VECTORS IN COMPONENT FORM - Uplift Education · Therefore P, Q and R are collinear. = 5 −1 −2, = −5 1 2 =−1× How can you check it: 1. Form two vectors with these three points.

VECTORS IN COMPONENT FORM

example:

𝑎 =27−3

=200

+070

+00−3

= 2100

+ 7010

− 3001

= 2 𝑖 + 7 𝑗 − 3 𝑘.

In Cartesian coordinates any 3 – D vector 𝑎 can be written as

𝑎 = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 ≡

𝑎𝑥

𝑎𝑦

𝑎𝑧

≡ 𝑎𝑥 𝑎𝑦 𝑎𝑧

where 𝑖, 𝑗 𝑎𝑛𝑑 𝑘 are unit vectors in x, y and z directions.

𝑖 =100

𝑗 =010

𝑘 =001

𝑎 = 𝑎𝑥2 + 𝑎𝑦

2 + 𝑎𝑧2 𝑎 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒, 𝑙𝑒𝑛𝑔𝑡ℎ,𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑟 𝑛𝑜𝑟𝑚

Unit vector

For a vector 𝑎 , a unit vector is in the

same direction as 𝑎 and is given by:

𝑎 =𝑎

𝑎

A unit vector is a vector whose length is 1.

Definition

It gives direction only!

𝑎 =𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧

𝑘

𝑎𝑥2 + 𝑎𝑦

2 + 𝑎𝑧2

=1

𝑎𝑥2 + 𝑎𝑦

2 + 𝑎𝑧2

𝑎𝑥

𝑎𝑦

𝑎𝑧

VECTOR BETWEEN TWO POINTS

𝐴𝐵 =

𝑥𝐵 − 𝑥𝐴

𝑦𝐵 − 𝑦𝐴

𝑧𝐵 − 𝑧𝐴= (𝑥𝐵 − 𝑥𝐴) 𝑖 + (𝑦𝐵 − 𝑦𝐴) 𝑗 + (𝑧𝐵 − 𝑧𝐴) 𝑘

𝐵𝐴 =

𝑥𝐴 − 𝑥𝐵

𝑦𝐴 − 𝑦𝐵

𝑧𝐴 − 𝑧𝐵= (𝑥𝐴 − 𝑥𝐵) 𝑖 + (𝑦𝐴 − 𝑦𝐵) 𝑗 + (𝑧𝐴 − 𝑧𝐵) 𝑘

𝑚𝑜𝑑𝑢𝑙𝑢𝑠 ≡ 𝑙𝑒𝑛𝑔𝑡ℎ: 𝐴𝐵 = 𝐵𝐴

= (𝑥𝐵 − 𝑥𝐴)2+(𝑦𝐵 − 𝑦𝐴)

2+(𝑧𝐵 − 𝑧𝐴)2

PARALLEL VECTORS

𝑎 = 𝑘𝑏 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟. (𝑤ℎ𝑒𝑟𝑒 𝑘𝜀𝑅)

𝑎 =693

𝑏 =231

693

= 3231

→ 𝑎 || 𝑏

Two vectors 𝑎 and 𝑏 are parallel if and only if

Definition

ARE 3 POINTS COLLINEAR ?

Show that P(0, 2, 4), Q(10, 0, 0) and R(5, 1, 2) are collinear.

𝑄𝑅 𝑎𝑛𝑑 𝑃𝑅 have a common direction and a common point.

Therefore P, Q and R are collinear.

𝑃𝑅 =5−1−2

, 𝑄𝑅 =−512

𝑄𝑅 = −1 × 𝑃𝑅

How can you check it:

1. Form two vectors with these three points.

They will definitely have one common point.

2. Check if these two vectors are parallel.

If two vectors have a common point and are parallel (or antiparallel)

∴ they are collinear.

