Vectors

Question 0

Modules of xi+yj+zk:

- - -

Answer 0

_______________

| x^2 + y^2 + z^2

Question 1

The distance between (x1, y1, z1) and (x2, y2, z2):

Answer 1

______________________________

|(x1-x2)^2 + (y1-y2)^2 + (z1-z1)^2

Question 2

The distance from the origin to the point (x1, y1, z1):

Answer 2

_______________

| x^2 + y^2 + z^2

Question 3

Column matrix xi + yj + zk =

- - -

Answer 3

(X)
(Y)
(Z)

Question 4

Scalar product a.b =

- -

Answer 4

= |a| |b| cosA

Where a and b are the lengths of vector a ans vector b

In coordinate form:
= a1b1 + a2b2 + a3b3

Question 5

The non-zero vectors a and b are perpendicular if: - -

Answer 5

a. b = 0

- -

Question 6

If a and b are parallel:

- -

Answer 6

a. b = |a| |b|
- - - -

a. a = |a|(^2)
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Question 7

Rearranged to find the angle cosAOB:

Answer 7

cosAOB = a.b
——-
|a| |b|

Question 8

Equation of a straight line passing through point A with position vector a and parallel to vector b is:

Answer 8

r = a + tb
- - -

In a diagram, tb is line 1. Essentially, this equation is vector a and something times vector b where vector b is parallel to line 1. To find b, we can also do -r+a.

Question 9

Equation of a straight line passing through points C and D with position vector c and parallel to vector CD is:

Answer 9

r = c + t(d - c)

- - - -

Question 10

Acute angle between straight lines is given by:

Answer 10

a.b |
| —— |
| |a| |b| |

Question 11

How to determine whether two straight lines intersect:

Answer 11

• Equate x components
• Equate y components
• Solve simultaneously to find t and s
• If the lines intersect, z components must also be equal so sub in t and s and see if they come out with the same answer

Question 12

Example modulus/magnitude:

It is basically pythagoras

Answer 12

If a = (6)
– (8)

|a| = |(6)^2+(8)^2 = 10

Question 13

Relationship between column vector and position:

Answer 13

(x) = xi+yz+zk
(y)
(z)

Question 14

If a vector has a column matrix (a b) then |vector| = |a^2 + b^2…

Answer 14

If the angle between it and the positive x axis, measured anti-clockwise, is A then

TanA = b / a

Question 15

a = (a1) b = (b1)

- - (a2) – (b2
a3) (b3)

Answer 15

Relationship because vector and points

Question 16

When having the lines
r=(ai+bj+ck)+t(di+ej+fk) [r=a+tb]
and r=(ai+bj+ck)+s(di+ej+fk) [r=a+sb] and working the acute angle…

Answer 16

You look at the direction vector, b so a.b would be (d) (d)

e) •(e
(f) (f)

Question 17

A unit vector is a vector with length (modulus) 1. How do we find a unit vector?

Answer 17

Find the length of the given vector then to make 1, we divide the vector by it.

E.g. Find a unit vector in the direction -12i+5j
Length is |12^2 + 5^2 = 13 so the required unit vector is 1/13 * 12i + 5j