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Problem 1

Sheet 1

Unsolved Problems 4d-20-32

𝐿𝑒𝑡 𝑒𝑎𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝐴⃑ = 5 𝑎 − 𝑎 + 3 𝑎 , 𝐵⃑ = −2 𝑎 + 2 𝑎 + 4 𝑎 , 𝑎𝑛𝑑 𝐶⃑ = 3 𝑎 − 4 𝑎 , 𝑒𝑥𝑡𝑒𝑛𝑑 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛 𝑜𝑓 𝑎 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑡𝑜 𝑝𝑜𝑖𝑛𝑡𝑠 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶, 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦. 𝑓𝑖𝑛𝑑 𝑎) 𝑡ℎ𝑒 𝑜𝑟𝑔𝑖𝑛 (𝑖. 𝑒. 𝑎

𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝐴 𝑡𝑜𝑤𝑎𝑟𝑑:

);

𝑏) 𝑃𝑜𝑖𝑛𝑡 𝐵 (𝑖. 𝑒. 𝑎 )

𝑐) 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑒𝑞𝑢𝑖𝑑𝑖𝑠𝑡𝑎𝑛𝑡 𝑓𝑟𝑜𝑚 𝐵 & 𝐶 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝐵𝐶 𝑖. 𝑒. 𝑎 ; 𝑑) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑒𝑟𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 Δ𝐴𝐵𝐶 Solution – Correct 𝑨𝒏𝒔.: (𝑎) − 0.845 𝑎 + 0.169 𝑎 − 0.507 𝑎 ; (𝑏) − 0.911 𝑎 + 0.391 𝑎 + 0.1302 𝑎 ; (𝑐) − 0.793 𝑎 + 0.463 𝑎 − 0.396 𝑎 ;

