IG: @musclemath_ 1

NAME:

TOPIC: Vectors II

Website: www.musclemathtuition.com

Instagram: musclemath_

Facebook: www.facebook.com/musclemathtuition

Email: musclemathtuition@gmail.com

Concepts Overview: - Forms of a Plane - Point of Intersection between line and plane - Acute angle between two planes - Acute angle between a line and a plane

- Parallel and Perpendicular planes - Foot of Perpendicular, shortest distance and reflection concepts in planes - Relationships between two planes - Direction Cosines

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Various forms of Equations of Planes

In lines, there are 3 different forms of equations that lines can take. They are vector equation form, parametric equations form and Cartesian equation form. Planes also have 3 different forms. They are called parametric form (vectors equation form), Scalar Product form and Cartesian form.

Parametric Form If a plane contains point A and is parallel to vectors 𝐦1 and 𝐦𝟐, then the equation of the plane may be written in the form 𝐫 = 𝐚 + 𝜆𝐦1 + 𝜇𝐦𝟐, where 𝜆, 𝜇 ∈ ℝ, where 𝐚 = position vector of a point on the plane 𝐦1 and 𝐦2 = vectors parallel to the plane.

Scalar Product Form

𝐫 ∙ 𝐧 = 𝐚 ∙ 𝐧 where 𝐧 = vector normal to the plane 𝐚 = Position vector of a point on the plane

OR

𝐫 ∙ 𝐧 = 𝑑 where 𝐧 = unit vector normal to the plane 𝑑 = Perpendicular distance from the plane to the origin Note: If the plane passes through the origin, then 𝑑 = 0. If a plane is parallel to two vectors 𝐛 and 𝐜, then 𝐛 × 𝐜 gives a vector normal to the plane (i.e. 𝐛 × 𝐜 = 𝐧).

Cartesian Form The Cartesian equation of a plane can be derived from the scalar product form. Given that 𝐫 ∙ 𝐧 = 𝐷 where 𝒏 = 𝑎𝐢 + 𝑏𝐣 + 𝑐𝐤, then

(𝑥𝑦𝑧) ∙ (

𝑎𝑏𝑐) = 𝐷 ⟹ 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝐷 .

Note that the equation of the:

- 𝑥𝑦 plane is 𝐫 ∙ (001) = 0 or 𝑧 = 0

- 𝑥𝑧 plane is 𝐫 ∙ (010) = 0 or 𝑦 = 0

- 𝑦𝑧 plane is 𝐫 ∙ (100) = 0 or 𝑥 = 0

Note: To uniquely define a plane, one needs one point and two directional vectors. The dot product will result in a scalar and the cross product will result in a vector.

IG: @musclemath_ 3

Example 1 (a) The plane 𝜋1 passes through the points 𝐴, 𝐵 and 𝐶 whose position vectors are given by

(110) , (

023) and (

203) respectively.

Find the parametric, scalar product and Cartesian equation of 𝜋1.

(b) The equation in scalar product form of 𝜋2 is given by 𝐫 ∙ (−11

−1) = 3.

Calculate the perpendicular distance from 𝜋2 to the origin. Find a parametric equation of 𝜋2.

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Point of Intersection of a Line and a Plane

If 𝑙: 𝐫 = 𝐚 + 𝜆𝐝 and 𝜋: 𝐫 ∙ 𝐧 = 𝐷, 1. Sub the line equation into the plane equation.

(𝐚 + 𝜆𝐝) ∙ 𝐧 = 𝐷 𝐚 ∙ 𝐧 + 𝜆𝐝 ∙ 𝐧 = 𝐷 𝜆 = _____

2. Sub 𝜆 = _____ back into the line equation to obtain the point of intersection. Note: Remember to convert the plane equation into the scalar product form before proceeding.

Example 2

The vector equation of the plane 𝜋 is given by 𝐫 = (600) + 𝜆 (

4−1−1

) + 𝜇 (6

−30

), where 𝜆, 𝜇 ∈ ℝ. Find the

position vector of the point of intersection of the line passing through the points (2,−1,−1) and (6, 1, 3).