THE DIVISION OF A LINE SEGMENT

X divides [AB]≡ AB in the ratio 𝑎: 𝑏 means 𝐴𝑋:𝑋𝐵 = 𝑎 ∶ 𝑏

INTERNAL DIVISION

P divides [AB] internally

in ratio 1:3. Find P

EXTERNAL DIVISION

X divide [AB] externally in ratio 2:1,

or

X divide [AB] in ratio –2:1

A = (2, 7, 8) B = ( 2, 3, 12)

𝑥 − 2𝑦 − 7𝑧 − 8

+1

4

0−44𝐴𝑃: 𝑃𝐵 = 1: 3 → 𝐴𝑃 =

1

4𝐴𝐵

𝑥 − 2𝑦 − 3𝑧 − 12

=0−44

point P is (2, 6, 9)

point Q is (2,– 1,16)

𝐴𝑄:𝑄𝐵 = −2: 1

𝐴𝑃 =1

4𝐴𝐵

𝐵𝑄 = 𝐴𝐵

DOT/SCALAR PRODUCT

The dot/scalar product of two vectors 𝑎 and 𝑏 is:

𝑎 • 𝑏 = 𝑏 • 𝑎 = 𝑎 𝑏 cos 𝜃

𝑎 • 𝑏 =

𝑎𝑥

𝑎𝑦

𝑎𝑧

𝑏𝑥

𝑏𝑦

𝑏𝑧

= 𝑎𝑥𝑏𝑥 + 𝑎𝑦𝑏𝑦 + 𝑎𝑧𝑏𝑧

or: Product of the length of one of them and

projection of the other one on the first one

Scalar: ± 𝑛𝑢𝑚𝑏𝑒𝑟

𝑎 𝑎

𝑏

𝑏

Definition

𝑎

𝑏

θ

Ex: Find the angle between

𝑎 = 5 𝑖 − 2 𝑗 + 𝑘 𝑎𝑛𝑑 𝑏 = 𝑖 + 𝑗 − 3 𝑘

𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠𝑎 • 𝑏

𝑎 𝑏=

0

30 11

𝜃 = 𝑎𝑟𝑐 𝑐𝑜𝑠 0 = 𝜋/2

Dot product of perpendicular

vectors is zero.

𝑎 • 𝑏 =

𝑎𝑥

𝑎𝑦

𝑎𝑧

𝑏𝑥

𝑏𝑦

𝑏𝑧

= 𝑎𝑥𝑏𝑥 + 𝑎𝑦𝑏𝑦 + 𝑎𝑧𝑏𝑧

𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 • 𝑏𝑥 𝑖 + 𝑏𝑦 𝑗 + 𝑏𝑧

𝑘 =

Properties of dot product

∎ 𝑎 • 𝑏 = 𝑏 • 𝑎

∎ 𝑖𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑡ℎ𝑒𝑛 𝑎 • 𝑏 = 𝑎 𝑏

∎ 𝑖𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑎𝑛𝑡𝑖𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑡ℎ𝑒𝑛 𝑎 • 𝑏 = − 𝑎 𝑏

∎ 𝑎 • 𝑎 = 𝑎 2

∎ 𝑎 • 𝑏 + 𝑐 = 𝑎 • 𝑏 + 𝑎 • 𝑐

∎ 𝑎 + 𝑏 • 𝑐 + 𝑑 = 𝑎 • 𝑐 + 𝑎 • 𝑑 + 𝑏 • 𝑐 + 𝑏 • 𝑑

∎ 𝑎 • 𝑏 = 0 𝑎 ≠ 0, 𝑏 ≠ 0 ↔ 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟

The magnitude of the vector 𝑎 × 𝑏 is equal

to the area determined by both vectors.

Direction of the vector 𝑎 × 𝑏 is given by right hand rule:

Point the fingers in direction of 𝑎; curl them toward 𝑏.

Your thumb points in the direction of cross product.