(𝑑) 25.5

∵ −𝐴⃑ = −5 𝑎 + 𝑎 − 3 𝑎 ∴𝑎

=

1 √5 + 1 + 3

−5 𝑎 + 𝑎 − 3 𝑎

∵ 𝑅 ⃑ = 𝐵⃑ − 𝐴⃑ = −2 𝑎 + 2 𝑎 + 4 𝑎 ∴𝑎 ⃑=

1 √7 + 3 + 1

𝑇ℎ𝑒 𝑃𝑜𝑖𝑛𝑡 𝑋 @

=

1 √35

−5 𝑎 + 𝑎 − 3 𝑎

− 5𝑎 − 𝑎 +3𝑎

−7 𝑎 + 3 𝑎 + 𝑎

=

𝟏 √𝟓𝟗

= −𝟎. 𝟖𝟒𝟓 𝒂𝒙 + 𝟎. 𝟏𝟔𝟗 𝒂𝒚 − 𝟎. 𝟓𝟎𝟕 𝒂𝒛

= −7 𝑎 + 3 𝑎 + 𝑎

−𝟕 𝒂𝒙 + 𝟑 𝒂𝒚 + 𝒂𝒛 = −𝟎. 𝟗𝟏 𝒂𝒙 + 𝟑𝟗 𝒂𝒚 + 𝟎. 𝟏𝟑 𝒂𝒛

−2 + 0 2 + 3 4 − 4 , , = (−1, 2.5, 0 ) 2 2 2

∴ 𝑟 ⃑ = −𝑎 + 2.5 𝑎 ∴ 𝑅 ⃑ = 𝑟 ⃑ − 𝑟⃑ = −𝑎 + 2.5 𝑎 ∴𝑎 ⃑=

1

− 5𝑎 − 𝑎 +3𝑎

−6 𝑎 + 3.5 𝑎 − 3 𝑎

=

= −6 𝑎 + 3.5 𝑎 − 3 𝑎 1

√6 + 3.5 + 3 √57.25 = −𝟎. 𝟕𝟗𝟑 𝒂𝒙 + 𝟎. 𝟒𝟔𝟑 𝒂𝒚 − 𝟎. 𝟑𝟗𝟔 𝒂𝒛

−6 𝑎 + 3.5 𝑎 − 3 𝑎

Perimeter of the triangle ABC

∵ 𝑟 ⃑ = 𝐵⃑ − 𝐴⃑ = −2 𝑎 + 2 𝑎 + 4 𝑎

− 5𝑎 − 𝑎 +3𝑎 𝐴𝐵⃑ =

∵ 𝑟 ⃑ = 𝐶⃑ − 𝐵⃑ = 3 𝑎 − 4 𝑎

− −2 𝑎 + 2 𝑎 + 4 𝑎 − 3𝑎 −4𝑎

7 + 3 + 1 = √59

=2𝑎 +𝑎 −8𝑎

𝐵𝐶⃑ = ∵ 𝑟 ⃑ = 𝐴⃑ − 𝐶⃑ = 5 𝑎 − 𝑎 + 3 𝑎

= −7 𝑎 + 3 𝑎 + 𝑎

2 + 1 + 8 = √69

=5𝑎 −4𝑎 +7𝑎 𝐶𝐴⃑ =

5 + 4 + 7 = √90

∴ 𝑃𝑒𝑟𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 Δ 𝐴𝐵𝐶 = 𝐴𝐵⃑ + 𝐵𝐶⃑ + 𝐶𝐴⃑ = √59 + √69 + √90 = 𝟐𝟓. 𝟒𝟕

Problem 2

Sheet 1

𝐼𝑓 𝐴⃑ = 2 𝑎 + 6 𝑎 − 3 𝑎 ,

𝐵⃑ = −3 𝑎 − 4 𝑎 − 5 𝑎 , 𝑓𝑖𝑛𝑑

(𝑎) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴 − 𝐵; (𝑏) 𝑡ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑖𝑑𝑒 𝑜𝑓 𝐴 + 𝐵; (𝑐)

𝐴 + 3𝐵 𝐵 |𝐴 + 3𝐵|

Solution (a) 𝐴 − 𝐵 = 2 𝑎 + 6 𝑎 − 3 𝑎 − −3 𝑎 − 4 𝑎 − 5 𝑎 = 5 𝑎 + 10 𝑎 + 2 𝑎 5 𝑎 + 10 𝑎 + 2 𝑎 5 𝑎 + 10 𝑎 + 2 𝑎 𝐴−𝐵 𝑎 = = = |𝐴 − 𝐵| √25 + 100 + 4 √129 𝒂𝑨 𝑩 = 𝟎. 𝟒𝟒 𝒂𝒙 + 𝟎. 𝟖𝟖 𝒂𝒚 + 𝟎. 𝟏𝟕𝟔 𝒂𝒛 (b) 𝐴 + 𝐵 = 2 𝑎 + 6 𝑎 − 3 𝑎 + −3 𝑎 − 4 𝑎 − 5 𝑎

=−𝑎 +2𝑎 −8𝑎

|𝐴 + 𝐵| = √1 + 4 + 64 |𝐴 + 𝐵| = √𝟔𝟗 (c) 𝐴 + 3𝐵 = 2 𝑎 + 6 𝑎 − 3 𝑎 + 3 −3 𝑎 − 4 𝑎 − 5 𝑎 |𝐴 + 3𝐵| = √49 + 36 + 144 = √229 𝐴 + 3𝐵 −7 6 12 = 𝑎 − 𝑎 − 𝑎 |𝐴 + 3𝐵| √229 √229 √229 𝐴 + 3𝐵 = |𝐴 + 3𝐵|

49 36 144 + + = 229 229 229

229 =1 229

𝐴 + 3𝐵 𝐵=𝐵 |𝐴 + 3𝐵| 𝑨 + 𝟑𝑩 ∴ 𝑩 = −𝟑 𝒂𝒙 − 𝟒 𝒂𝒚 − 𝟓 𝒂𝒛 |𝑨 + 𝟑𝑩|

= −7 𝑎 − 6 𝑎 − 12 𝑎

Problem 3

Sheet 1

𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑅 ⃑ 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑓𝑟𝑜𝑚 𝐴 (1, 2, 3) 𝑡𝑜 𝐵. 𝐼𝑓 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑅 ⃑ 𝑖𝑠 10 𝑢𝑛𝑖𝑡𝑠 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝑎⃑ = 0.6 𝑎 + 0.64 𝑎 + 0.38 𝑎 ,

𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝐵.

