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LINES AND PLANES 
IN 
3-DIMENSIONS 
Compulsary question in paper 2, section A, 
question 4 , and full marks given is 3 marks. 
( paper 1, question 14.)
ANALYSIS OF SPM MATHEMATICS 
TOPIC : LINES AND PLANES IN 3-D 
PAPER 2 : SECTION A (3 MARKS) 
YEAR : 
2004 (N) 
Type 
PYRAMID (2) 
CUBOID (1) 
2005 (N) 
Type : 
RIGHT 
PRISM (2) 
RIGHT 
PYRAMID(1) 
2006 (N) 
Type : 
2007 (N) 
Type : 
RIGHT 
PRISM(2) 
2008 (N) 
Type : 
CUBOID(2) 
2009 (N) 
Type : 
CUBOID(2) 
PYRAMID (1) 
BIL Topic / 
Content K1 K2 J K1 K2 J K1 K2 J K1 K2 J K1 K2 J K1 K2 T 
1 
LINES 
AND 
PLANES 
IN 3-D 
1 4 5 1 3 4 1 3 4 1 3 4 1 3 4 1 3 4
PYRAMID WITH RIGHT PRISM 
RECTANGULAR BASE 
CUBOID 
PYRAMID 
CUBOID 
CUBOID
S
No Steps Solutions 
1 Draw the line UP and 
shade the plane RPQ U 
S 
P 
R 
Q 
T 
Question 1 
The Diagram shows a triangular prism. Identify the angle 
between the line UP and the plane RPQ
No Steps Solutions 
2. 1. Arrange the line and plane 
in two rows 
2. Find out the same alphabet – point P 
(identify the point of line UP that touches the plane RPQ, ) 
3. Draw 3 boxes, 
4. Write point P in the middle and point U in 
the first box (point does not touch the plane) 
5. Look at U, choose Which One is the Nearest 
to U (WON) (Non-slashed alphabets) 
Between point R and Q, point R will be 
chosen and write it in the third box. 
6. Angle between line UP to the plane RPQ 
is 
U P 
R P Q 
U P 
R P Q 
U P R 
U 
S 
P 
R 
Q 
T 
ÐUPR
No Steps Solutions 
1 Draw the line QT and 
shade the plane PQRS 
T U 
P 
V 
S 
R 
W 
8 cm 
Q 
16 cm 
6 cm 
Question 2 
The Diagram shows a cuboid with a horizontal rectangular 
base PQRS. Calculate the angle between the line QT and 
the base PQRS
No Steps Solutions 
2. 1. Arrange the line and plane 
in two rows 
2. Find out the same alphabet – point Q 
(identify the point of line QT that touches the plane PQRS, ) 
3. Draw 3 boxes, 
4. Write point Q in the middle and point T in 
the first box. (point does not touch the plane) 
5. Look at T, choose Which One is the 
Nearest to T (WON) (Non-slashed alphabets- 
Between point P, R or S, point S will be 
chosen and write it in the third box. 
6. Angle between line UP to the plane RPQ 
is 
Q T 
P Q R S 
Q T 
P Q R S 
T Q S 
T U 
ÐTQS P 
V 
S 
W 
Q 
R 
16 cm 
6 cm 
8 cm
No Steps Solutions 
3 Refer to the points you 
have obtained in step 2. 
(point T,Q,S) Draw ΔTQS. 
Mark the right angle. 
P 
V 
S 
W 
Q 
U 
R 
T 
16 cm 
6 cm 
8 cm 
T 
S Q
No Steps Solutions 
4 With the information 
given in the question, 
label the length of the 
sides of ΔTQS. At least 
two sides of the length 
must be known. Use 
Pythagoras Theorem if 
necessary 
P 
V 
S 
W 
Q 
U 
R 
T 
16 cm 
6 cm 
8 cm 
T 
S Q 
8 cm 
162 +62 =17.09
No Steps Solutions 
5 Mark 
the opposite side,TS 
the adjacent side,SQ 
P 
V 
S 
W 
Q 
U 
R 
T 
16 cm 
6 cm 
8 cm 
T 
S Q 
8 cm 
162 +62 =17.09
ÐTQS 
tanÐTQS = 8 
ÐTQS = 25.08°or25°5' 
P 
V 
S 
W 
Q 
U 
R 
16 cm 
6 cm 
8 cm 
T 
17.09 
T 
S Q 
8 cm 
are 162 +62 =17.09
No Steps Solutions 
1 Shade the plane PBC the 
plane BCRQ 
9 cm 
S 
P Q 
U 
T 
R 
12 cm 
C 
D 
15 cm 
A B 
Question 4 
The diagram shows a prism with cross section BCRQ. 
