MCV4U1
Part 1.
5.2 Vector Operations in Component/Coordinate Form
Vector Addition/Subtraction
v
v
1. Given the vectors a = (2, 4) and b = (6, 1) :
v
v
(a) Graph a and b . (use the grid below)
v
r v
r v
r v
(c) Express a + b and a − b as ordered pairs.
v
r
v
(b) Find a + b and a − b by translating the initial point of b to the terminal point of a .
2. Use the results of #1 to complete the general statement below.
v
(
)
v
If a = a1 , a2 and b =
v v
a+b = (
(b , b ) , then
1
2
,
)&
v v
a−b = (
,
)
3. Prove generally using components.
r
v
r
v
4. Use the general rule above to find a + b and a − b for each pair of vectors given below.
v
v
a = (3,7), b = ( − 1, 6)
v v
(a) a + b =
v v
a−b =
v
v
a = (2,− 11), b = ( − 7, − 1)
v v
(b) a + b =
v v
a−b =
Part 2.
v
v
a = ( − 1,− 1), b = (0, − 3)
v v
(c) a + b =
v v
a−b =
Scalar Multiplication
v
1. Given the vector c = (4, − 2) :
v
(a) Graph c .
(b) Draw a position vector (O is origin) collinear with and
v
in the same direction as c but 2 times as long.
(c) Express this vector as an ordered pair.
(d) Draw a position vector collinear with but in the
v
opposite direction of c but 1 2 as long.
(e) Express this vector as an ordered pair.
2. Use the above results to complete the following
general statement.
v
If c =
(c , c ) , and k ∈ R , then
1
2
v
kc = (
,
)
3. Use the box above to write the following as single vectors.
(a) 3(7, 11)
(b) −
1
(0, 9)
3
(c) 0(-6, 1)
Part 3.
Linear Combinations (i.e. add/subt scalar multiples of vectors)
1. Use Parts 1 and 2 to express the following linear combinations as single vectors.
(a) 2(3, -1) - (-4, 7)
(b) (3, 6) + 3(-1, -2) - (6, -1)
(c) 5[(-2, 7) - (3, 0)]
Part 4.
Position Vectors and Vectors Joining Two Points (i.e., point-to-point vectors)
1. Given the position vectors OP =
(4,1) and OQ = (8,6) .
(a) Graph OP and OQ (use the grid above or your own grid)
(b) Draw the vector PQ
(c) Translate PQ such that it too is a position vector {i.e., initial
point (0, 0)}
(d) Write PQ as an ordered pair.
2. Repeat #1 for OP =
(3,− 1) and OQ = (− 2,− 5)
3. Use the results from #1 and #2 above to complete the following general statement.
(
If OP = p1 , p2
) and OQ = (q , q ) , then PQ = (
1
2
,
)
Quick Proof Using Vector Addition...
4. Use the general rule above to find both PQ and QP for each of the following.
OP = (3, − 7) and OQ = ( − 1, 2)
OP = (1, − 1) and OQ = (5, − 2)
(a) PQ = (
(b) PQ = (
QP = (
QP = (
Part 5.
More Linear Combinations
1. Given three points in the plane, P(2, 7) , Q(-3, 0), and R(-5, -1), use Parts 1 to 4 to express the following
as a single vector.
2 PQ − RQ + 3( RP − PR)
Part 6.
Magnitudes
of Vectors and Unit Vectors
v
1. Given the vector v = (3,4) ,
(a) Graph it.
v
(b) Use Pythagoras to determine v
v
2. Repeat #1 for v = ( − 7,− 1) .
3. Use the results above to complete the following general
statement.
v
v
If v = (a , b) ,
then v =
4. Use the box above to find the magnitude of each of the following vectors.
v
p = (3,− 2)
(a) v
p=
v
t = (1,0)
(c) v
t =
v
s = (− 1,− 1)
(b) v
s =
v
v = 21 , 23
(d)
v
v =
(
)
Recall, a unit vector is a vector having magnitude 1.
v
v
Definition: To normalize a vector v means to find a unit vector parallel to and in the same direction as v .
1 v
v
Recall, v$ = v v
5. Normalize the following vectors. (Also check that v$ = 1 , t$ = 1 and p$ = 1 .)
r
(a) v =
v
(3,7)
(
v
)
(b) t = − 2,− 1
(c) p =
Let’s extend all of these operations into space (3-D)....
v
v
a = a1 , a2 , a3 and b = b1 , b2 , b3 and k is a scalar.
v v
v v
a+b =
a−b =
(
)
(
)
v
ka =
v
a =
(
If P x1 , y1 , z1
PQ = (
) and Q( x , y , z ) are 2 points in space then
2
2
,
( 21 , 21 )
2
,
)