13.1 Vectors: Displacement Vectors
Some physical quantities can be completely defined by magnitudes (speed, mass, length, time, etc.)
These are called scalars. Other quantities need a magnitude and direction to be completely defined
(displacement, velocity, force, etc.) These are called vectors.
The displacement vector 𝑣⃗ from one point to another is an arrow with its tail at the first point and its tip
at the second point. The magnitude of the displacement vector 𝑣⃗ , denoted by |𝑣⃗| , is the distance
between the points and represented by the length of its arrow. The direction of the displacement vector
is the direction of its arrow, and it is the angle measured from the positive x-axis.
Properties of Vectors
1.
Two vectors are equal 𝑣⃗ = 𝑢
⃗⃗, if they have the same magnitude and direction.
2.
The negative of the vector 𝑣⃗ is a vector with same magnitude but opposite direction.
3.
The product of a vector 𝑣⃗ with a constant k is a vector with magnitude k times the magnitude of
𝑣⃗ in, the direction of 𝑣⃗.
4.
The zero-vector is a vector with no length and no specified direction.
5.
We add two vectors geometrically by using the parallelogram law. Place the tail of the second
vector at the tip of the first vector, and connect the tail of the first vector at the tip of the second
vector with another vector. This new vector is called the resultant vector.
See the vector addition applet at http://comp.uark.edu/~jgeabana/java/VectorCalc.html
6.
To subtract two vectors geometrically place the two tails of the vectors together and connect the
two tips with another vector. This vector is in the direction tip minus tail.
See vector subtraction applet at http://www.frontiernet.net/~imaging/vector_calculator.html
Vectors in Coordinate Systems on the Plane (2-space)
1.
A position vector has its initial point at the origin of the coordinate system and the final point at
some other point on the coordinate plane.
2.
A position vector can be expressed by listing its components (coordinates of the tip of a position
vector). 𝑣⃗ =  v1 , v2 
eg. The vector < 2, 4 > can be interpreted as a position vector that points to the point (2, 4) on
the plane.
1
3.
Any vector can be made a position vector. If the vector

w
= < w1, w2 > from the point
w = < x2 – x1, y2 – y1 > where P1 is
P1 = (x1 y1) to the point P2 = (x2, y2) is not at the origin, then 
the initial point, and P2 is the final point (tip minus tail). This is the same as subtracting two
position vectors one at the tip the other at the tail. 𝑤
⃗⃗⃗ = 𝑢
⃗⃗ − 𝑣⃗ where 𝑢
⃗⃗ is the tip vector and 𝑣⃗ is
the tail vector.
e.g. The vector from point (2, 3) to point (5, - 2), can be represented as < 3, - 5 >.
If you subtract the position vector 𝑣⃗ =< 2,3 > from vector 𝑢
⃗⃗ =< 5, −2 >, the vector 𝑢
⃗⃗ − 𝑣⃗ =
< 3, − 5 >.

4.
The zero-vector is expressed 0 = < 0, 0 >.
5.
To add or subtract two vectors algebraically, add the respective components. If 𝑢
⃗⃗ = < u1, u2 >,
and 𝑣⃗ = < v1, v2 >, then 𝑢
⃗⃗  𝑣⃗ = < u1  v1, u2  v2 >
e.g. Draw the vectors 𝑢
⃗⃗ = < 3, 1 >, 𝑣⃗ = < 2, 4 > and 𝑢
⃗⃗ +𝑣⃗.
6.
If a vector is multiplied by a constant k (scalar multiplication) then each component is multiplied
by k. k 𝑣⃗ = < kv1 , kv2 >
Vectors in coordinate systems in space (3-space)
1)
2)
3)
4)
5)
6)
A position vector has its initial point at the origin of the coordinate system.
A position vector can be expressed by listing its components (coordinates of the tip of a position
vector). 𝑣⃗ = < 𝑣1 , 𝑣2 𝑣3 >.
Any vector can be made a position vector. If the vector 𝑤
⃗⃗⃗ = < 𝑤1 , 𝑤2 , 𝑤3 > is not at the origin,
but from the point P1 = (x1, y1, z1) to the point P2 = (x2, y2, z2), then
𝑤
⃗⃗⃗ = < 𝑥2 − 𝑥1 , 𝑦2 − 𝑦1 , 𝑧2 − 𝑧1 > where P1 is the initial point, and P2 is the final point. (tip
minus tail). This is the same as subtracting two position vectors one at the tip the other at the
tail. 𝑤
⃗⃗⃗ = 𝑢
⃗⃗ − 𝑣⃗ where 𝑢
⃗⃗ is the tip vector and 𝑣⃗ is the tail vector.
The zero-vector is expressed ⃗0⃗ = < 0,0,0 >
To add or subtract two vectors add the respective components. If 𝑢
⃗⃗ = < 𝑢1 , 𝑢2 , 𝑢3 >, and
𝑣⃗ = < 𝑣1 , 𝑣2 , 𝑣3 >, then 𝑢
⃗⃗ ± 𝑣⃗ = < 𝑢1 ± 𝑣1 , 𝑢2 ± 𝑢2 , 𝑢3 ± 𝑣3 >.
If a vector is multiplied by a constant k (scalar multiplication) then each component is multiplied
by k. 𝑘𝑣⃗ = < 𝑘𝑣1 , 𝑘𝑣2 , 𝑘𝑣3 >.
Norm or magnitude (length) of a vector
The norm of a vector from the point P1 = (x1, y1) to the point P2 = (x2, y2), in 2-space, is given by the
distance formula as
|𝑣⃗| = 𝑣 = √ 𝑣12 + 𝑣22 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
2
The norm of a vector from the point P1 = (x1, y1, z1) to the point P2 = (x2, y2, z2), in 3-space, is given by
the distance formula as |𝑣⃗| = 𝑣 = √𝑣12 + 𝑣22 + 𝑣32 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧1 )2
Unit vectors on the plane
A unit vector is a vector with magnitude of one unit. 𝑣̂ =
⃗⃗
𝑣
|𝑣
⃗⃗|
. A unit vector gives direction.
Vectors of importance are the unit vectors in the direction of the coordinate axis
̂𝑖 = < 1, 0 >; ̂𝑗 = < 0, 1 >.
Any vector can be expressed in terms of the unit vectors ̂,
𝑖 ̂𝑗 .
eg1 Express < 2,5 > as a vector in terms of ̂,
𝑖 ̂.
𝑗
< 2, 5 > = 2 ̂𝑖 +5 ̂𝑗
Check: 2 ̂𝑖 + 5̂𝑗 = 2 < 1, 0 > + 5 < 0, 1 > = < 2, 5 >.
eg2 Express < 2,5 > as a unit vector.
2
5
│< 2, 5 >│ = 29 , so <
> is a unit vector.
,
29
29

eg3 Find a vector 3 units long in the direction of A = < 2, −1 >.
3  2,1   6,3 
3𝐴̂ =
.

5
5
Any vector can be expressed as a product of its magnitude and its direction.
r
= | r | r̂
r = |r |
|r |
eg4 Express r = < 2,5 > as a product of its magnitude and its direction.
2
5
,
 | r | rˆ, where 29 is the magnitude of the vector < 2,5 >, and
r = < 2,5 > = 29 <
29 29
2
5

,
 is the unit vector or direction of the vector < 2,5 >.
29 29
Unit vectors in space:
⃗⃗
𝑣
Vector with magnitude of one unit 𝑣̂ = |𝑣⃗⃗| . The vectors in the direction of the coordinate axis are
̂𝑖 = < 1, 0, 0 > ; ̂𝑗 = < 0, 1, 0 > ; 𝑘̂ = < 0, 0, 1 >.
eg5 Give the direction of the vector 𝑣
⃗⃗⃗⃗ = < 4, −4, 2 >
Since 𝑣⃗ = < 4. −4,2 > = 2 < 2. −2, 1 > , 𝑣̂ =
2<2.−2,1>
2√9
=<
2
3
.