Given T and U are the midpoint of AD and BC respectively, 
P and Q are right above T and U respectively and PQRS 
is a square.
No Steps Solutions 
2. 1. Arrange the plane PCB and plane 
BCRQ in two rows 
2. Find out the same alphabet 
(identify the points of PCB that touch the plane BCRQ, ) 
3. Draw 3 boxes, 
4. Write point CB in the middle and point P in 
the first box. 
P C B 
B C R Q 
P C B 
B C R Q 
P CB 
U 
T 
C 
9 cm 
S 
D 
A B 
R 
P Q 
15 cm 
12 cm
No Steps Solutions 
2. 5. Look at P, choose WON – Slashed 
alphabet 
BU = CU, SO CHOOSE THE MIDPOINT OF 
BC (U) 
6. Look at P choose Which One is the 
Nearest to P (WON) (Non-slashed 
alphabets - point at the plane BCRQ, 
point Q is the nearest to P, Write point 
Q in the third box. 
7. Angle between plane PCB to the plane 
BCRQ is 
P C B 
B C R Q 
P BC 
P U Q 
U 
T 
C 
9 cm 
S 
D 
A B 
R 
P Q 
15 cm 
12 cm 
ÐPUQ
No Steps Solutions 
3 Refer to the points you 
have obtained in step 2. 
(point P,U,Q) Draw 
ΔPUQ. Mark the right 
angle. 
P 
Q 
U 
U 
T 
C 
9 cm 
S 
D 
A B 
R 
P Q 
15 cm 
12 cm
No Steps Solutions 
4 With the information 
given in the question, 
label the length of the 
sides of ΔPUQ. At least 
two sides of the length 
must be known. Use 
Pythagoras Theorem if 
necessary 
P 
Q 
U 
U 
T 
9 cm 
C 
9 cm 
S 
D 
A B 
R 
P Q 
15 cm 
12 cm 
12 cm
No Steps Solutions 
5 Mark 
the opposite side PQ 
the adjasent side UQ 
P 
Q 
U 
U 
T 
9 cm 
C 
9 cm 
S 
D 
A B 
R 
P Q 
15 cm 
12 cm 
12 cm
ÐPUQ 
Q 
tanÐPUQ = 9 
12 
ÐPUQ = 36.87°or36°52' 
P 
U 
9 cm 
12 cm 
U 
T 
C 
9 cm 
S 
D 
A B 
R 
P Q 
15 cm 
12 cm 
are
No Steps Solutions 
1 Shade the plane PRV and 
the plane QRVU 
4 cm 
T R 
U 
R 
Q 
S 
P 
5 cm 
W V 
12 cm 
Question 5 
The diagram shows a cuboid with base TUVW, Calculate the 
angle between PRV and the plane QRVU
No Steps Solutions 
2. 1. Arrange the plane PRV and plane 
QRVU in two rows 
2. Find out the same alphabet 
(identify the points of PRV that touch the plane QRVU, ) 
3. Draw 3 boxes, 
4. Write point RV at the middle and point P in 
the first box. 
P R V 
Q R V U 
P R V 
Q R V U 
P RV 
R 
Q 
S 
P 
T U 
5 cm 
W V 
12 cm 
4 cm 
R
No Steps Solutions 
2. 5. Look at P, choose WON – Slashed 
alphabet (Between point R and V, point 
which is nearer to point P. Point R is chosen 
and write it in the middle box. 
6. Look at P choose Which One is the 
Nearest to P (WON) (Non-slashed 
alphabets – Between point Q and and U, 
point Q is chosen and write it in the third 
box, 
7. Angle between plane PRV to the plane 
QRUV is 
P R V 
Q R V U 
P RV 
P R Q 
ÐPRQ 
R 
Q 
S 
P 
T U 
5 cm 
W V 
12 cm 
4 cm 
R
No Steps Solutions 
3 Refer to the points you 
have obtained in step 2. 
(point P,R,Q) Draw 
ΔPRQ. Mark the right 
angle. 
P 
Q 
R 
R 
Q 
S 
P 
T U 
5 cm 
W V 
12 cm 
4 cm 
R
No Steps Solutions 
4 With the information 
given in the question, 
label the length of the 
sides of ΔPRQ. At least 
two sides of the length 
must be known. Use 
Pythagoras Theorem if 
necessary 
P 
Q 
R 
12 cm 
5 cm 
R 
Q 
S 
P 
T U 
5 cm 
W V 
12 cm 
4 cm 
Diagram 5 
R
No Steps Solutions 
5 Mark 
the opposite side,PQ 
the adjacent side, RQ 
P 
Q 
R 
12 cm 
5 cm 
R 
Q 
S 
P 
T U 
5 cm 
W V 
12 cm 
4 cm 
R
ÐPRQ 
Q 
tanÐPRQ = 12 
5 
ÐPRQ = 67.38°or67°22' 
P 
U 
12 cm 
5 cm 
R 
Q 
S 
P 
T U 
5 cm 
W V 
12 cm 
4 cm 
R 
are
Try this.. 
