Vectors2

VECTORS II

1 Vector Product

The vector product can be seen in 3 different forms, used for different types of questions.

(1) Mathematical calculation of vector product:Vector (cross) product

a1a2a3

×

b1b2b3

=

a2b3 − a3b2−(a1b3 − a3b1)a1b2 − a2b1

(2) 3D picture of vector product:

a

b

n

Π

Let Π be the plane that a and b lies. The cross product of a and b produces a vector n that isperpendicular to Π. A vector that is perpendicular to a plane is also called a normal vector tothe plane.

(3) Vector product in terms of sine:

a× b = (|a||b| sin θ)n̂|a× b| =

∣∣|a||b| sin θ∣∣where θ is the angle between the 2 vectors.

Example 1.Let a = 2i+ j− 3k and b = −i+ 2j+ 3k, find the vectors a× b and b× a. What is the relationshipbetween a× b and b× a?

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2 Equation of Planes

m1

m2

n

Πb

A

A plane Π is defined by a point A which it passes through, and two direction vectors m1 and m2

that lies on the plane.Parametric form or Vector form

r =−→OA+ λm1 + µm2, λ, µ ∈ R

where m1,m2 are parallel to the plane and A is a point on the plane.

Scalar-Product form

Let r, a be points on the plane, n be a normal vector to the plane. Then,

r · n = a · n where n = m1 ×m2.

Proof: For any position vector −−→OR on the plane, since −→AR is perpendicular to n,

−→AR · n = 0(−−→

OR−−→OA

)· n = 0

−−→OR · n−

−→OA · n = 0−−→OR · n =

−→OA · n

Cartesian Form

We must first have the equation in scalar-product form.

r · n = D

r ·

abc

= D

xyz

·

abc

= D

ax+ by + cz = D (Cartesian form)


Example 2.t

(a) Find the equation of the following, in parametric, scalar-product and cartesian form.

Plane Π which passes through the point (1, 0, 2) and is parallel to the vectors

101

and

010

.

Solution:

Parametric: r =

102

+ λ

101

+ µ

010

, λ, µ ∈ R.

Scalar-Product:

101

×

010

=

−101

r · n = a · n

r ·

−101

=

102

·

−101

r ·

−101

= −1 + 2 = 1

r ·

−101

= 1

Cartesian: xyz

·

−101

= 1

−x+ z = 1

*Note that the following equations refer to the same plane.

r ·

−101

= 1 r ·

10−1

= −1 r ·

−202

= 2


(b) The plane Π contains the point A(2, 2,−2) and the line r =

13−3

+ λ

−201

, λ ∈ R.

Find the equation of Π1 in scalar product form.

[r ·

132

= 4]

(c) Find the cartesian equation of the following planes:

(i) x− y plane

x

y

z

(ii) x− z plane

x

y

z

(iii) y − z plane

x

y

z

(d) Given the following equation,

r ·

132

= 4,

how do we create points that lies on the plane?How do we check if a point lies on a plane?


3 Foot of perpendicular from a point to a plane

b

bP

F

n

Π

Finding foot of perpendicular from point to plane

Strategy:Given equation of the plane Π : r · n = D (1)

Step 1: Form the equation of the line ℓ that passes through P and is perpendicular to Π.

ℓ : r =−−→OP + λn, λ ∈ R. (2)

Step 2: Intersect ℓ with Π to get the foot of the perpendicular. That is, substitute (2) into (1).

(−−→OP + λn) · n = D

Substitute λ back into (2) to get the foot of the perpendicular.

Example 3.

Find the foot of perpendicular from the point P (2,−1, 3) to Π : r ·

−11−1

= 3.

Solution:Let ℓ be the line that passes through P and perpendicular to Π.

ℓ : r =

2−13

+ λ

−11−1

, λ ∈ R.

Intersect ℓ and Π, 2− λ−1 + λ3− λ

·

−11−1

= 3

−2 + λ− 1 + λ− 3 + λ = 3

−6 + 3λ = 3

λ = 3

∴ Foot of perpendicular: −−→OF =

2−13

+ 3

−11−1

=

−120

.