−2 1
3
, 3 > is the direction of the vector 𝑣⃗.
Any vector in space can be expressed in terms of the unit vectors ̂,
𝑖 ̂,
𝑗 ̂𝑘 .
3
eg6 < 2,5,-1 > = 2 ̂𝑖 + 5 ̂𝑗 − ̂𝑘 = 2 < 1, 0, 0 > + 5 < 0, 1, 0 > - < 0, 0, 1 > =
< 2, 0, 0 > + < 0, 5, 0 > + < 0, 0, - 1 > = < 2,5,-1 >
Any vector in space can be expressed as a product of its magnitude and its direction.
⃗⃗⃗⃗⃗ 𝑟⃗ = |𝑟|
⃗⃗⃗⃗⃗ r̂
𝑟⃗ = |𝑟|
|𝑟⃗|
Vectors in Polar Coordinates
A vector 𝑟⃗ = < 𝑥, 𝑦 > can be expressed in polar coordinates as 𝑟⃗⃗⃗ =  rcos rsin   where r = |𝑟⃗| and
𝜃 is the angle with the positive x-axis.
𝑟⃗
The unit vector 𝑟̂ becomes 𝑟̂ = |𝑟⃗| =
<𝑟𝑐𝑜𝑠𝜃,𝑟𝑠𝑖𝑛 𝜃>
𝑟
= < cos 𝜃, sin 𝜃 >.

eg7 Find the position vector v with magnitude of 2 and angle of inclination of 

v  2 cos(
4
4
), 2sin(
)     1,
3
3
4
.
3
3 .
eg8 Find the unit vector that makes an angle of 3π/4 with the horizontal axes.
𝑟̂ = < cos 𝜃, sin 𝜃> = < cos3π/4, sin3π/4 > = <
−√2 √2
, 2
2
eg9 Give the angle of inclination of the vector < coordinates.
>=
−√2
2
̂𝑖 +
√2
2
𝑗̂ .
3 , - 1 >, and represent the vector in polar
Since this vector is a position vector in the third quadrant, the angle will be tan (𝜃) =
7𝜋
1
√3
,𝜃 =
7𝜋
6
in
7𝜋
QIII. Since the magnitude of the vector is 2, so the desired vector is < 2 cos ( 6 ) , 2 sin ( 6 ) >.
Homework 13.1
1. The figure below shows force vectors F, G, and H and their x- and y- components. Calculate
their resultant F + G + H and add it to the drawing.
Ans: 6,11
4
2. Give the two vectors of length 10 that are perpendicular to 3, 4 . Then draw the three vectors.
Ans: 8, 6 , 8, 6
3
3. Find an angle of inclination of 5, 3 . Give an exact answer. Ans:   tan 1  
5
4. Find numbers a and b such that a 3, 1 + b 1, 2 = 1, 12 . Ans: (a = 2, b = -5)
5. What vector of length 7 has the same direction as PQ where P = (-5, -3) and Q = (4, -8)?
7 9, 5
106
Ans:
6. Three ropes are supporting a 10 lb. weight. Two of the ropes exert forces 𝑢
⃗⃗ = < 2, 3, 4 >lb and
𝑣⃗ = < −1, −2, 3 > lb in the xyz space with the positive z-axis pointing up. What is the tension
in the third rope? Ans:[ < -1, -1, 3 >, √11].
7. Consider the vector v  i  2 j  7k . Find a vector that
a. Is parallel but not equal to v .
b. Points in the opposite direction of v .
c. Has unit length and is parallel to v
8. What would you get if you drew all possible unit vectors in 3-space with tails at the origin?
Ans: a unit sphere.
9. What object in three space is traced by the tips of all vectors starting at the origin that arc of the
form
a) v  i  j  bk where b is any real number.
b) v  i  aj  bk where a and b are any real numbers.
Ans Line parallel to z-axis through point (1,1), plane x=1
10. Describe the object created by all scalar multiples of v  i  j with tail at the point (0, 0, 1).
Ans: Line through point (0,0,1) parallel to v  i  j
11. Decide if each of the following statements is true or false.
a) The length of the sum of two vectors is always strictly larger than the sum of the lengths of
the two vectors.
b) v  v1  v2  v3 , where v  v1i  v2 j  v3k .
c)  i ,  j ,  k are the only unit vectors.
d) v and w are parallel if v   w for some scalar λ .
e) Any two parallel vectors point in the same direction.
f) 2v has twice the magnitude as v .
Ans: Only d and f are true.
5
13.2 Vectors In General
eg 10 Give the components of the velocity vector of an airplane moving at 45mph in the direction 30ο
east of south.
The components are< 45 cos(60°), −45 sin(60°) >= < 22.5 , −22.5√3 >.
eg 11 An airplane with airspeed of 200 mph is headed N60oW. A wind is blowing directly south. If the
true direction of the airplane is west, find the true speed of the airplane and the speed of the wind.
The true speed of the plane is the x-component 200sin60o = 100 3 mph.
The speed of the wind is the y-component 200cos 60o = 100 mph.
eg 12 A ship is traveling at a speed of 22.0 knots at a compass heading of 95.0o (measured clockwise
from the north). The current is flowing due south at a speed of 5.00 knots. Find the actual speed υ and
compass heading 𝜃 of the ship.
If we use trigonometry, since the angle between the direction of the ship and the current is 95.0o, by law
of cosines, the actual speed of the ship |𝑣⃗| =
sig. If we use the law of sines,
222  52  2* 22*5*cos(95o )  22.98  23.0 knots to 3
sin(95o ) sin( )

, giving   72.5o, so the actual heading of the ship is
22.98
22.0
107.5o.
If we use vectors, the ship vector 𝑠⃗ = < 22.0 cos(5.0o), − 22.0 sin(5.0o) > and the current vector
𝑐⃗ = < 0, −5.00 >. If we add the two vectors, we have 𝑠⃗ + 𝑐⃗ = < 22.0 cos(5o), − 22.0 sin(5.0o) – 5.00 >.
The magnitude of the vector 23.0 knots is the actual speed. The direction of the vector tan (  ) =
𝑦
𝑥
, 𝑠𝑜 𝛾 ≈ - 17.5o is the reference angle, so the actual heading of the ship is 107.5o.
eg 13 Find the resultant 𝐹⃗ force at a point, if the angle of 𝐹⃗1 with magnitude 8 lb. is 135o, and the angle of
𝐹⃗2 with magnitude 6 lb. is 30o at that point.
3𝜋
3𝜋
𝜋
Since 𝐹⃗1 = < 8 cos 4 , 8 sin 4 > ; 𝐹⃗2 = < 6 cos 6 , 6𝑠𝑖𝑛
𝜋
6
>,
3𝜋
𝜋
3𝜋
𝜋
+ 6 cos , 8 sin
+ 6 sin > =
4
6
4
6
4 2 3
y
, or θ ≈ -87o + 180o = 93o
  4 2  3 3, 4 2  3  with direction tan (θ) = =
x 4 2 3 3
𝐹⃗1 + 𝐹⃗2 = < 8 cos
6
eg 14 An object is pulled by a force 𝐹⃗ in the direction of < 1, 1 > and force 𝐺⃗ in the direction of
< 3, - 2 >. Find the two forces if the resultant force is < 50, 0 >.
Since 𝐹⃗ + 𝐺⃗ = 𝑎 < 1, 1 > +𝑏 < 3, −2 > = < 50, 0 > ⇒ 𝑎 + 3𝑏 = 50 𝑎𝑛𝑑 𝑎 − 2𝑏 = 0. By solving
the system, b = 10 and a = 20, so 𝐹⃗ = 20 < 1,1 > and 𝐺⃗ = 10 < 3, - 2 >.