H 
E 
U G 
T 
L M 
N 
R 
F 
P 
Name the angle between the plane LUM 
with the plane LPNM 
Answer : URT
Find the angle between the plane JFE 
and the plane DEF. 
L 
F D 
E 
J 
M 
5 
5 
13 
5
M 
L 
E 
D 
13 
5 
5 
F 
J 
5 
Identify the angle ÐJMD
5 
- 
132 52 
tan JMD = 
Ð JMD 
22.62 
o 
o 
22 37 ' 
= 
OR
Linenplanebengkelseminar

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Linenplanebengkelseminar

  • 1. LINES AND PLANES IN 3-DIMENSIONS Compulsary question in paper 2, section A, question 4 , and full marks given is 3 marks. ( paper 1, question 14.)
  • 2. ANALYSIS OF SPM MATHEMATICS TOPIC : LINES AND PLANES IN 3-D PAPER 2 : SECTION A (3 MARKS) YEAR : 2004 (N) Type PYRAMID (2) CUBOID (1) 2005 (N) Type : RIGHT PRISM (2) RIGHT PYRAMID(1) 2006 (N) Type : 2007 (N) Type : RIGHT PRISM(2) 2008 (N) Type : CUBOID(2) 2009 (N) Type : CUBOID(2) PYRAMID (1) BIL Topic / Content K1 K2 J K1 K2 J K1 K2 J K1 K2 J K1 K2 J K1 K2 T 1 LINES AND PLANES IN 3-D 1 4 5 1 3 4 1 3 4 1 3 4 1 3 4 1 3 4
  • 3. PYRAMID WITH RIGHT PRISM RECTANGULAR BASE CUBOID PYRAMID CUBOID CUBOID
  • 4. S
  • 5.
  • 6. No Steps Solutions 1 Draw the line UP and shade the plane RPQ U S P R Q T Question 1 The Diagram shows a triangular prism. Identify the angle between the line UP and the plane RPQ
  • 7. No Steps Solutions 2. 1. Arrange the line and plane in two rows 2. Find out the same alphabet – point P (identify the point of line UP that touches the plane RPQ, ) 3. Draw 3 boxes, 4. Write point P in the middle and point U in the first box (point does not touch the plane) 5. Look at U, choose Which One is the Nearest to U (WON) (Non-slashed alphabets) Between point R and Q, point R will be chosen and write it in the third box. 6. Angle between line UP to the plane RPQ is U P R P Q U P R P Q U P R U S P R Q T ÐUPR
  • 8. No Steps Solutions 1 Draw the line QT and shade the plane PQRS T U P V S R W 8 cm Q 16 cm 6 cm Question 2 The Diagram shows a cuboid with a horizontal rectangular base PQRS. Calculate the angle between the line QT and the base PQRS
  • 9. No Steps Solutions 2. 1. Arrange the line and plane in two rows 2. Find out the same alphabet – point Q (identify the point of line QT that touches the plane PQRS, ) 3. Draw 3 boxes, 4. Write point Q in the middle and point T in the first box. (point does not touch the plane) 5. Look at T, choose Which One is the Nearest to T (WON) (Non-slashed alphabets- Between point P, R or S, point S will be chosen and write it in the third box. 6. Angle between line UP to the plane RPQ is Q T P Q R S Q T P Q R S T Q S T U ÐTQS P V S W Q R 16 cm 6 cm 8 cm
  • 10. No Steps Solutions 3 Refer to the points you have obtained in step 2. (point T,Q,S) Draw ΔTQS. Mark the right angle. P V S W Q U R T 16 cm 6 cm 8 cm T S Q
  • 11. No Steps Solutions 4 With the information given in the question, label the length of the sides of ΔTQS. At least two sides of the length must be known. Use Pythagoras Theorem if necessary P V S W Q U R T 16 cm 6 cm 8 cm T S Q 8 cm 162 +62 =17.09
  • 12. No Steps Solutions 5 Mark the opposite side,TS the adjacent side,SQ P V S W Q U R T 16 cm 6 cm 8 cm T S Q 8 cm 162 +62 =17.09
  • 13. ÐTQS tanÐTQS = 8 ÐTQS = 25.08°or25°5' P V S W Q U R 16 cm 6 cm 8 cm T 17.09 T S Q 8 cm are 162 +62 =17.09
  • 14.