IG: @musclemath_ 5

Point of Intersection between two lines

If 𝑙1: 𝐫 = 𝐚 + 𝜆𝐝𝟏 and 𝑙2: 𝐫 = 𝐛 + 𝜇𝐝𝟐 1. Equate the line equations and form three equations. 2. Use the equations to find the unknowns 𝜆 and 𝜇. 3. Use the values found and substitute them back into either 𝑙1 or 𝑙2 to find the point of intersection.

Example 3 [SRJC Prelim/2011/P2/Q2(b)] (2016 MJC Extra Questions 2) The equation of line 𝑙2 is 𝐫 = (−2 − 𝜆)𝐢 + (3 + 3𝜆)𝐣 + (1 + 4𝜆)𝐤, 𝜆 ∈ ℝ. The points 𝐵 and 𝐶 have position vectors 4𝒊 + 3𝐣 − 𝐤 and 11𝒊 + 𝑎𝒌 respectively. The line 𝑙2 intersects the line which passes through 𝐵 and 𝐶, at 𝑃. Find the value of 𝑎. Hence find the coordinates of 𝑃. [4]

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Acute Angle between Two Planes

2D Version:

3D Version:

𝜋1: 𝐫 ∙ 𝐧1 = 𝐷1 𝜋2: 𝐫 ∙ 𝐧2 = 𝐷2 Concept: The angle between two planes = The angle between the two normal of the planes Let 𝜃 be the acute angle between 𝜋1 and 𝜋2. From the dot product,

|𝒏𝟏 ∙ 𝒏𝟐| = |𝒏𝟏||𝒏𝟐| cos 𝜃

∴ cos 𝜃 =|𝒏𝟏 ∙ 𝒏𝟐|

|𝒏𝟏||𝒏𝟐|

Example 4 The vector equation of the planes 𝜋1 and 𝜋2 are given as follows: 𝜋1: 𝑥 + 2𝑦 − 2𝑧 = −4

𝜋2: 𝐫 ∙ (1

−1−8

) = 3

Find the acute angle between the two planes.

𝜃

𝐧𝟏

𝐧𝟐 𝜃 𝐧2

𝐧1

θ

θ

θ

π1

π2

IG: @musclemath_ 7

Finding Acute Angle between a Line and a Plane

𝑙 ∶ 𝐫 = 𝐚 + 𝜆𝐝 𝜋 ∶ 𝐫 ∙ 𝐧 = 𝐷 Let 𝜃 be the acute angle between the line 𝑙 and the plane 𝜋. Given that the direction vector of 𝑙 is 𝐝 and the normal to the plane 𝜋 = 𝐧, then

cos(90 − 𝜃°) =|𝐧 ∙ 𝐝|

|𝐧||𝐝|

∴ sin 𝜃 = |𝐧 ∙ 𝐝|

|𝐧||𝐝|

Example 5 The Cartesian equation of the plane 𝜋 is given by 2𝑥 − 5𝑦 − 2𝑧 = 8. Find the acute angle between the plane 𝜋 and the line passing through the points (0, 0, 3) and (−7, 1, −4), giving your answer to the nearest degree.

𝜋

𝜃

n d

𝑙 𝛼 = (90° − 𝜃)

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Parallel and Perpendicular Lines and Planes

(A) Parallel and Perpendicular Lines 𝑙1: 𝐫 = 𝐚1 + 𝜆𝐝1 𝑙2: 𝐫 = 𝐚2 + 𝜇𝐝2 • 𝑙1 and 𝑙2 are parallel ⟺ 𝐝1 is parallel to 𝐝2

⟺ 𝐝1 = 𝑘𝐝2for some constant 𝑘 • 𝑙1 and 𝑙2 are perpendicular ⟺ 𝐝1 is perpendicular to 𝐝2