𝜋
𝜋
⃗⃗ = 2 ∠
eg 15 An object is pulled by the forces 𝐹⃗ = 2 ∠ 3 , 𝐺⃗ = 2√2 ∠ - 4 and 𝐻
𝜋
6
,
Find the resultant force, with its magnitude and direction.
If we represent the angles as unit vectors, the resultant force will be

R 2
 1, 3 
,1,1 
 3,1 
2 2
2
  3  3, 3  1  with magnitude 16  4 3 and direction
2
2
2
θ = tan-1
3 1
, since θ is in the first quadrant.
3 3
eg 16 A 100 lb. weight hangs from two wires. The wire to the left has a tension T1 and an angle of
depression of 50o and the one to the right has a tension T2 and an angle of depression of 32o. Find the
two tensions.


Let T1  T 1 and T2  T 2

T 1 = T1 < - cos(50o), sin(50o) > and

T 2 = T2 < cos(32o), sin(32o) >.

Since the two tensions counterbalance the weight w = < 0, - 100 >, the sum of the x-components is zero
and the sum of the y-components is 100.
So; – T1 cos(50o) + T2 cos(32o) = 0 and T1 sin(50o) + T2 sin(32o) = 100.
By solving the system, T1 =
100𝑐𝑜𝑠(32°)
𝑠𝑖𝑛(82°)
≈ 85.64lb and T2
100 𝑐𝑜𝑠(50°)
𝑠𝑖𝑛 (82°)
We have used the identity sin(a + b) = sin(a)cos(b) + sin(b)cos(a).
7
= ≈ 64.91 lb.
Homework 13.2
1. A skater skates in the direction of the vector 4, 3 from point P to point R and then in the
direction of 4, 2 to Q, where PQ = 0, 50 meters (figure below). How far is R from P
and how far is R from Q? Ans: PR  50 , QR  20 5
2. The force of the wind on a rubber raft is twenty-five pounds toward the southwest. What force
must be exerted by the raft’s motor for the combined force of the wind and motor to be twenty
25 25
pounds toward the south? Ans:
,
 20 .
2 2
3. If weight of 500 lb is supported as shown in the figure, draw a force diagram to find the exact
magnitude of the forces in the members CB and AB.
3 m.
A
1m.
B
500
l
b
C
Ans: CB: 1000lb, AB: 500 3 lb.
8
4.
A 75lb. weight is suspended from two wires. If 𝐹⃗1 makes an angle of 55° with the horizontal,
and 𝐹⃗2 makes an angle of 40◦ with the horizontal, find the magnitude of the two forces.
Ans: 𝐹⃗1 = 57.67lb. , 𝐹⃗2 = 43.18lb.
5. The boat in the figure below is being pulled into a dock by two ropes. The force F by the upper
rope has the direction of the vector 2, 2 with the usual orientation of coordinate axes, and the
force G by the lower rope is in the direction of 6, 4 . The total force on the boat by the two
ropes is 100, 0 pounds. What are the magnitudes of the forces by the two ropes?
Ans: upper 40 2 , lower 20 13
6. Two wires suspend a twenty-five-pound weight. One wire makes an angle of 45° with the
vertical and the tension in it (the magnitude of the force it exerts) is twenty pounds. What is the
tension in the other rope and what angle does it make with the vertical? (Notice that the angle
with the vertical is not the angle of inclination.) Ans: 17.83 lb 52.48 .
7. A sailor tacks toward the northeast from point P to point R and then tacks toward the northwest
to Q (figure below). The point Q is 300 meters north and 100 meters east of P. How far does the
sailor travel on each of the tacks? Ans: PR  200 2 , QR  100 2
8. The figure below shows four forces, measured in Newtons that are applied to a ring. Find a
magnitude and angle of inclination of their resultant.
Ans: 17. 3N 8 .
9
9. A plane is heading due east and climbing at the rate of 80 km/hr. If its airspeed is 480 km/hr. and
there is a wind blowing 100 km/hr. to the northeast, what is the ground speed of the plane?
Ans: 549 km/hr.
10. An airplane heads northeast at an airspeed of 700 km/hr, but there is a wind blowing from the
west at 60 km/hr. In what direction does the plane end up flying? What is its speed relative to the
ground? Ans: 48.3° relative to North, 744 km/hr.
13.3 The Dot Product
Dot Product: (Scalar Product)
The dot product is defined geometrically by
|𝑢
⃗⃗| |𝑣⃗|𝑐𝑜𝑠𝜃
𝑢
⃗⃗ ∙ 𝑣⃗ = {0
𝑢 ≠ 0, 𝑣 ≠ 0
𝑢 = 0 𝑜𝑟 𝑣 = 0
where θ is the angle between the vectors 𝑢⃗⃗ and 𝑣⃗ (0 ≤ 𝜃 ≤ π).
If 𝑢
⃗⃗= < u1, u2 > and 𝑣⃗ = < υ1, υ2 > then the dot product is defined in the Cartesian coordinate system by
𝑢
⃗⃗ ∙ 𝑣⃗ = u1 υ1 + u2 υ2.
Proof: Consider 𝑢
⃗⃗ = < u1, u2 > and 𝑣⃗ = < υ1, υ2 >.
Since the vector 𝑢
⃗⃗ − 𝑣⃗ = < u1 − υ1, u2 − υ2> , the magnitude squared of this vector will be |𝑢
⃗⃗ − 𝑣⃗|2 =
(u1 – υ1)2 + (u2 – υ2)2. This last expression can be expanded and rearranged as (u12 + u22) + (v12 + v22) −
2(u1υ1 + u2υ2), so |𝑢
⃗⃗ − 𝑣⃗|2 = |𝑢
⃗⃗|2 + |𝑣⃗|2 - 2 (u1v1 + u2v2).
If we use the law of cosines with θ the angle between 𝑢
⃗⃗ and 𝑣⃗, we can say
2
2
2
|𝑢
⃗⃗ − 𝑣⃗| = |𝑢
⃗⃗| + |𝑣⃗| - 2 |𝑢
⃗⃗||𝑣⃗| cos(𝜃).
By comparing the last two expressions we can conclude that
|𝑢
⃗⃗||𝑣⃗|cos (θ) = u1v1 + u2υ2.
In space, if 𝑢
⃗⃗ = < 𝑢1 , 𝑢2 𝑢3 > and 𝑣⃗ = < 𝑣1 𝑣2 , 𝑣3 > then the dot product is defined in the Cartesian
coordinate system by 𝑢
⃗⃗ ∙ 𝑣⃗ = 𝑢1 𝑣1 + 𝑢2 𝑣2 + 𝑢3 𝑣3 .
𝜋
eg 17 If 𝑢
⃗⃗ = < -3,0 > and 𝑣⃗ = < - 1, - 1 > then 𝑢
⃗⃗ ⋅ 𝑣⃗ = |𝑢
⃗⃗||𝑣⃗|𝑐𝑜𝑠𝜃 = (3)(√2)𝑐𝑜𝑠 4 = 3; since θ is
Also, by using the definition, 𝑢
⃗⃗ ⋅ 𝑣⃗ = (-3) (-1) + (0) (-1) = 3.
Note: The dot product gives a number.
10
𝜋
4
.
Angle Between Two Vectors
The angle between two non-zero vectors in 2-space can be found by using the dot product.
acute angle if 𝑢
⃗⃗ ⋅ 𝑣⃗ > 0
𝑢
⃗⃗ ∙ 𝑣⃗
⃗⃗ ⋅ 𝑣⃗ < 0
𝑐𝑜𝑠(𝜃) =
; 𝜃 = { obtuse angle if 𝑢
|𝑢
⃗⃗||𝑣⃗|
𝜋⁄ 𝑖𝑓 𝑢
2 ⃗⃗ ⋅ 𝑣⃗ = 0
Two non-zero vectors are orthogonal (perpendicular) if 𝑢
⃗⃗ ⋅ 𝑣⃗ = 0.