CROSS / VECTOR PRODUCT

𝑎 × 𝑏 = 𝑎 𝑏 𝑠𝑖𝑛 𝜃

𝑏 × 𝑎 = − 𝑎 × 𝑏

Definition

∎ 𝑎 × 𝑏 = − 𝑏 × 𝑎

∎ 𝑖𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟, 𝑡ℎ𝑒𝑛 𝑎 × 𝑏 = 𝑎 𝑏

∎ 𝑎 × 𝑏 + 𝑐 = 𝑎 × 𝑏 + 𝑎 × 𝑐

∎ 𝑎 + 𝑏 × 𝑐 + 𝑑 = 𝑎 × 𝑐 + 𝑎 × 𝑑 + 𝑏 × 𝑐 + 𝑏 × 𝑑

∎ 𝑎 × 𝑏 = 0 𝑎 ≠ 0, 𝑏 ≠ 0 ↔ 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙

𝐹𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑖𝑠 0.

=>i × i = j × j = k × k = 0 i × j = k j × k = i k × i = j

⇒ 𝑎 × 𝑏 =

𝑎𝑦𝑏𝑧 − 𝑎𝑧𝑏𝑦

𝑎𝑧𝑏𝑥 − 𝑎𝑥𝑏𝑧

𝑎𝑥𝑏𝑦 − 𝑎𝑦𝑏𝑥

=

𝑖 𝑗 𝑘𝑎𝑥 𝑎𝑦 𝑎𝑧

𝑏𝑥 𝑏𝑦 𝑏𝑧

Properties of vector/cross product

𝑎 = 5 𝑖 − 2 𝑗 + 𝑘

𝑏 = 𝑖 + 𝑗 − 3 𝑘

(a) Find the angle between them

(b) Find the unit vector perpendicular to both

(a) 𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛𝑎 × 𝑏

𝑎 𝑏

𝑎 × 𝑏 = 𝑖 𝑗 𝑘

5 −2 11 1 −3

= 5 𝑖 + 16 𝑗 + 7 𝑘

𝑎 × 𝑏 = 5 𝑖 + 16 𝑗 + 7 𝑘 = 330

𝑎 = 30 𝑏 = 11

𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛 1 = 𝜋/2

(b) 𝑛 =𝑎 ×𝑏

𝑎 ×𝑏=

1

330

5167

Find all vectors perpendicular to both

𝑎 =123

𝑎𝑛𝑑 𝑏 =321

𝑎 × 𝑏 = 𝑖 𝑗 𝑘

1 2 33 2 1

=−48−4

⤇ 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑘 𝑖 − 2 𝑗 + 𝑘

𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑛𝑜𝑛 − 𝑧𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟.

Find the area of the triangle with vertices A(1,1,3), B(4,-1,1), and C(0,1,8)

It is one-half the area of the parallelogram determined by the vectors

𝐴𝐵 =3−2−2

and 𝐴𝐶 =−105

1

2𝐴𝐵 × 𝐵𝐶 =

1

2

𝑖 𝑗 𝑘3 −2 −2−1 0 5

⤇1

2

−10−13−2

=1

2(−10)2+(−13)2+(−2)2

= 8.26 𝑢𝑛𝑖𝑡𝑠2

• To find angle between vectors the easiest way is to use dot product,

not vector product.

• Angle between vector can be positive or negative

• Angle between lines is by definition acute angle between them, so

Dot product of perpendicular vectors is zero.

For perpendicular vectors the dot/scalar product is 0.

𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠 𝑎 • 𝑏

𝑎 𝑏

𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠 𝑎 • 𝑏

𝑎 𝑏

Conclusions:

• To show that two lines are perpendicular usethe dot product with line direction vectors.

• To show that two planes are perpendicularuse the dot product on their normal vectors.

• To find the angle between two lines

cos 𝜃 =𝑎 • 𝑏

𝑎 𝑏

𝑎 𝑎𝑛𝑑 𝑏 are direction vectors

acute angle !