⟺ 𝐝1 ∙ 𝐝2 = 0

(B) Parallel and Perpendicular Planes 𝜋1: 𝐫 ∙ 𝐧1 = 𝐷1 𝜋2: 𝐫 ∙ 𝐧2 = 𝐷2 • 𝜋1 and 𝜋2 are parallel ⇔ 𝐧1 is parallel to 𝐧2

⇔ 𝐧1 = 𝑘𝐧2 for some constant 𝑘 • 𝜋1 and 𝜋2 are perpendicular ⟺ 𝐧1 is perpendicular to 𝐧2

⟺ 𝐧1 ∙ 𝐧2 = 0

(C) Parallel Line and Plane • 𝑙 and 𝜋 are parallel ⟺ 𝑙 is perpendicular to 𝒏

⟺ 𝐝 ∙ 𝐧 = 0

Condition for a line 𝒍 to lie on a plane 𝝅

Method 1: For a line 𝑙 to lie on a plane 𝜋, the following two conditions must both be satisfied.

1. 𝑙 must first be parallel to the plane 𝜋, which means 𝐝 ∙ 𝐧 = 0 2. One point on the line 𝑙 must also lie on the plane 𝜋, which means 𝐚 ∙ 𝐧 = 𝐷

Method 2: Show that (𝐚 + 𝜆𝐝) ∙ 𝐧 = 𝐷, for all 𝜆 ∈ ℝ

(D) Perpendicular Line and Plane

• 𝑙 is perpendicular to the plane 𝜋 ⟺ 𝐧 is parallel to 𝐝 ⟺ 𝐧 = 𝑘𝐝 for some constant 𝑘

d

n

l: 𝐫 = 𝐚 + λ𝐝

𝜋: 𝐫 ∙ 𝐧 = 𝐷

d

n

l: 𝐫 = 𝐚 + λ𝐝

𝜋: 𝐫 ∙ 𝐧 = 𝐷

IG: @musclemath_ 9

Example 6 The plane 𝜋 contains the points 𝐴(1,−4, 1), 𝐵(2, −1, 0) and 𝐶(0, 1, 2).

The lines 𝑙1 and 𝑙2 have equations 𝒓 = (122) + 𝜆 (

−3−23

) and 𝒓 = (141) + 𝜇 (

10

−1) respectively.

(i) Show that 𝜋 and 𝑙1 are parallel. (ii) Show that 𝑙2 lies on 𝜋.

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Various Problems involving Points, Lines and Planes

Foot of perpendicular from a given point P to a plane 𝝅

2D Version: 3D Version: Step 1: Let 𝐹 be the foot of perpendicular from 𝑃 to 𝜋.

Step 2: Introduce a new line: 𝑙𝑃𝐹: 𝒓 = 𝑂𝑃⃗⃗ ⃗⃗ ⃗ + 𝜆𝒏 Step 3: 𝐹 is the point of intersection of 𝑙𝑃𝐹 and 𝜋 (i.e. solve simultaneously to find 𝐹)

𝑙𝑃𝐹 : 𝒓 = 𝑂𝑃⃗⃗ ⃗⃗ ⃗ + 𝜆𝒏… (1) 𝜋: 𝒓 ∙ 𝒏 = 𝐷…(2)

Sub (1) into (2), → (𝑂𝑃⃗⃗ ⃗⃗ ⃗ + 𝜆𝒏) ∙ 𝒏 = 𝐷

→ 𝜆 = _____

Step 4: Sub 𝜆 = _____ back into (1) to obtain 𝑂𝐹⃗⃗⃗⃗ ⃗. Perpendicular (shortest) Distance from a Point to a Plane 2D Version: 3D Version:

Method 1:

Find the position vector of the foot of perpendicular 𝑂𝐹⃗⃗⃗⃗ ⃗. From there, calculate the perpendicular

distance |𝑃𝐹⃗⃗⃗⃗ ⃗|.