The vectors ̂,
𝑖 ̂𝑗 and ̂𝑘 are orthogonal since ̂𝑖 ∙̂𝑗 = 0, ̂𝑗 ∙ 𝑘̂ = 0, 𝑘̂ ∙̂𝑖 = 0.
eg 18 Find the angle between the vectors < 0, 1 > and < -1, - √3 >.
cos (𝜃) =
⃗⃗∙𝑣
⃗⃗
𝑢
|𝑢
⃗⃗||𝑣
⃗⃗|
=
−√3
2
,𝜃 =
5𝜋
6
.
Properties of Dot Products:
For any vectors 𝑢
⃗⃗, 𝑣⃗, 𝑤
⃗⃗⃗ and any scalar k,
a)
𝑢
⃗⃗ ∙ 𝑣⃗ = 𝑣⃗ ∙ 𝑢
⃗⃗
b)
𝑢
⃗⃗ ∙ (𝑣⃗ + 𝑤
⃗⃗⃗) = 𝑢
⃗⃗ ∙ 𝑣⃗ + 𝑢
⃗⃗ ∙ 𝑤
⃗⃗⃗
c)
k (𝑢
⃗⃗ ⋅ 𝑣⃗ ) = (𝑘 𝑢
⃗⃗ ⋅ 𝑣⃗ )= (𝑢
⃗⃗ ∙ 𝑘 𝑣⃗)
d)
𝑣⃗ ∙ 𝑣⃗ = |𝑣⃗||𝑣⃗| = |𝑣⃗|2 or |𝑣⃗| = √𝑣⃗ ⋅ 𝑣⃗
See a Dot Product applet at http://falstad.com/dotproduct/
See a Dot Product calculator at http://hyperphysics.phy-astr.gsu.edu/Hbase/vsca.html
Work
⃗⃗ is given by 𝑊 = 𝐹⃗ ∙ 𝐷
⃗⃗ =
The work W done by a constant force 𝐹⃗ in the direction of motion 𝐷
⃗⃗| cos 𝜃 where 𝜃 is the angle between the force and the direction of motion.
|𝐹⃗ | |𝐷
𝜋
eg 19 Find the work needed to move an object 3m if a 6N force is applied to the object at an angle of 6
with the direction of motion.
⃗⃗ = (3) (6) cos𝜋 = 9√3 𝑁𝑚 = 9√3𝐽 (𝑗𝑜𝑢𝑙𝑒𝑠).
W = 𝐹⃗ ∙ 𝐷
6
or, < 3, 0 > ∙ < 6cos (π⁄6), 6sin (𝜋/6) > = 9√3𝐽
eg. 20 A constant force with vector representation F = i + 2j moves an object along a straight line from
the point (2, 4) to the point (5, 7). Find the work done in foot-pounds if force is measured in pounds and
distance is measured in feet. Ans. [9 ft-lb]
11
Directional Cosines
Finding angles between vectors in 3-space and the coordinate axis
Consider the unit vector 𝑢̂ =
⃗⃗
𝑢
|𝑢
⃗⃗|
<𝑢1 ,𝑢2 ,𝑢3 >
|𝑢
⃗⃗|
=
𝑢1
|𝑢
⃗⃗|
= <
𝑢
𝑢
, |𝑢⃗⃗|2 |𝑢⃗⃗|3 > .
Any unit vector can be expressed in terms of its directional cosines ∝, 𝛽, 𝛾, since 𝑐𝑜𝑠 ∝ =
𝑐𝑜𝑠𝛽 =
⃗⃗⃗⃗∙ ̂𝑗
𝑢
|𝑢
⃗⃗|
𝑢2
|𝑢
⃗⃗|
=
⃗⃗⃗⃗∙ ̂𝑘
𝑢
|𝑢
⃗⃗|
, 𝑐𝑜𝑠𝛾 =
𝑢
= |𝑢⃗⃗|3 and 𝑢̂ =
⃗⃗
𝑢
|𝑢
⃗⃗|
= <
𝑢1
|𝑢
⃗⃗|
𝑢 𝑢3
, |𝑢⃗⃗|2
⃗⃗
𝑢
⃗⃗ ∙̂𝑖
𝑢
|𝑢
⃗⃗|
=
𝑢1
|𝑢
⃗⃗|
> = <cos ∝, 𝑐𝑜𝑠𝛽, 𝑐𝑜𝑠𝛾 >. From the
components of any unit vector in 3-space, we can obtain the directional cosines where
𝑐𝑜𝑠 2 ∝ + 𝑐𝑜𝑠 2 β + 𝑐𝑜𝑠 2 γ =
u21
|u
⃗⃗|2
+
u22
|u
⃗⃗|2
+
u23
|u
⃗⃗|2
u21 + u22 + u23
=
2
= 1.
(√u21 + u22 + u23 )
eg 21 Find the directional cosines of 𝑢
⃗⃗ = < 2, - 1, -2 >.
⃗⃗
𝑢
𝑢̂ = |𝑢⃗⃗| = 𝑢̂ =
<2.−1,−2>
‖<2.−1.−2>‖
2
=<3,
−1 −2
,
3
3
>. So 𝑐𝑜𝑠 ∝ =
2
3
, cos 𝛽 =
−1
3
, 𝑐𝑜𝑠𝛾 =
−2
3
.
The directional cosines apply also to vectors in 2D. A unit vector in 2D can be represented as
𝜋
< cos ∝, cos 𝛽 > = < cos ∝, cos ( 2 − ∝) > = < cos ∝ 𝑠𝑖𝑛 ∝ > since ∝ and 𝛽 are complementary
angles.
Projections
The component of the vector 𝑏⃗⃗ along 𝑎⃗ is |𝑏⃗⃗| cos 𝜃 where 𝜃 is the angle between 𝑏⃗⃗ and 𝑎⃗.
It is expressed as comp→ 𝑏⃗⃗ =
⃗⃗| cos 𝜃
|𝑎⃗⃗||𝑏
|𝑎⃗⃗|
𝑎
=
⃗⃗
𝑎⃗⃗∙𝑏
|𝑎⃗⃗|
= |𝑏⃗⃗| cos(𝜃).
The projection (also called parallel projection) vector of 𝑏⃗⃗ along 𝑎⃗ and is expressed as
𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ = |𝑏⃗⃗| 𝑐𝑜𝑠 𝜃 𝑎̂ =
⃗⃗⃗⃗| cos 𝜃
|𝑎⃗⃗||𝑏
|𝑎⃗⃗|
𝑎
𝑎̂ =
⃗⃗
⃗⃗⃗⃗𝑏
𝑎∙
|𝑎⃗⃗|
𝑎̂ =
⃗⃗
𝑎⃗⃗∙𝑏
𝑎⃗.
|𝑎⃗⃗|2
The projection vector of 𝑏⃗⃗ orthogonal to 𝑎⃗ (also called perpendicular projection) is
𝑐⃗ = orth→ 𝑏⃗⃗ = 𝑏⃗⃗ − 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ = 𝑏⃗⃗ −
𝑎
𝑎
⃗⃗
𝑎⃗⃗∙𝑏
|𝑎⃗⃗|2
𝑎⃗.
eg. 22 Let 𝑏⃗⃗ = < 3, 2 > ; 𝑎
⃗⃗⃗⃗ = < 2, 1 >;
𝑐𝑜𝑚𝑝 → 𝑏⃗⃗ =
𝑎
⃗⃗
𝑎⃗⃗∙𝑏
|𝑎|
=
8
√5
; 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ =
orth→ 𝑏⃗⃗ = < 3, 2 > −
𝑎
𝑎
8
5
< 2, 1 > =
⃗⃗⃗⃗∙⃗⃗⃗⃗
𝑎
𝑏
|𝑎
⃗⃗⃗⃗|2
1
5
⃗⃗⃗⃗ =
𝑎
8
5
< 2, 1 > ;
< −1, 2 >.
eg 23 Let 𝑏⃗⃗ = < 3, 2, −1 > ; 𝑎⃗ = < 2, 1, 1>;
12
,
⃗⃗
𝑎⃗⃗ ∙ 𝑏
6+2−1
𝑐𝑜𝑚𝑝 → 𝑏⃗⃗ = |𝑎⃗⃗| =
=
√6
𝑎
𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ =
𝑎
⃗⃗
𝑎⃗⃗ ∙ 𝑏
𝑎⃗
|𝑎⃗⃗|2
=
7
√6
6+2−1
6
.