Volume of a parallelepiped = scalar triple product

𝑉 = 𝑐 ● 𝑎 × 𝑏 =

𝑐𝑥 𝑐𝑦 𝑐𝑧𝑎𝑥 𝑎𝑦 𝑎𝑧

𝑏𝑥 𝑏𝑦 𝑏𝑧

𝑢𝑛𝑖𝑡𝑠3

Volume of a tetrahedron = 1

6scalar triple product

𝑉 =1

6 𝑐 ● 𝑎 × 𝑏 =

1

6

𝑐𝑥 𝑐𝑦 𝑐𝑧𝑎𝑥 𝑎𝑦 𝑎𝑧

𝑏𝑥 𝑏𝑦 𝑏𝑧

𝑢𝑛𝑖𝑡𝑠3

TEST FOR FOUR COPLANAR POINTS

If the volume of the tetrahedron is zero points are coplanar.

LINE EQUATION IN 2 – D and 3 – D COORDINATE SYSTEM

● Vector equation of a line

The position vector 𝒓 of any general point P on the line passing through

point A and having direction vector 𝑏 is given by the equation

𝑟 = 𝑎 + 𝑡 𝑏 𝑡 ∈ 𝑅IB Convention: 𝑢𝑠𝑒 𝒕 𝑓𝑜𝑟 2 − 𝐷 𝑙𝑖𝑛𝑒

𝝀 𝑓𝑜𝑟 3 − 𝐷 𝑙𝑖𝑛𝑒

𝑟 = 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 + 𝜆 𝑏1 𝑖 + 𝑏2 𝑗 + 𝑏3

𝑘 𝑜𝑟𝑥𝑦𝑧

=

𝑎1

𝑎2

𝑎3

+ 𝜆

𝑏1

𝑏2

𝑏3

● Parametric equation of a line – λ is called a parameter λ ∈ 𝑅

𝑥𝑦𝑧

=

𝑎1

𝑎2

𝑎3

+ 𝜆

𝑏1

𝑏2

𝑏3

⇒

𝑥 = 𝑎1 + 𝜆𝑏1

𝑦 = 𝑎2 + 𝜆𝑏2

𝑧 = 𝑎3 + 𝜆𝑏3

● Cartesian equation of a line

𝑥 = 𝑎1 + 𝜆𝑏1 ⟹ 𝜆 = (𝑥 − 𝑎1)/𝑏1

𝑦 = 𝑎2 + 𝜆𝑏2 ⟹ 𝜆 = (𝑦 − 𝑎2)/𝑏2

𝑧 = 𝑎3 + 𝜆𝑏3 ⟹ 𝜆 = (𝑧 − 𝑎3)/𝑏3

⟹𝑥−𝑎1

𝑏1=

𝑦−𝑎2

𝑏2=

𝑧−𝑎3

𝑏3(= 𝜆)

Find a) vector b) parametric c) Cartesian and d) general line

equation of a line passing through 𝐴 =12

& 𝐵 =36

.

choosing point 𝐴 =12

and direction 𝐴𝐵 =24

𝑎)𝑥𝑦 =

12

+ 𝑡24

𝑡 ∈ 𝑅

𝑏)𝑥 = 1 + 2𝑡𝑦 = 2 + 4𝑡

𝑡 ∈ 𝑅

𝑐)𝑥−1

2=

𝑦−2

4

𝑑) 4𝑥 − 4 = 2𝑦 − 4 ⟹ 2𝑦 − 4𝑥 = 0

Shortest distance from a point to a line

Point P is at the shortest distance from the line

when PQ is perpendicular to 𝑏

∴ 𝑃𝑄 • 𝑏 = 0

Find the shortest distance between 𝑟 =131

+ 𝜆232

and point P (1,2,3).