Example 4.The point P has position vector i + 2j - k and a plane Π has equation Π : x + 2y + 3z = 16. Let Abe the point with position vector 9i+ 2j+ k.

(i) Find the foot of perpendicular from P to Π.

(ii) Hence, find the shortest distance between P and Π.

(iii) Show that A lies on the plane Π.

(iv) Let ℓ be the line that passes through A and P . Find the vector equation of the reflection of theline ℓ in the plane Π.

[(i) −−→OF =

242

(ii)√14]


(iii) Since

921

·

123

= 9 + 4 + 3 = 16, A lies on the plane.

(iv)

Π

b

b

A

P

Fb

b

P ′

ℓ

−−→OF =

−−→OP +

−−→OP ′

2

2−−→OF =

−−→OP +

−−→OP ′

−−→OP ′ = 2

−−→OF −

−−→OP

= 2

242

−

12−1

=

365

Direction vector of ℓ =−−→OP ′ −

−→OA =

365

−

921

=

644

∴ r =

921

+ µ

644

, µ ∈ R.


4 Angles

In this section, denote m as the direction vector of the line, n as the normal vector of the plane.

θ

ℓ1

ℓ2

m1

m2

Acute angle between two lines

cos θ =

∣∣∣∣ m1 ·m2

|m1||m2|

∣∣∣∣ .

Π1 Π2

θ

θ

n2 n1

Acute angle between 2 planes

cos θ =

∣∣∣∣ n1 · n2

|n1||n2|

∣∣∣∣ .

ϕ

θ

Π

ℓ

m

n

Acute angle between line and a plane

cosϕ =

∣∣∣∣ m · n|m||n|

∣∣∣∣ .∴ θ = 90◦ − ϕ

Alternative Method

sin θ =

∣∣∣∣ m · n|m||n|

∣∣∣∣ .


Example 5.Find the ACUTE angle between the following two lines:

ℓ1 : r = (λ− 3)i+ (2λ− 1)j+ (2λ− 1)k, λ ∈ R.

ℓ2 : r = (3µ− 1)i+ (2µ− 1)j+ (µ− 1)k, µ ∈ R.

[36.7◦]

Example 6.Find the acute angle between the line r = (i+ 2j− k) + λ(i− j+ k) and plane 2x− y + z = 4.

[70.5◦]

Example 7.Find the acute angle between the planes r · (i+ j+ k) = 3 and r · (2i+ 2j− k) = 1. [54.7◦]


5 Relationship between a line and a plane

Are the line and plane parallel?

YES

The line lies in the plane. The line does not lie in the plane.

NO

The line intersects the plane at a point.

m

n

Π : r · n = d

ℓ

Recall that if two vectors a and b are perpendicular, then

a · b = 0.

Let m be a direction vector for the line ℓ and n be a normal for the plane Π.

m is parallel to the plane Π

⇔ m is perpendicular to n⇔ m · n = 0

3 different ways to check if ℓ lies on Π

Let the equations of the line ℓ and plane Π be given by

ℓ : r = a+ λm, λ ∈ R and Π : r · n = d.

(1) Check if ℓ and Π are parallel. If they are, then

m · n = 0.

If we have checked that the above holds, to check if ℓ lies in the plane Π, check that

(Any point on ℓ) ·n = d

(2) If (a+ λm) · n = d, then line lies in plane.

(3) Show that 2 points on ℓ lie on Π.

What about the case where ℓ does not lie on Π?


Example 8.

(a) Determine the relationship between pairs of lines and planes:

(i) ℓ1 : r =

122

+ λ

2−1−1

and Π1 : r ·

111

= 5

(ii) ℓ2 : r = (i+ j+ k) + λ(i− 2j+ k) and Π2 : r · (i+ j+ k) = 5

(b) Show that the line ℓ1 : r =

122

+ λ

2−1−1

intersects the plane r ·

101

= 2 at only 1 point.

[(ii) Parallel but does not lie in plane ]

Solution:

(i) m · n =

2−1−1

·

111

= 2− 1− 1 = 0. Therefore, ℓ1 and Π1 are parallel.

Furthermore, since

122

·

111

= 1 + 2 + 2 = 5, ℓ1 lies in Π1.