7
7 7
< 2, 1, 1 > = < 3 , 6, 6 >
𝑜𝑟𝑡ℎ → 𝑏⃗⃗ = 𝑏⃗⃗ − 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ = < 3, 2, −1 > − <
𝑎
𝑎
7
7
7
2
5
, , >=<3,6,
3 6 6
−13
6
>
Any vector can be expressed as the sum of its parallel and perpendicular projection to another vector.
⃗⃗⃗⃗ = 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ + (𝑏
⃗⃗⃗⃗ − 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗) = 𝑏⃗⃗.
Since 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ + 𝑜𝑟𝑡ℎ → 𝑏
𝑎
𝑎
𝑎
𝑎
eg. 24 Let 𝑏⃗⃗ = < 3, 2 > ; 𝑎
⃗⃗⃗⃗ = < 2, 1 >. Express 𝑏⃗⃗ as the sum of its parallel and perpendicular projection
to 𝑎
⃗⃗⃗⃗
⃗⃗⃗⃗ = 8 < 2, 1 > + 1 < - 1, 2 >.
𝑏⃗⃗ = <3, 2> = 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ + 𝑜𝑟𝑡ℎ → 𝑏
𝑎
𝑎
5
5
eg 25 Let 𝑏⃗⃗ = < 3, 2, −1 > ; 𝑎⃗ = < 2, 1, 1>.>. Express 𝑏⃗⃗ as the sum of its parallel and perpendicular
projection to 𝑎
⃗⃗⃗⃗
7 7 7
2 5 −13
𝑏⃗⃗ = < 3, 2, −1 > = 𝑝𝑟𝑜𝑗 → 𝑏⃗⃗ + 𝑜𝑟𝑡ℎ → 𝑏⃗⃗ = < , , > + < , ,
>.
𝑎
𝑎
3
6
6
The angle between the vectors is also given by𝑐𝑜𝑠(𝜃) =
3
6
⃗⃗|𝑐𝑜𝑠𝜃
|𝑎⃗⃗||𝑏
⃗⃗|
|𝑎⃗⃗||𝑏
=
6
⃗⃗
𝑐𝑜𝑚𝑝→𝑏
eg 26 The angle between 𝑏⃗⃗ = < 3, 2 > and 𝑎
⃗⃗⃗⃗ = < 2, 1 > is cos(𝜃) =
𝑎
⃗⃗|
|𝑏
.
8√5
5
√13
eg 27 The angle between 𝑏⃗⃗ = < 3, 2, −1 > and 𝑎
⃗⃗⃗⃗ = < 2, 1,1 > is cos(𝜃) =
√21
6
=
8√5
5√13
√6
√14
=
, 𝑜𝑟 𝜃 ≈ 7.12°.
7
√84
=
, 𝑜𝑟 𝜃 ≈ 40.2°.
Planes
The equation of a plane is determined by a point on the plane and a normal to the plane.
Let 𝑟⃗ = < x, y, z > be a position vector in space that describes a plane. If ⃗⃗⃗⃗
𝑟0 = <x0, y0, z0 > is the position
vector of the point (x0, y0, z0 ) on the plane and 𝑛⃗⃗ = < a, b, c > is the normal to the plane, the equation of
the plane will be given by (𝑟⃗ - ⃗⃗⃗⃗)
𝑟0 ∙ 𝑛⃗⃗ = 0. So a(x – x0) + b(y – y0) + c(z – z0) = 0 or ax + by + cz –
(ax0 + by0 + cz0) = 0. We can write the equation as ax + by + cz + d = 0 where d = - (ax0 – by0 + cz0).
Eg28 Find the equation of the plane through (2, 4, - 1) with 𝑛⃗⃗ = < 2, 3, 4> .
2x + 3y + 4z – ( (2) (2) + (3) (4) + (4) ( -1)) = 0, So 2x + 3y + 4z – 12 = 0 is the equation of the plane.
Eg29 Give the equation of the plane through (1, - 1, 2) parallel to the plane 3x – 5y + 6z = 10.
Since the normal of both planes are the same, 𝑛⃗⃗ = < 3, - 5, 6 >, the equation of the parallel plane
becomes 3x – 5y + 6z = 3 + 5 + 12 or 3x – 5y + 6z = 20
13
Perpendicular Distance Between a point and a Line in 2-space
If p(x1, y1) is a fixed point, show that the minimum distance between the point and the line ax + by + c =
0 is given by d =
|𝑎𝑥1 +𝑏𝑦1 +𝑐|
.
√𝑎2 +𝑏2
Method 1. Use Calculus to minimize the distance between the point and the line.
s = d2 = (x – x1)2 + (y - y1)2. If we substitute the line y =
s = d2 = (x – x1)2 + (
𝑑𝑠
𝑑𝑥
−𝑐−𝑎𝑥
= 2(𝑥 − 𝑥1 ) + 2 (
−𝑐−𝑎𝑥
2
− 𝑦1 ) . To minimize we need
𝑏
−𝑐−𝑎𝑥
𝑏
x and y into s, we obtain s =
into the distance formula we have
𝑏
𝑑𝑠
𝑑𝑥
= 0.
𝑎
− 𝑦1 ) (− 𝑏) = 0. Solving for x we have x =
(𝑎𝑥1 +𝑏𝑦1 + 𝑐)2
𝑎2 +𝑏 2
𝑜𝑟 𝑑 =
|𝑎𝑥1 +𝑏𝑦1 + 𝑐|
√𝑎2 +𝑏2
𝑥1 𝑏 2 −𝑐𝑎−𝑦1 𝑏𝑎
𝑎2 +𝑏 2
. If we substitute
.
Method 2. Use Vectors
First show that the vector < a, b > is perpendicular to the line ax + by + c = 0.
One way to show this is to let the points (x1, y1) and (x2, y2) be on the line. A vector on the line is given
by 𝑣
⃗⃗⃗⃗ = < 𝑥2 − 𝑥1 , 𝑦2 − 𝑦1 >. If we assume a normal vector 𝑛
⃗⃗⃗⃗ = < 𝑎. 𝑏 >, then 𝑣⃗ ∙ 𝑛⃗⃗ = a(x2 – x1) +
b(y2 – y1) = (ax2 + by2) – (ax1 + by1) . Since the two points satisfy the line, we have
(ax2 + by2) – (ax1 + by1) = (- c) – ( - c) = 0 thus 𝑣⃗ and 𝑛⃗⃗ are perpendicular since the dot product is zero.
Another way is to use slopes. We can show that 𝑛
⃗⃗⃗⃗ = < 𝑎. 𝑏 >, is perpendicular to the line
𝑎
ax + by + c = 0. Since the slope of the line is m = − 𝑏, and 𝑚𝑝𝑒𝑟 =
𝑏
𝑎
, a vector perpendicular to the line
will be < 𝑎. 𝑏 >.
Let (x0, y0) be a point on the line and (x1, y1) a fixed point. The distance between (x1, y1) and the line
will be given by D = |𝑐𝑜𝑚𝑝 → ⃗⃗⃗|
𝑟 =
𝑛
|𝑟
⃗⃗⃗⃗∙𝑛
⃗⃗⃗⃗|
|𝑛
⃗⃗⃗⃗|
=
|𝑎(𝑥1 −𝑥0 )+𝑏(𝑦1 −𝑦0 )|
√𝑎2 +𝑏 2
=
|𝑎𝑥1 +𝑏𝑦1 −𝑎𝑥0 − 𝑏𝑥0 |
√𝑎2 +𝑏2
and 𝑟⃗⃗⃗ = < (𝑥1 − 𝑥0 ), (𝑦1 − 𝑦0 ) >.