(The goal is to find Q first, and then 𝑃𝑄 )

Point Q is on the line, hence its coordinates must satisfy line equation:

𝑥𝑄

𝑦𝑄

𝑧𝑄=

1 + 2𝜆3 + 3𝜆1 + 2𝜆

⇒ 𝑃𝑄 =2𝜆

1 + 3𝜆−2 + 2𝜆

⇒2𝜆

1 + 3𝜆−2 + 2𝜆

•232

= 0

⇒ 4𝜆 + 3 + 9𝜆 − 4 + 4𝜆 = 0 ⇒ 17 𝜆 = 1 ⇒ 𝜆 =1

17

⇒ 𝑃𝑄 =

2/1720/1732/17

⇒ 𝑓𝑖𝑛𝑑 𝑃𝑄

Relationship between lines

2 – D:

3 – D:

● the lines are coplanar (they lie in the same plane). They could be:

▪ intersecting ▪ parallel ▪ coincident

● the lines are not coplanar and are therefore skew(neither parallel nor intersecting)

𝑟1 =123

+ λ456

and 𝑟2 = −12−1

+ 𝜇203

.

Are the lines

∙ the same?…….check by inspection

∙ parallel?………check by inspection

∙ skew or do they have one point in common?

solving 𝑟1 = 𝑟2 will give 3 equations in and µ.

Solve two of the equations for and µ.

if the values of and µ do not satisfy the third equation then the lines are skew, and they do not intersect.

If these values do satisfy the three equations then substitute the value of or µ into the appropriate line and find the point of intersection.

Line 1: 𝑥 = −1 + 2𝑠, 𝑦 = 1 − 2𝑠, 𝑧 = 1 + 4𝑠Line 2: 𝑥 = 1 − 𝑡, 𝑦 = 𝑡, 𝑧 = 3 − 2𝑡Line 3: 𝑥 = 1 + 2𝑢, 𝑦 = −1 − 𝑢, 𝑧 = 4 + 3𝑢

a) Show that lines 2 and 3 intersect and find angle between them

b) Show that line 1 and 3 are skew.

a) 1 − 𝑡 = 1 + 2𝑢 ⇒ 𝑡 = −2𝑢, 𝑡 = −1 − 𝑢 ⇒ −2𝑢 = −1 − 𝑢 ⇒ 𝑢 = 1 & 𝑡 = −2

𝑐ℎ𝑒𝑐𝑘𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑧: 3 − 2𝑡 = 4 + 3𝑢 ⇒ 3 − 2 −2 = 4 + 3 1

𝑐𝑜𝑛𝑓𝑖𝑟𝑚𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 (3,−2, 7)

𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑓𝑜𝑟 𝑙𝑖𝑛𝑒 2 𝑎𝑛𝑑 𝑙𝑖𝑛𝑒 3 𝑎𝑟𝑒: 𝑏 =−11−2

𝑎𝑛𝑑 𝑑 =2−13

cos 𝜃 =𝑏 • 𝑑

𝑏 𝑑=

−2−1−6

1+1+4 4+1+9=

9

84⇒ 𝜃 ≈ 10.9𝑜

𝑏) − 1 + 2𝑠 = 1 + 2𝑢 ⇒ 2𝑠 − 2𝑢 = 2, 1 − 2𝑠 = −1 − 𝑢 ⇒ −2𝑠 + 𝑢 = −2, ⇒ 𝑢 = 0 & 𝑠 = 1

𝑐ℎ𝑒𝑐𝑘𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑧: 1 + 4𝑠 = 4 + 3𝑢 ⇒ 1 + 4 1 = 4 + 3 0 ⇒ 5 ≠ 4

⇒ 𝑛𝑜 𝑠𝑖𝑚𝑢𝑙𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑡𝑜 𝑎𝑙𝑙 3 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠

∴ 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑑𝑜 𝑛𝑜𝑡 𝑚𝑒𝑒𝑡, 𝑎𝑛𝑑 𝑎𝑠 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑏 ≠ 𝑘 𝑑 , 𝑘𝜖𝑅 𝑡ℎ𝑒𝑦 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑠𝑘𝑒𝑤.