Alternative Method:

Since

1 + 2λ2− λ2− λ

·

111

= 1 + 2λ+ 2− λ+ 2− λ = 5, ∴ ℓ1 lies in Π1.


6 Relationship between 2 planes

6.1 Parallel Planes

Two planes are parallel to each other ⇐⇒ Their normals are scalar multiple of each other.

Q: What is the relationship between the following 2 planes?

p1 : r ·

1−12

= 2, p1 : r ·

−22−4

= 3.


6.2 Line of intersection between 2 planes

Any 2 non parallel, non identical planes will intersect in a line.n1

n2

ℓ

Π1

Π2

Method 1: Cross the 2 normalsFinding line of intersection between 2 planes

The direction vector, m, of the line of intersection between Π1 and Π2 is given by

m = n1 × n2,

where n1, n2 are the normal vectors of Π1 and Π2 respectively.

We will use a GC if there are no unknowns in n1 and n2.

Method 2: Using GCExample 9.The equations of two planes p1, p2 are

2x− 4y + z = 6

x+ y − 2z = 6

respectively. The planes p1 and p2 meet in a line l. Find a vector equation of l.

Question: Is there an alternative method to find a common point on both planes?


6.3 Relationship of two perpendicular planes

p1

p2

n2n1

If 2 planes, p1 and p2 are perpendicular, then the following occurs:

(a) n1 is parallel to p2 =⇒ p2 contains the direction vector n1,

(b) n2 is parallel to p1 =⇒ p1 contains the direction vector n2.

Example 10 (2012/RVHS/Prelim/I/10modified).The equations of two planes p1 and p2 are

2y − z = 0,

2x+ z = 2.

Find the equation of the plane p3 that passes through the point (1, 1, 1) and is perpendicular to bothp1 and p2.

Solution: p1

p2

p1

p2

p3

p3 is perpendicular to p1 =⇒ p3 is parallel to n1.p3 is perpendicular to p2 =⇒ p3 is parallel to n2.∴ p3 is parallel to both n1 and n2.

n3 = n1 × n2 =

02−1

×

201

=

2−2−4

Therefore,

p3 : r ·

2−2−4

=

111

·

2−2−4

= 2− 2− 4

= −4


6.4 Reflection of a plane in another plane.π1

π2

l

2 direction vectors of the reflected plane

Suppose we want to find the reflection of plane π1 in π2. Lets call it π′1.

We also know that π1 and π2 intersect at the line l.

1) l will also lie on the reflected plane π′1. Hence π′

1 also contains direction vector of l.

2) Take a point P from π1 and reflect it in π2. This reflected point P ′ will be on π′1.

Now take a point A from l (which is also on π′1). π′

1 will contain direction vector−−→AP ′.


Example 11 (2011/YJC/Prelim/I/10modified).The planes π1 and π2 have equations r · (2i− k) = 3 and r · (−i+ 3j) = 2 respectively. The 2 planesintersect in a line l.

(i) Find a vector equation of l. [2]

(ii) Find the coordinates of the point Q which is the reflection of the point P (5,−2, 7) in π2. [4]

(iii) It is given that P lies on π1. Hence or otherwise, find a vector equation of the reflection of theplane π1 in π2. [2]

[(i) r =

32760

+ t

316

, t ∈ R (ii) (2.4, 5.8, 7) (iii) r =

32760

+ s

316

+ t

− 910

−13930−7

, s, t ∈ R]


7 Distance involving plane

bA

B

n

ΠbF

θ∣∣∣−−→AB · n̂∣∣∣

∣∣∣−−→AB × n̂∣∣∣

bA

B

Π

bF

ℓ bA

BΠ2

bF

Π1

∣∣∣−−→AB × n̂∣∣∣ ∣∣∣−−→AB × n̂

∣∣∣

∣∣∣−−→AB · n̂∣∣∣ ∣∣∣−−→AB · n̂

∣∣∣

Proof:

Dist between the point and plane

|−−→AB · n̂| =

∣∣∣|n̂||−−→AB| cos θ∣∣∣

=∣∣|−−→AB| cos θ

∣∣= |

−→AF |

Length of projection onto plane

|−−→AB × n̂| =

∣∣∣|n̂||−−→AB| sin θ∣∣∣

=∣∣|−−→AB| sin θ

∣∣= |

−−→FB|

8 Distance involving line

ℓ : r = a+ λm, λ ∈ R

b

A

B

b θ

|−−→AB × m̂|

|−−→AB · m̂|

Distance from point to line:|−−→AB| sin θ = |

−−→AB × m̂|

Projection of −−→AB to line:|−−→AB| cos θ = |

−−→AB · m̂|


Example 12.The equation of a plane π is shown below:

π : r.