Since the point (x0, y0) satisfies the line then D =
|𝑎𝑥1 +𝑏𝑦1 + 𝑐|
√𝑎2 + 𝑏2
.
eg 30 Find the distance between the point (1,3) and the line 3x – 4y + 1 = 0.
D=
|𝑎𝑥1 +𝑏𝑦1 + 𝑐|
√𝑎2 + 𝑏2
=
|3(1)−4(3)+ 1|
√32 + 42
8
= 5.
14
where 𝑛
⃗⃗⃗⃗ = < 𝑎, 𝑏 >
eg 31 Find the distance between the two lines l1: (2x – 3y = 1) and l2: (- 4x + 6y = 1).
Since the two lines are parallel, we can choose any point in one of the lines, and a vector normal to the
lines. A point in l2 (x1, y1) = (0, 1/6). So D =
|2(0)−3(1⁄6)−1|
√22 +32
=
3/2
.
√13
Parallel and Perpendicular Vectors to the Tangent Line of a Curve
The parallel (tangent) vector to the tangent line of y = f (x) at x0 can be found by finding the slope m1 of
the tangent line. The perpendicular (normal) vector can be found from the slope of the normal line m2,
since m1m2 = -1 for perpendicular lines.
eg 32 Find two unit tangent and two unit normal vectors of f (x) =
Since f ´ (x) =
±
3𝑥 2
2
3
│x=1 = 2 =
∆𝑦
, the unit tangent vectors will be ±
∆𝑥
𝑥3
2
1
+ at (1, 1).
2
<2,3>
√13
, and the unit normal vectors
<3,−2>
√13
.
Homework 13.3
1. Give the two unit vectors in an xy-plane that are parallel to the line y = 2x + 1.
Ans:  1, 2 / 5
2. Give the two unit vectors in an xy-plane that are normal to y = sin x at x =  / 3 .
Ans:  1, 2 / 5
3. Find numbers A such that 3, 4 and A, 2 (a) are parallel and (b) are perpendicular.
Ans: a) A=3/2 b) A=-8/3
4. (a) Calculate A • B for A = i + 5j and B= 6i + 4j. (b) What is the component of A in the
direction of B? (c) Find the projection of A on a line through B. (d) What is the length of
proj A B (e) Find orthA B f) What vector do you obtain when you add proj A B + orthA B ?
Ans: a) 26; b) 13 ; c)
1,5
; d)
26 e)
5, 1
; f) B
5. Find exact and approximate decimal values of the angles (a) between 1, 3, 4 and 2, 0, 1
and (b) between 2i + 3j + 4k and -3i + 4j – 5k. Ans: cos 1 (2 / 130) , cos1 (14 / 1450)
6. Find the direction cosines and the exact and approximate decimal values of the direction angles
2
1
2
of A= 2, 1, 2 . Ans:   cos1 ,   cos 1 ,  cos1
3
3
3
15
7. Three ropes attached to a hook on it support a box. The ropes exert forces
F1 = 20i + 10j + 15k, F2 = –30i – 4j + 6k, and F3 = 10i – 6j + 8k pounds, relative to xyzcoordinates with the positive z-axis pointing up. How much does the box weigh?
Ans: 29lb
8. What is the component of i + 2j – 3k in the direction from (2, 5, 8) toward (–2, 3, 9)?
Ans: 11/ 21
9. Find the constant k such that the projection of 6, 10, k on a line parallel to –3, 2, 1 is
3 2
1
, – , – . Ans: k = -4
7 7
7
10. Find the vector whose component in the direction of i is 1, whose component in the direction of j
+ k is 2 2 , and whose component in the direction of i – 3j is 10 . Ans:  1, 3, 7 
11. Find a vector 6 unit long in the direction of 𝐴⃗ = < 2, 2, - 1 >. Ans:[6 𝐴̂ = < 4, 4, - 2 > ]
12. Show 𝑢
⃗⃗ = < 1, 0, 1 > is perpendicular to 𝑣⃗ = < - 1, 1, 1 >.
2 −1 −2
13. Find the directional cosines of < 2, - 1, - 2 >. [{3 ,
3
,
3
}].
14. Find the angle the position vector < 1, 2, 3 > makes with the y axis. [𝛽 = 𝑐𝑜𝑠 −1
2
]
√14
15. Use the property of the dot product 𝑢
⃗⃗ ∙ 𝑣⃗ = |𝑢
⃗⃗||𝑣
⃗⃗⃗⃗| cos𝜃 to derive a formula for the magnitude
⃗⃗⃗⃗⃗⃗ 𝑐𝑜𝑠𝜃 |𝑣⃗⃗| = 𝑢⃗⃗ ∙ 𝑣⃗⃗
of the projection (component) of vector 𝑢
⃗⃗ into vector 𝑣⃗ . Ans: comp → 𝑢
⃗⃗ = |𝑢|
|𝑣
|𝑣
⃗⃗|
⃗⃗|
𝑣
16. Find the component of 𝑢
⃗⃗ = < 1, 0, −1 > parallel to 𝑣⃗ = < 1, 1, 1 >.Ans: [2/√3]
⃗⃗ = < 1,2,3 > as the sum of a vector parallel to 𝐴⃗ = < 1,1,0 > and a vector orthogonal to 𝐴⃗.
17. Write 𝐵
33
Ans: [B = < 2 2 ,0 > + <
−1
2
1
,2,3 > ]
⃗⃗ = < 0, 3, 4 > as the sum of a vector parallel to 𝐴⃗ = < 1,1,0 > and a vector orthogonal to
18. Write 𝐵
3 3
3 3
𝐴⃗ . Ans: [< , ,0 > + < , , 4 >]
2
2
2
2
⃗⃗ <2, 1, - 3 > as the sum of a vector parallel to 𝐴⃗ = < 3, -1, 0 > and a vector orthogonal to
19. Write 𝐵
3 −1
1 3
𝐴⃗. Ans: [< 2 , 2 ,0 > + < 2 , 2 , - 3 >]
3
20. Find the distance from the point (2,4) to the line y = − 4 𝑥 + 1. Ans: [18/5units]
21. Use slopes to show that the vector < 2, 3> is normal to the line 2x + 3y + 4 = 0.
22. Use the dot product to show that the vector < 3, - 2 > is normal to the line 3x – 2y + c = 0.
23. Find a unit tangent and a unit normal vector to the curves in 2-space.
a) y = cos (x) at x = 𝜋/3.
b) y = sin -1 (x) at x = √2/2 .
c) y = tan -1 (x) at x = √3/3.
Ans: [ ±
Ans: [ ±
<1.√2>
√3
Ans: [ ±
16
<2,− √3 >
√7
,±
,±
<√3,2>
√7
<√2 ,−1>
√3
<4,3>
5
,±
]
<3,−4>
5
]
]
13.4 The Cross Product
Cross Product: (Vector Multiplication)
The cross product is an operation on two vectors in a three-dimensional space that results in another
vector which is perpendicular to the plane containing the two input vectors. The direction of the new
vector that results from the cross product of the vectors 𝑎⃗ and 𝑏⃗⃗ is given by the right hand rule
The cross product 𝑢
⃗⃗ x 𝑣⃗ is defined
̂𝑖 ̂𝑗 ̂𝑘
𝑢2 𝑢3
𝑢1 𝑢3
𝑢1 𝑢2
𝑢
⃗⃗ x 𝑣⃗ = |𝑢1 𝑢2 𝑢3 | = ̂𝑖 |𝑣 𝑣 | -̂𝑗 |𝑣 𝑣 | + ̂𝑘 |𝑣 𝑣 | =
2
3
1
3
1
2
𝑣1 𝑣2 𝑣3
̂𝑖 (u2v3 – v2u3) - ̂𝑗 (u1v3 – v`1u3) + ̂𝑘 (u1v2 – v1u2). We can see that the cross product is a vector. The
direction of this vector is perpendicular to the plane that contains the vectors 𝑢
⃗⃗ and 𝑣⃗.
eg33. Find the cross product of 𝑢
⃗⃗ = < 2, 1, 3 > and 𝑣⃗ < 0, 2, 1 >.