Distance between two skew lines

𝑟 = 𝑎 + 𝜆 𝑏 𝑎𝑛𝑑 𝑟 = 𝑐 + 𝜇 𝑑

The cross product of 𝑏 and 𝑑 is perpendicular

to both lines, as is the unit vector:

𝑛 =𝑏×𝑑

𝑏×𝑑

The distance between the lines is then

𝑑 = 𝑛 • ( 𝑐 − 𝑎)

(sometimes I see it, sometimes I don’t)

PLANE EQUATION

● Vector equation of a plane 𝑟 = 𝑎 + 𝜆 𝑏 + µ 𝑐

A plane is completely determined by two intersecting lines, what can

be translated into a fixed point A and two nonparallel direction vectors

The position vector 𝑟 of any general point P on the plane passing

through point A and having direction vectors 𝑏 and 𝑐 is given by the equation

𝑟 = 𝑎 + 𝜆 𝑏 + µ 𝑐 𝜆, µ ∈ 𝑅 𝐴𝑃 = 𝜆 𝑏 + µ 𝑐

● Parametric equation of a plane: λ , μ are called a parameters λ,μ ∈ 𝑅

𝑥𝑦𝑧

=

𝑎1

𝑎2

𝑎3

+ 𝜆

𝑏1

𝑏2

𝑏3

+ 𝜇

𝑐1𝑐2𝑐3

⇒

𝑥 = 𝑎1 + 𝜆𝑏1 + 𝜇𝑐1𝑦 = 𝑎2 + 𝜆𝑏2 + 𝜇𝑐2𝑧 = 𝑎3 + 𝜆𝑏3 + 𝜇𝑐3

● Normal/Scalar product form of vector equation of a plane

𝑛 • 𝑟 = 𝑛 • 𝑎 + 𝜆 𝑏 + µ 𝑐 ⇒ 𝑟 • 𝑛 = 𝑎 • 𝑛 𝑜𝑟 𝑛 • 𝑟 − 𝑎 = 0

● Cartesian equation of a plane

𝑟 • 𝑛 = 𝑎 • 𝑛 ⤇ 𝑛1𝑥 + 𝑛2𝑦 + 𝑛3𝑧 = 𝑛1𝑎1 + 𝑛2𝑎2 + 𝑛3𝑎3 = 𝑑

𝑛1𝑥 + 𝑛2𝑦 + 𝑛3𝑧 = 𝑑

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑜𝑟𝑖𝑔𝑖𝑛:

𝐷 = 𝑟 • 𝑛 = 𝑎 • 𝑛

=𝑎•𝑛

𝑛12+𝑛2

2+𝑛32

=𝑛1𝑎1+𝑛2𝑎2+𝑛3𝑎3

𝑛12+𝑛2

2+𝑛32

Find the equation of the plane passing through the

three points P1(1,-1,4), P2(2,7,-1), and P3(5,0,-1).

𝑏 = 𝑃1𝑃2 =18−5

𝑐 = 𝑃1𝑃3 =41−5

𝑜𝑛𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑃1 =1−14

vector form:

𝑛 = 𝑖 𝑗 𝑘

1 8 −54 1 −5

=−35−15−31

Any non-zero multiple of 𝑛 is also

a normal vector of the plane. Multiply by -1.

𝑛 =351531

𝑟 =1−14

+ 𝜆18−5

+ µ41−5

351531

•1−14

= 144

𝐶𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑓𝑜𝑟𝑚:35𝑥 + 15𝑦 + 31𝑧 = 144

Find the equation of the plane with normal vector

135

containing point (-2, 3, 4) .

135

• −2,3,4 = −2 + 9 + 20 = 27

𝑥 + 3𝑦 + 5𝑧 = 27

Find the distance of the plane 𝑟●32−4

= 8

from the origin, and the unit vector

perpendicular to the plane.

32−4

= 29

1

29 𝑟●

32−4

=8

29

𝐷 =8

29 𝑛 =

1

29

32−4

It's fairly straightforward to convert a vector equation into a Cartesian equation, as you simply find the

cross product of the two vectors appearing in the vector equation to find a normal to the plane and use

that to find the Cartesian equation. But this process can't exactly be reversed to go the other way.