21−2

= 6

(a) The point A has position vector 3i − j + 4k. B is another point such that −−→OB = 3i − 6j + 2k.

Find the length of the projection of AB onto the plane π.

(b) Find the distance between point A and the plane.

Solution:

(a)

bA

21−2

π

bB

θ

−−→AB =

−−→OB −

−→OA

=

3−62

−

3−14

=

0−5−2

|−−→AB × n̂| =

∣∣∣∣∣∣ 0−5−2

× 1√22 + 1 + 22

21−2

∣∣∣∣∣∣=

1

3

∣∣∣∣∣∣ 0−5−2

×

21−2

∣∣∣∣∣∣=

1

3

∣∣∣∣∣∣12−410

∣∣∣∣∣∣=

1

3

√122 + 42 + 102

=2√65

3

(b)

bA

π

θ

C

The point C =

060

lies on the plane since060

·

21−2

= 6.

−→AC =

−−→OC −

−→OA =

060

−

3−14

=

−374

|−→AC · n̂| =

∣∣∣∣∣∣−3

74

· 13

21−2

∣∣∣∣∣∣=

1

3| − 6 + 7 + 8|

= 3


Example 13.Show the the line ℓ1 : r = 2i− 2j+ 3k+ λ(i− j+ 4k) is parallel to the plane Π : r · (i+ 5j+ k) = 5.Find the distance between them. [ 10√

27units]


Example 14 (2011/VJC/Prelim/I/12).The position vectors of points A,B,C and D with respect to the origin O are −2i+ 5j+ 5k, 3i+ j+4k, 3i− 4j+ k and i+ j− 5k respectively.

(i) Obtain an equation of the plane Π, containing A,B and C in the r · n = p form.

(ii) Calculate the angle between Π and the line joining A and D. Hence, or otherwise, find theshortest distance of D from Π.

[(a) r.

715−25

= −64 (b) 39.0◦, 7.04 units ]


9 Alternative method to find distance between two planes

Distance from origin to plane

If r · n̂ = d, then,Distance from origin to plane = |d|

Example 15.

Find the distance between the following plane and the origin: Π1 : r ·

151

= 3

Solution:

Π1 : r ·

1√27

151

=3√27

Therefore, distance between Π1 and origin is 3√27.

b

Π1

Π2

O

d1

d2

n

Distance between 2 parallel planes

Π1 : r · n̂ = d1 Π2 : r · n̂ = d2

Distance between the two planes = |d1 − d2|.

Note: If d1 and d2 are of the same signs, then the planes Π1 and Π2 lie on the same side of the origin.If d1 and d2 are of opposite signs, then the planes Π1 and Π2 lie on different sides of the origin.

Example 16.Find the distance between the following planes and determine if they lie on the same side of the origin.

Π1 : r ·

151

= 3 Π2 : r ·

151

= −2

Solution:

Π1 : r ·

1√27

151

=3√27

Π2 : r ·

1√27

151

=−2√27

Therefore, distance between Π1 and Π2 is∣∣∣ 3√

27− −2√

27

∣∣∣ = 5√27.

They lie on different sides of the origin since 3√27

and −2√27

have different signs.


Distance from origin to plane

If r · n̂ = d, then distance from origin to plane = |d|.

Proof: (Optional)

bO

A

n

Π : r · n̂ = db

θ

|a · n̂|

Let A be a point on Π such that its position vector is a. Then,

length of projection of −→OA onto n = |a · n̂|.

In addition, since A lies on Π, we also have

a · n̂ = d.

Hence,length of projection of −→OA onto n = |a · n̂| = |d|.


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