Method 2:

Find a point 𝑂𝐴⃗⃗⃗⃗ ⃗ on 𝜋. From there, find 𝐴𝑃⃗⃗⃗⃗ ⃗. In triangle 𝐴𝑃𝐹, 𝑃𝐹 is the adjacent side to angle 𝜃. The perpendicular distance 𝑃𝐹 is

given by 𝐴𝑃 cos 𝜃 = |𝐴𝑃⃗⃗⃗⃗ ⃗ ∙ 𝐧| where 𝐧 =𝐧

|𝑛|.

P 𝑙𝑃𝐹: 𝐫 = 𝑂𝑃⃗⃗ ⃗⃗ ⃗ + 𝜆𝐧

F

𝜋: 𝐫 ∙ 𝐧 = 𝐷

n

𝜋

n

𝑃

𝐹

A

P

n

𝜋: 𝐫 ∙ 𝐧 = 𝐷

Distance from P to 𝜋

F

𝜃

𝜋

𝑃

𝐹 𝐴

n

𝜃

IG: @musclemath_ 11

Various Problems involving Points, Lines and Planes

Reflection of a Point P in a plane 𝝅

2D Version: 3D Version: Let 𝑄 be the point of reflection of 𝑃 in 𝜋, then 𝐹 is the midpoint of 𝑃𝑄.

Step 1: Notice that 𝑃𝑄⃗⃗ ⃗⃗ ⃗ = 2𝑃𝐹⃗⃗⃗⃗ ⃗

Step 2: ∴ 𝑂𝑄⃗⃗⃗⃗⃗⃗ can then be deduced using 𝑂𝑄⃗⃗⃗⃗⃗⃗ − 𝑂𝑃⃗⃗ ⃗⃗ ⃗ = 2𝑃𝐹⃗⃗⃗⃗ ⃗.

Length of Projection of 𝑷𝑸⃗⃗⃗⃗⃗⃗ onto 𝝅 and the normal of 𝝅

2D Version: 3D Version:

Length of Projection of 𝑃𝑄⃗⃗ ⃗⃗ ⃗ onto 𝜋

= ||𝑃𝑄⃗⃗ ⃗⃗ ⃗| sin 𝜃|

= ||𝑃𝑄⃗⃗ ⃗⃗ ⃗||𝐧| sin𝜃|

= |𝑃𝑄⃗⃗ ⃗⃗ ⃗ × 𝐧|

Length of Projection of 𝑃𝑄⃗⃗ ⃗⃗ ⃗ onto the normal of 𝜋

= ||𝑃𝑄⃗⃗ ⃗⃗ ⃗| cos 𝜃|

= ||𝑃𝑄⃗⃗ ⃗⃗ ⃗||𝐧| cos 𝜃|

= |𝑃𝑄⃗⃗ ⃗⃗ ⃗ ∙ 𝐧|

𝜋: 𝐫 ∙ 𝐧 = 𝐷

P

F

Q

𝜋

n

𝑃

𝐹

𝑄

𝑃

𝑄

𝜃

𝐧

𝜋: 𝐫 ∙ 𝐧 = 𝐷 Length of Projection of PQ onto 𝜋

Length of Projection of PQ onto the normal of 𝜋

𝜋

𝑄

𝑃 n

𝜃 Length of Projection

of 𝑃𝑄⃗⃗ ⃗⃗ ⃗ onto the plane

𝜋

𝑄

𝑃 n

𝜃 Length of Projection of

𝑃𝑄⃗⃗ ⃗⃗ ⃗ onto normal of plane

𝜋

𝑄

𝑃 n

𝜃 Length of Projection

of 𝑃𝑄⃗⃗ ⃗⃗ ⃗ onto the plane

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Example 7 The equation of the plane 𝜋 is given by the equation 2𝑧 + 𝑥 − 2 = 0. The coordinates of points 𝐴 and 𝐵 are (7, 2, 5) and (16,3, 8) respectively. The line 𝑙 passes through the points 𝐴 and 𝐵. (a) Find the position vector of the foot of perpendicular from point 𝐴 to 𝜋.

Hence, or otherwise, find the shortest distance from 𝐴 to 𝜋.

