0.
*NOTE |V| does not represent the absolute value of V, but the norm (or length) of V. ||CV|| represents the norm (or length) of the product of CV.

Front 24

Find the magnitude and distance of the resultant R of scalar C times vector V. C=3 V=(2,6) @ 45deg

Back 24

22+62=|V|2 4+36=|V|2 40=|V|2 |V|=6.325 ||CV||=3*6.325 ||CV||=18.974 Magnitude R=18.974, Direction R=45deg because c>0