  • 15. No Steps Solutions 1 Shade the plane PBC the plane BCRQ 9 cm S P Q U T R 12 cm C D 15 cm A B Question 4 The diagram shows a prism with cross section BCRQ. Given T and U are the midpoint of AD and BC respectively, P and Q are right above T and U respectively and PQRS is a square.
  • 16. No Steps Solutions 2. 1. Arrange the plane PCB and plane BCRQ in two rows 2. Find out the same alphabet (identify the points of PCB that touch the plane BCRQ, ) 3. Draw 3 boxes, 4. Write point CB in the middle and point P in the first box. P C B B C R Q P C B B C R Q P CB U T C 9 cm S D A B R P Q 15 cm 12 cm
  • 17. No Steps Solutions 2. 5. Look at P, choose WON – Slashed alphabet BU = CU, SO CHOOSE THE MIDPOINT OF BC (U) 6. Look at P choose Which One is the Nearest to P (WON) (Non-slashed alphabets - point at the plane BCRQ, point Q is the nearest to P, Write point Q in the third box. 7. Angle between plane PCB to the plane BCRQ is P C B B C R Q P BC P U Q U T C 9 cm S D A B R P Q 15 cm 12 cm ÐPUQ
  • 18. No Steps Solutions 3 Refer to the points you have obtained in step 2. (point P,U,Q) Draw ΔPUQ. Mark the right angle. P Q U U T C 9 cm S D A B R P Q 15 cm 12 cm
  • 19. No Steps Solutions 4 With the information given in the question, label the length of the sides of ΔPUQ. At least two sides of the length must be known. Use Pythagoras Theorem if necessary P Q U U T 9 cm C 9 cm S D A B R P Q 15 cm 12 cm 12 cm
  • 20. No Steps Solutions 5 Mark the opposite side PQ the adjasent side UQ P Q U U T 9 cm C 9 cm S D A B R P Q 15 cm 12 cm 12 cm
  • 21. ÐPUQ Q tanÐPUQ = 9 12 ÐPUQ = 36.87°or36°52' P U 9 cm 12 cm U T C 9 cm S D A B R P Q 15 cm 12 cm are
  • 22. No Steps Solutions 1 Shade the plane PRV and the plane QRVU 4 cm T R U R Q S P 5 cm W V 12 cm Question 5 The diagram shows a cuboid with base TUVW, Calculate the angle between PRV and the plane QRVU
  • 23. No Steps Solutions 2. 1. Arrange the plane PRV and plane QRVU in two rows 2. Find out the same alphabet (identify the points of PRV that touch the plane QRVU, ) 3. Draw 3 boxes, 4. Write point RV at the middle and point P in the first box. P R V Q R V U P R V Q R V U P RV R Q S P T U 5 cm W V 12 cm 4 cm R
  • 24. No Steps Solutions 2. 5. Look at P, choose WON – Slashed alphabet (Between point R and V, point which is nearer to point P. Point R is chosen and write it in the middle box. 6. Look at P choose Which One is the Nearest to P (WON) (Non-slashed alphabets – Between point Q and and U, point Q is chosen and write it in the third box, 7. Angle between plane PRV to the plane QRUV is P R V Q R V U P RV P R Q ÐPRQ R Q S P T U 5 cm W V 12 cm 4 cm R
  • 25. No Steps Solutions 3 Refer to the points you have obtained in step 2. (point P,R,Q) Draw ΔPRQ. Mark the right angle. P Q R R Q S P T U 5 cm W V 12 cm 4 cm R
  • 26. No Steps Solutions 4 With the information given in the question, label the length of the sides of ΔPRQ. At least two sides of the length must be known. Use Pythagoras Theorem if necessary P Q R 12 cm 5 cm R Q S P T U 5 cm W V 12 cm 4 cm Diagram 5 R
  • 27. No Steps Solutions 5 Mark the opposite side,PQ the adjacent side, RQ P Q R 12 cm 5 cm R Q S P T U 5 cm W V 12 cm 4 cm R
  • 28. ÐPRQ Q tanÐPRQ = 12 5 ÐPRQ = 67.38°or67°22' P U 12 cm 5 cm R Q S P T U 5 cm W V 12 cm 4 cm R are
  • 29. Try this.. H E U G T L M N R F P Name the angle between the plane LUM with the plane LPNM Answer : URT
  • 30. Find the angle between the plane JFE and the plane DEF. L F D E J M 5 5 13 5
  • 31. M L E D 13 5 5 F J 5 Identify the angle ÐJMD
  • 32. 5 - 132 52 tan JMD = Ð JMD 22.62 o o 22 37 ' = OR