(b) Find the point of intersection between 𝑙 and 𝜋.

(c) 𝑙′ is the reflection of the line 𝑙 in 𝜋. Find the position vector of the point of reflection of 𝐴 in 𝜋. Hence, find the equation of 𝑙′.

(d) Find the length of projection of 𝐴𝐵 onto

(i) the plane 𝜋, (ii) a normal of the plane 𝜋.

IG: @musclemath_ 13

Example 8 [HCI J2 BlockTest2/2014/P1/Q9] (JJC Topical 2017) The point 𝐴 has coordinates (3, 0,−2) and the plane Π has equation 2𝑥 − 𝑦 + 3𝑧 = 7. The line through

𝐴 parallel to the line 3−𝑥

2= 𝑦 =

𝑧+2

2 meets Π at the point 𝐵. The perpendicular from 𝐴 to Π meets Π at

point 𝐶. (i) Show that the coordinates of 𝐵 are (−11, 7, 12) and find the coordinates of 𝐶. (ii) Find the equation of the image of the line 𝐴𝐵 after a reflection in Π.

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Example 9 [CJC Promo/2017/Q8] The line 𝑙 has equation 𝐫 = 4𝐢 − 9𝐣 + 9𝐤 + 𝜆(−2𝐢 + 𝐣 − 2𝐤), 𝜆 ∈ ℝ. (i) The line 𝑙 lies in the plane with cartesian equation 𝑎𝑥 + 𝑏𝑦 − 3𝑧 + 1 = 0, where 𝑎 and 𝑏 are

constants. Find the values of 𝑎 and 𝑏. [4] The point 𝑃 has position vector 3𝐢 + 8𝐣 + 5𝐤. (ii) Find the foot of perpendicular from 𝑃 to 𝑙. [3] (iii) Hence or otherwise, find the shortest distance from 𝑃 to 𝑙. [2]

IG: @musclemath_ 15

Example 10 [HCI Promo/2014/Q8]

A point 𝑃(1, 4,−3) lies on the line L with equation 1 − 𝑥 =𝑧+3

2, 𝑦 = 4. The plane p1 has equation

2𝑥 − 𝑦 − 2𝑧 = 16. (i) Find Q, the point of intersection between L and p1. (ii) Find the exact value of the sine of the acute angle between L and p1 . Hence find the length of

projection of 𝑃𝑄⃗⃗ ⃗⃗ ⃗ onto p1. (iii) Find, in the form 𝐫 ⋅ 𝐧 = d , the equation of the plane p2 that contains L and is perpendicular to

p1 .

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Relationship between 2 planes 2 planes can have one of 3 relationships:

Parallel Intersecting (at a line) Coincidental

• Parallel Planes

• 𝒏𝟏 ∥ 𝒏𝟐 ⟺ 𝒏𝟏 = 𝑘𝒏𝟐, 𝑘 ∈ ℝ • To find distance between 2 planes:

Step 1: Find the scalar product forms, 𝐫 ∙ 𝐧 = 𝑑, of both planes. Step 2: Since 𝑑1 and 𝑑2 will represent the shortest distance from each plane to the origin, we take |𝑑1 − 𝑑2| as the distance between two planes. Note: 𝐧�̂� and 𝐧�̂� must be the same.

• Coincidental Planes

Essentially, the same plane. However, one plane can have different forms of equations. • Intersecting Planes

Two intersecting planes must intersect at a line. To find the equation of the line:

Method 1 (Use when there are no unknowns present in the Plane equations)

Step 1: Firstly, obtain the two Cartesian equations of the planes.

For example, {𝑥 − 𝑦 + 𝑧 = −1

3𝑥 − 2𝑦 − 3𝑧 = −8

Step 2: Use the PlySmlt2 function in the GC. “apps” → “PlySmlt2” → “Simultaneous Eqn Solver". Step 3: Key in the number of equations and unknowns.