∙ To convert Cartesian -> vector form, you need either two vectors or three points that lie on the plane!

Convert the Cartesian equation of the plane x -2y + 2z = 5 into

(a) a vector equation of the form 𝑟 • 𝑛 = 𝐷, where 𝑛 is a unit vector.

(b) vector form

(c) State the perpendicular distance of the plane from the origin.

(a) 𝑛 =1

1+4+4

1−22

=1

3

1−22

⟹𝑥𝑦𝑧

●

1/3−2/32/3

= 5/3

5/3 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛

(b) choose three arbitrary random non-collinear points: 𝐴(0, 1/2, 3) 𝐵(0, 3/2, 4) 𝐶(3, 0, 1)

𝐴𝐵 =011

𝐴𝐶 =3

−1/2−2

⟹ 𝑟 =0

1/23

+ 𝜆011

+ 𝜇3

−1/2−2

(c) 5/3

ANGLES

● The angle between a line and a plane

𝑠𝑖𝑛 𝜃 = 𝑐𝑜𝑠 𝜙 =𝑛 • 𝑑

𝑛 𝑑

take acute angle

𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛𝑛• 𝑑

𝑛 𝑑

● The angle between two planes

The angle between two planes is the same

as the angle between their 2 normal vectors

𝑐𝑜𝑠 𝜃 =𝑛 • 𝑚

𝑛 𝑚

𝜃 = 𝑎𝑟𝑐 𝑐𝑜𝑠𝑛 • 𝑚

𝑛 𝑚

INTERSECTION of a LINE and a PLANE

First check that the line is not contained in the plane, nor parallel to it.

i.e456

•123

= 32

Substitute the line equation into the plane equation to obtain the value of the

line parameter, µ. Substitute for µ into the equation of the line to obtain the

co-ordinates of the point of intersection.

i.e. Solve

1 + 4𝜇−2 + 5𝜇−1 + 6𝜇

123

= 5 1 + 4µ - 4+10µ -3 + 18µ = 5.

Solve for µ and substitute into the equation of the line to get the point of intersection.

If this equation gives you something like 0 = 5, then the line will be parallel and not in the plane,

and if the equation gives you something like 5 = 5 then the line is contained in the plane.

Line L: 𝑟 =1−2−1

+ 𝜇456

and plane Π: 𝑥 + 2𝑦 + 3𝑧 = 5

therefore the line and the plane are not parallel and the line will intersect the plane in one point.

INTERSECTION of TWO PLANES and the EQUATION of the LINE of INTERSECTION

Solve between the two plane equations in terms of a parameter say, λ,

plane Π1: 𝑥 + 2𝑦 + 3𝑧 = 5 and plane Π2: 2𝑥 − 2𝑦 − 2𝑧 = 2

First, check by inspection, that the planes are not parallel (normal vectors are not parallel).

Find intersection:

0 0 01 2 32 −2 −2

052

~ 𝑅2 = 𝑅2 +3

2𝑅3 ~

0 0 04 −1 02 −2 −2

082

⤇ 𝑥 = 𝑡 𝑦 = 4𝑥 + 8 𝑧 = −3𝑥 + 7

⤇ 𝑡 = 𝑥 =𝑦 + 8

4=

𝑧 − 7

−3

which is the equation of the common line, which in vector form is 𝑟 =0−87

+ 𝑡14−3

(Equally you can write t = y as a function of x and z, or t = z as a function of x and y).

OR Find the vector product of both normals to give the direction of the line.

Then you need a point on the line – do it in the future

● INTERSECTION OF TWO or MORE PLANES

What does the equation 3x + 4y = 12 give in 2 and 3 dimensions?

ID: [email protected]: kokakoka11

http://www.globaljaya.net/secondary/IB/Subjects%20Report/May%202012%20subject%20report/Maths%20HL%20subject%20report%202012%20TZ1.pdf

https://www.osc-ib.com/ib-videos/default.asp