̂
̂𝑖 ̂𝑗 𝑘
2 3 ̂ 2 1
1 3
𝑢
⃗⃗ x 𝑣⃗ = |2 1 3 | = ̂𝑖 |
| -̂𝑗 |
|+𝑘|
| = < -5, - 2, 4 >.
0 1
0 2
2 1
0 2 1
The magnitude of the cross product is given by |𝑢
⃗⃗ x 𝑣⃗| = |𝑢
⃗⃗||𝑣
⃗⃗⃗⃗| sin(𝜃), where 𝜃 is the angle between 𝑢
⃗⃗
and 𝑣⃗, in the direction of the unit vector ̂𝑛 parallel to the vector 𝑢
⃗⃗ x 𝑣⃗. The vector ̂𝑛 is perpendicular to
the plane that contains the vectors 𝑢
⃗⃗ and 𝑣⃗.
Another way of writing the cross product vector is 𝑢
⃗⃗ x 𝑣⃗ = |𝑢
⃗⃗||𝑣
⃗⃗⃗⃗| sin(𝜃) ̂𝑛 .
Two vectors are parallel if |𝑢
⃗⃗ x 𝑣⃗| = 0.
Properties of Cross Products
For any vectors 𝑢
⃗⃗, 𝑣⃗, 𝑤
⃗⃗⃗ in 3-space and any scalar k,
a)
𝑢
⃗⃗ x 𝑣⃗ = − 𝑣⃗ x 𝑢
⃗⃗
b)
𝑢
⃗⃗ x (𝑣⃗ + 𝑤
⃗⃗⃗) = 𝑢
⃗⃗ x 𝑣⃗ + 𝑢
⃗⃗ x 𝑤
⃗⃗⃗
c)
k (𝑢
⃗⃗ x 𝑣⃗) = (k 𝑢
⃗⃗ x 𝑣⃗) = (𝑢
⃗⃗ x k𝑣⃗)
d)
𝑣⃗ x 𝑣⃗ = ⃗0⃗
17
𝑢
⃗⃗ ∙ 𝑢
⃗⃗ x 𝑣⃗ = ⃗0⃗ since 𝑢
⃗⃗ and 𝑢
⃗⃗ x 𝑣⃗ are perpendicular.
𝑣⃗ ∙ 𝑢
⃗⃗ x 𝑣⃗ = ⃗0⃗ since 𝑣⃗ and 𝑢
⃗⃗ x 𝑣⃗ are perpendicular.
1)
2)
Areas
The area of a parallelogram defined by the vectors 𝑢
⃗⃗ and 𝑣⃗ is given by
A = base∗height = |𝑢
⃗⃗||𝑣
⃗⃗⃗⃗| sin (𝜃) = |𝑢
⃗⃗ x 𝑣⃗|, where 𝜃 is the angle between 𝑢
⃗⃗ and 𝑣⃗.
The area of a triangle defined by the vectors 𝑢
⃗⃗ and 𝑣⃗ is given by
1
A = 2 base*height =
1
2
|𝑢
⃗⃗||𝑣
⃗⃗⃗⃗| 𝑠𝑖𝑛(𝜃) =
1
2
|𝑢
⃗⃗ x 𝑣⃗|.
eg35 Find the area of the parallelogram and the triangle defined by 𝑢
⃗⃗ = < 1, 0, 1 > and 𝑣⃗ = < 2, 1, 0 >.
̂𝑖
𝑢
⃗⃗ 𝑥 𝑣⃗ = |1
2
̂𝑗
0
1
̂𝑘
1 | = < - 1, 2, 1 >. So the area of the parallelogram is
0
|𝑢
⃗⃗ x 𝑣⃗| = √6 sq units and the area of the triangle is
√6
2
sq units,
The cross product of the unit vectors are ̂𝑖 x ̂𝑗 = ̂𝑘; , ̂𝑗 x ̂𝑘 = ̂;
𝑖
eg 36
̂𝑖
̂𝑖 x ̂𝑗 = |1
0
̂𝑗
0
1
̂𝑘 x ̂𝑖 = ̂𝑗 .
̂𝑘
0 | = < 0, 0, 1 > = ̂𝑘.
0
Triple Scalar Product
𝑎1
⃗
⃗
⃗
⃗
The triple scalar product is defined as [𝑎⃗ 𝑏 𝑐⃗] = 𝑎⃗ ∙ (𝑏 x 𝑐⃗) = |𝑏1
𝑐1
Proof:
̂𝑖 ̂𝑗 ̂𝑘
𝑎⃗ ∙ (𝑏⃗⃗ x 𝑐⃗) = < a1, a2, a3 >. |𝑏1 𝑏2 𝑏3 | =
𝑐1 𝑐2 𝑐3
𝑎2
𝑏2
𝑐2
𝑎3
𝑏3 |
𝑐3
< a1, a2, a3 > ∙ < (b2c3 – c2b3), - (b1c3 – c1b3), (b1c2 – c1b2) > =
𝑎1
a1 (b2c3 – c2b3), - a2 (b1c3 – c1b3) + a3 (b1c2 – c1b2) = |𝑏1
𝑐1
𝑎2
𝑏2
𝑐2
𝑎3
𝑏3 | .
𝑐3
Since interchanging two rows of a determinant only changes the sign of the determinant, it can be shown
that 𝑎⃗ ∙ (𝑏⃗⃗ x 𝑐⃗) = 𝑐⃗ ∙ (𝑎⃗ x 𝑏⃗⃗) = 𝑏⃗⃗ ∙ (𝑐⃗ x 𝑎⃗) .
18
Volume of a Parallelepiped
The volume of a parallelepiped formed by the vectors 𝑎⃗, 𝑏⃗⃗ 𝑎𝑛𝑑 𝑐⃗ is given by |𝑎⃗ ∙ (𝑏⃗⃗ x 𝑐⃗)| =
|𝑎⃗||𝑏⃗⃗ x 𝑐⃗| cos(𝜃) where 𝜃 is the angle between 𝑎⃗ and 𝑏⃗⃗ x 𝑐⃗ .
If |𝑎⃗ ∙ (𝑏⃗⃗ x 𝑐⃗)| = 0, 𝑎⃗ and 𝑏⃗⃗ x 𝑐⃗ are perpendicular so 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are coplanar.
eg37 Show that the vectors 𝑎⃗ = < 1, 4, - 7 >, 𝑏⃗⃗ = <2, -1,4 > and 𝑐⃗ = < 0, -9, 18 > are coplanar.
1 4 −7
𝑎⃗ ∙ (𝑏⃗⃗ x 𝑐⃗) = |2 −1 4 | = 1(18) – 4(36) – 7 ( - 18) = 0.
0 −9 18
1
The volume of the tetrahedron formed by the vectors 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ is given by 6 |𝑎⃗ ∙ (𝑏⃗⃗ x 𝑐⃗)|
Planes
Recall from a previous section that the equation of a plane is determined by a point on the plane and a
normal to the plane.
Let 𝑟⃗ = < x, y, z > be a position vector in space that describes a plane. If ⃗⃗⃗⃗
𝑟0 = <x0, y0, z0 > is the position
vector of the point (x0, y0, z0 ) on the plane and 𝑛⃗⃗ = < a, b, c > is the normal to the plane, the equation of
the plane will be given by (𝑟⃗ - ⃗⃗⃗⃗)
𝑟0 ∙ 𝑛⃗⃗ = 0. So a(x – x0) + b(y – y0) + c(z – z0) = 0 or ax + by + cz –
(ax0 + by0 + cz0) = 0. We can write the equation as ax + by + cz + d = 0 where d = - (ax0 – by0 + cz0).
eg38 Find the equation of the plane that contains the points (1, 3, 2), (3, - 1, 6), (5, 2, 0).