In the example above, we have 2 equations and 3 unknowns. Step 4: Key in information into the system matrix. In the example above,

[1 −1 13 −2 −3

−1−8

]

Then press solve (graph). Step 5: Deduce Solution. The GC will give a ‘solution set’. In the example above,

{𝑥1 = −6 + 5𝑥3

𝑥2 = −5 + 6𝑥3

𝑥3 = 𝑥3

Hence the answer is where 𝑥3 = 𝜆, 𝒓 = (−6−50

) + 𝜆 (561) , 𝜆 ∈ ℝ.

𝐧𝟏 𝐧𝟐

𝐧𝟏

𝐧𝟐

𝐧𝟏

𝐧𝟐

𝐵

𝐴

IG: @musclemath_ 17

Method 2 (Use when there are unknowns present) Step 1: Suppose 𝑙 is the line of intersection between two non-parallel planes 𝜋1: 𝐫 ∙ 𝐧𝟏 = 𝐷1 and 𝜋2: 𝐫 ∙ 𝐧𝟐 = 𝐷2, then 𝑙 is perpendicular to both 𝐧1 and 𝐧2. Step 2: Therefore, directional vector, 𝐝, of the line 𝑙 can be obtained by 𝐧𝟏 × 𝐧𝟐. i.e. 𝐝 = 𝐧𝟏 × 𝐧𝟐 or some scalar multiple of 𝐧𝟏 × 𝐧𝟐. Step 3: Find a point 𝐚 on the line 𝑙. Note that this point is sometimes easily obtainable via inspection. Step 4: Equation of the line of intersection is 𝐫 = 𝐚 + λ𝐝.

Example 11 [RIJC Year 6 CT Term 3/2014/Q6(i)-(iii)] The plane 𝑝1 has equation 𝐫 ∙ (𝐢 + 𝐤) = 7, and the points 𝐴 and 𝐵 have position vectors 𝐢 + 6𝐤 and 2𝐢 +α𝐣 + 2𝐤 respectively, where 𝛼 ∈ ℝ. (i) Find, in terms of 𝛼, the position vector of 𝑁, the foot of perpendicular from 𝐵 to 𝑝1. The plane 𝑝2 contains the points 𝐴, 𝐵 and 𝑁. (ii) Show that the equation of 𝑝2 is 𝐫 ∙ (−α𝐢 + 5𝐣 + α𝐤) = 5α. (iii) Find, in terms of 𝛼, the equation of the line of intersection between 𝑝1 and 𝑝2.

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Example 12 [CJC Promo/2015/Q10 (i)-(iv)] The points A, B and C have position vectors given respectively by

𝐚 = 𝐢 + 2𝐤, 𝐛 = 2𝐢 + 𝐣 and 𝐜 = 2𝐢 + 2𝐣 + 3𝐤. It is known that the plane 1 contains A, B and C. (i) Show that the cartesian equation of plane 1 is 5𝑥 – 3𝑦 + 𝑧 = 7. [3] (ii) The point P has position vector 5i + j – k. Find the position vector of the foot of perpendicular

from P to plane 1 and hence find the exact distance from P to the plane 1. [4]

(iii) The point P is the reflection of point C about another plane 2. Show that the cartesian

equation of plane 2 is 3𝑥 – 𝑦 – 4𝑧 = 5. [3]

(iv) Find the equation of the line of intersection between planes 1 and 2. [2]

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Example 13 [PJC J2 CommonTest1/2014/P1/Q12(i)-(iii)]

The planes 𝑝1 and 𝑝2 have equations 𝐫 ∙ (−2−15

) = 3 and 𝐫 ∙ (𝑎𝑏1) = 𝑐 respectively, where 𝑎, 𝑏 and 𝑐 are

constants. 𝑝1 is perpendicular to 𝑝2 and the two planes meet in a line with an equation given by 𝐫 =

(021) + 𝜆 (

1−20

) , 𝜆 ∈ ℝ.

(i) Find the values of 𝑎, 𝑏 and 𝑐. (ii) A plane intersects 𝑝1 and 𝑝2 at a unique point. State the 𝑧 −coordinate of this point.

The plane 𝑝3 has equation 𝐫 ∙ (1

−2𝑑

) = 4, where 𝑑 is a positive constant.

(iii) If 𝑝3 makes an angle of 60° with 𝑝1, show that 𝑑 = √15

7.

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Example 14 [HCI Prelim/2008/P1/Q12(b)] (2016 MJC Extra Questions 2) Referring to the origin 𝑂, two planes 𝜋1 and 𝜋2 are given by

𝜋1: 𝐫 ∙ (12

−4) = 13 and 𝜋2: 𝐫 ∙ (

133) = −8.

(i) Given that a point 𝐴(1, 7,−10) lies on 𝜋2, show that the perpendicular distance from 𝐴 to 𝜋1 is

2√21. [2]

(ii) Hence or otherwise find 𝑂𝐵⃗⃗⃗⃗ ⃗ where 𝐵 is the image of 𝐴 when reflected in the plane 𝜋1. [2] (iii) Write down the Cartesian equations of both 𝜋1 and 𝜋2. [1]

Find a vector equation of the line of intersection of 𝜋1 and 𝜋2. [1] (iv) Find a vector equation of the plane which is the image of 𝜋2 when 𝜋2 is reflected in 𝜋1. [3]

IG: @musclemath_ 21

Example 15 [NJC Prelim/2018/P2/Q2]

The planes 𝑝1 and 𝑝2, have equations 2𝑥 + 3𝑦 + 6𝑧 = 0 and 𝐫 ∙ (122) = 6 respectively.

(i) Find a vector equation of the line of intersection, 𝑙, between 𝑝1 and 𝑝2. [2] The line 𝑚 passes through the points 𝐴(2, 1, 1) and 𝐵(5, 4, 2). (ii) Verify that 𝐴 lies on 𝑝2. [1] (iii) Find the coordinates of the points on 𝑚 that are equidistant from planes 𝑝1 and 𝑝2. [5]

www.musclemathtuition.com 22

Direction Cosines The direction cosines of a vector are the cosines of the angles between the vector and the positive direction of the three coordinates axes.

If 𝛼, 𝛽 and 𝛾 are the angles that 𝑂𝑃⃗⃗ ⃗⃗ ⃗ makes with the positive direction of 𝑥-axis, 𝑦-axis and 𝑧-axis

respectively, then the direction cosines 𝑙, 𝑚 and 𝑛 of 𝑂𝑃⃗⃗ ⃗⃗ ⃗ are given by

𝑙 = cos 𝛼 =𝑎

√𝑎2 + 𝑏2 + 𝑐2

𝑚 = cos 𝛽 =𝑏

√𝑎2 + 𝑏2 + 𝑐2

𝑛 = cos 𝛾 =𝑐

√𝑎2 + 𝑏2 + 𝑐2

where 𝑙2 + 𝑚2 + 𝑛2 = 1 and 𝑂𝑃⃗⃗ ⃗⃗ ⃗ = (𝑎, 𝑏, 𝑐).

From these definitions, the unit vector of 𝑂𝑃⃗⃗ ⃗⃗ ⃗ =1

√𝑎2+𝑏2+𝑐2(𝑎𝑏𝑐) = (

cos 𝛼cos 𝛽cos 𝛾

).

Example 16 [Mathematics-The Core Course for A level (1985)/pg491/Q6 (modified)]

The line passes through the point (2, 7,−1) and has direction cosines 4

5, 0,

3

5 .

(i) State the angle between the line and the positive 𝑦-axis. (ii) Find the vector equation of the line.

IG: @musclemath_ 23

Example 17 (a) A vector 𝐩 is inclined at 60° to the 𝑥-axis and 45° to the 𝑦-axis. Find its inclination to the 𝑧-axis.

(b) Point 𝐴 has coordinates (2, 1, 4). Find the direction cosines of 𝑂𝐴⃗⃗⃗⃗ ⃗.

END