We can make the vectors 𝑢
⃗⃗ = < 2, -4, 4 > from the first two points and 𝑣⃗ = <2, 3, - 6 > from the last two.
Since the normal to the plane is 𝑛⃗⃗ = 𝑢
⃗⃗ x 𝑣⃗ = < 12, 20, 14 > = < 6, 10, 7 >, the plane is 6x + 10y + 7z –
(6 + 30 + 14) = 0 or 6x + 10y + 7z – 50 = 0, where the point (1, 3, 2) was used as the point on the plane.
eg39 Find the smallest angle between the planes x + y + z = 1 and x – 2y + 3z = 1.
The angle between two planes will be the angle between their normal vectors ⃗⃗⃗⃗⃗
𝑛1 = < 1, 1, 1 > and
𝑛2 =< 1, − 2, 3 >.
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
𝑛 ∙ ⃗⃗⃗⃗⃗⃗
𝑛
1
2
By using the dot product, 𝑐𝑜𝑠(𝜃) = |𝑛⃗⃗⃗⃗⃗⃗|
=
|𝑛
⃗⃗⃗⃗⃗⃗|
1
2
1−2+3
√3 √14
=
2
√42
so 𝜃 ≈ 72o.
eg40 Find a vector 𝑣⃗ parallel to the intersection between the planes x + y + z = 1 and x - 2y + 3z = 1.
A parallel vector will be the cross product of the normal of each plane, so 𝑣⃗ = < 5, - 2, - 3 >.
Distance From a Point to a Plane
Let P1 be the point (x1, y1, z1), and P0 = (x0, y0, z0) be any point on the plane ax + by + cz + d = 0. The
distance D from P1 to the plane will be D = |𝑐𝑜𝑚𝑝𝑛⃗⃗⃗⃗ 𝑢
⃗⃗| =
⃗⃗ ∙ 𝑢
⃗⃗
𝑛
|𝑛
⃗⃗⃗⃗|
where 𝑢
⃗⃗ = < x1 – x0, y1 – y0, z1 – z0 >, and
𝑛⃗⃗ = < a, b, c > is the normal to the plane.
D=
⃗⃗ ∙ 𝑢
⃗⃗
𝑛
|𝑛
⃗⃗⃗⃗|
=
|<𝑎,𝑏,𝑐>∙<𝑥1 − 𝑥0 ,𝑦1 − 𝑦0 ,𝑧1 −𝑧0 >|
√𝑎2 +𝑏2 + 𝐶 2
=
|𝑎𝑥1 +𝑏𝑦1 +𝑐𝑧1 − (𝑎𝑥0 +𝑏𝑦0 +𝑐𝑧0 )|
√𝑎2 +𝑏2 + 𝑐 2
19
Since the point (x0, y0, z0) satisfies the plane ax + by + cz + d = 0, d = - (ax0 + by0 + cz0), and
D=
⃗⃗ ∙ 𝑢
⃗⃗
𝑛
|𝑛
⃗⃗⃗⃗|
=
|𝑎𝑥1 +𝑏𝑦1 +𝑐𝑧1 + 𝑑|
√𝑎2 +𝑏2 + 𝑐 2
eg42 Find the distance from the point P1 (2, - 3, 4) to the plane x + 2y + 2z = 13.
D=
|𝑎𝑥1 +𝑏𝑦1 +𝑐𝑧1 + 𝑑|
√𝑎2 +𝑏2 + 𝑐 2
|1(2)+2(−3)+2(4)−13|
=
√12 +22 +22
=
|−9|
3
=3
eg41 Find the distance D from the point P1 = (2, 1, 5) to the plane that contains the point P0 = (1, - 1, 4)
with normal 𝑛⃗⃗ = < 2, 4, 1 >.
n
.P
.P
1
d
0
Let 𝑢
⃗⃗ be the vector from P0 to P1. d is the | comp → 𝑢
⃗⃗|
D=
⃗⃗ ∙ 𝑢
⃗⃗
𝑛
|𝑛
⃗⃗⃗⃗|
=
|<2,4,1>∙<2−1,1+1,5−4>|
√22 +42 +12
=
𝑛
|<2,4,1>∙<1,2,1>|
√21
=
11
√21
.
Homework 13.4
1. Find exact and approximate decimal values of the angles (a) between 1, 3, 4 and 2, 0, 1 and
(b) between 2i + 3j + 4k and -3i + 4j – 5k. Ans: cos 1 (2 / 130) , cos1 (14 / 1450)
2. Find the direction cosines and the exact and approximate decimal values of the direction angles of
2
1
2
A= 2, 1, 2 . Ans:   cos1 ,   cos 1 ,  cos1
3
3
3
3. Three ropes attached to a hook on it support a box. The ropes exert forces F1 = 20i + 10j + 15k lb.,
F2 = –30i – 4j + 6k lb., and F3 = 10i – 6j + 8k lb, relative to xyz-coordinates with the positive z-axis
pointing up. How much does the box weigh?
Ans: 29lb
4. What is the component of i + 2j – 3k in the direction from (2, 5, 8) toward (–2, 3, 9)?
Ans: 11/ 21
20
5. Find the constant k such that the projection of 6, 10, k on a line parallel to –3, 2, 1 is
3 2
1
, – , – . Ans: k = -4
7 7
7
6. Find the vector whose component in the direction of i is 1, whose component in the direction of j +
k is 2 2 , and whose component in the direction of i – 3j is 10 . Ans:  1, 3, 7 
7. Find a unit vector orthogonal to both 𝑢
⃗⃗ = < 1, 0, 1 > and 𝑣⃗ = < - 1, 1, 1 >.
⃗⃗ x 𝑣
⃗⃗
𝑢
⃗⃗⃗⃗
|𝑢 x 𝑣|
Ans: [⃗⃗⃗⃗⃗
= <-1,-2,-1>/√6]
8. For which values of t are the vectors ⃗A⃗ = < t + 2, t, t > and ⃗B⃗ = < t − 2, t + 1, t > are orthogonal?
−4
Ans: [t = { 3 , 1}]
9. Find the area of the triangle with vertices P(1,-1,0), Q(2, 1,-1), R(-1,1,2). Ans: [3√2]
⃗⃗ = < - 2, 0, 3 > and
10. Find the volume of the parallelepiped determined by 𝐴⃗ = < 1, 2, - 1 >, 𝐵
𝐶⃗ = < 0, 7, - 4 >. Ans: [23cu-units]
11. Consider the parallelepiped with sides 𝑢
⃗⃗ = < 1, 1, 1 > , 𝑣⃗ = < - 1, 1, 1 > , 𝑤
⃗⃗⃗ = < – 1, 0,1 >. Find the
angle between the plane containing the vectors 𝑢
⃗⃗ and 𝑣⃗, and the vector 𝑤
⃗⃗⃗. Ans: π/6
17
12. Find the distance from the point S(1,1,3) to the plane 3x + 2y + 6z = 6. Ans: [ 7 ]
2
13. Find the distance from the point (2, - 3, 4) to the plane x + 2y + 2z = 6. Ans; [3]
14. Find an equation for the plane through (0, 1, 0) and parallel to i + j and to j – k, Ans: x - y – z = -1.
15. Find an equation for the plane through (5, -1, -2) and perpendicular to the planes y – z = 4
and x + z = 3.
Ans: x - y – z = 8.
16. Find an equation for the plane through (2, 2, 4), (5, 6, 4), and (1, 3, 5),
Ans: 4x - 3y + 7z = 30.
17. Find an equation for the plane through (1, 2, 3) and parallel to the plane 4x – y + 3z = 0
Ans: 4 x  y  3 z  11
21