Solution - Correct 𝑨𝒏𝒔. : 7, 8.4, 7.8

∵ 𝑅 ⃑ = 𝑅 ⃑ 𝑎⃑ = 10 𝑎⃑ ∴ 𝑅 ⃑ = 10 0.6 𝑎 + 0.64 𝑎 + 0.38 𝑎

= 6 𝑎 + 6.4 𝑎 + 3.8 𝑎

𝒃𝒖𝒕 𝑅 ⃑ = 𝐵⃑ − 𝐴⃑ ∴ 6 𝑎 + 6.4 𝑎 + 3.8 𝑎 = (𝐵 − 1) 𝑎 + 𝐵 − 2 𝑎 + (𝐵 − 3) 𝑎 𝑤𝑒 𝑔𝑒𝑡 𝐵 −1=6

𝐵 =7

𝐵 − 2 = 6.4

𝐵 = 8.4

𝐵 − 3 = 3.8

𝐵 = 7.8

∴ 𝑩(𝟕,

𝟖. 𝟒,

𝟕. 𝟖)

Problem 4

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑡𝑤𝑜 𝑝𝑜𝑖𝑛𝑡𝑠 𝑀(2, 5, −3) 𝑎𝑛𝑑 𝑁(−3, 1, 4),

𝑓𝑖𝑛𝑑

𝑎) 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒𝑖𝑟 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛𝑠 (𝑖. 𝑒. 𝑀𝑁⃑) 𝑏) 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑀𝑁; 𝑐) 𝑓𝑖𝑛𝑑 𝑎 𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅

.

𝑑) 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑀𝑁 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑋 = 0 𝑝𝑙𝑎𝑛𝑒. Solution ∵ 𝑀𝑁⃑ = 𝑁⃑ − 𝑀⃑ = −3 𝑎 + 𝑎 + 4 𝑎

− 2𝑎 +5𝑎 −3𝑎 𝑀𝑁⃑ =

= −5 𝑎 − 4 𝑎 + 7 𝑎

5 + 4 + 7 = √𝟗𝟎

(b) 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑀𝑁 𝑎𝑡 𝑇ℎ𝑒 𝑃𝑜𝑖𝑛𝑡 𝑃 @

−3 + 2 1 + 5 4 − 3 , , = (−0.5, 3, 0.5 ) 2 2 2

∴ 𝑟 ⃑ = −0.5 𝑎 + 3 𝑎 + 0.5 𝑎

𝑃⃑ =

0.5 + 3 + 0.5 = √𝟗. 𝟓

(c) ∴𝑎 ⃑= (d) ???

𝑟 ⃑ 1 = −5 𝑎 − 4 𝑎 + 7 𝑎 |𝑟 ⃑| √5 + 4 + 7

=

𝟏 √𝟗𝟎

−𝟓 𝒂𝒙 − 𝟒 𝒂𝒚 + 𝟕 𝒂𝒛

Problem 5

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑖𝑒𝑙𝑑 𝐺 = 2𝑥 𝑦 𝑎 − 2(𝑧 − 𝑥) 𝑎 + 3𝑥𝑦𝑧 𝑎 ,

𝑓𝑖𝑛𝑑:

𝑎) 𝐺 𝑎𝑡 𝑃(2, −3, 4) 𝑏) 𝑎 𝑢𝑛𝑖𝑡 𝑉𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺 𝑎𝑡 𝑃 𝑐) 𝑇ℎ𝑒 𝑠𝑐𝑎𝑙𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑛 𝑤ℎ𝑖𝑐ℎ |𝐺| = 100 𝑑) 𝑡ℎ𝑒 𝑦 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑄(−3, 𝑦, 5) 𝑖𝑓 𝐺

= 100 & 𝑦 > 0

𝑒) 𝑇ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑃 & 𝑄. Solution - Correct 𝐴𝑛𝑠.: (𝑎) − 24 𝑎 − 4 𝑎 − 72 𝑎 ; (𝑏) − 0.316 𝑎 − 0.0526 𝑎 − 0.947 𝑎 ; (𝑐) 4𝑥 𝑦 + 4 (𝑧 − 𝑥) + 9 𝑥 𝑦 𝑧 = 10 ; (𝑑) 2.04 (𝑒)7.17

𝐺 ⃑ = −𝟐𝟒 𝒂𝒙 − 𝟒 𝒂𝒚 − 𝟕𝟐 𝒂𝒛

𝒂𝒕 𝑃(2, −3, 4) (b) ∴ 𝐺⃑ =

𝐺⃑ 1 = −24 𝑎 − 4 𝑎 − 72 𝑎 𝐺⃑ √24 + 16 + 72

∴ 𝐺⃑ = −

6 1 18 𝑎 − 𝑎 − 𝑎 19 19 19

=

1 −24 𝑎 − 4 𝑎 − 72 𝑎 76

𝑮𝑷⃑ = −𝟎. 𝟑𝟏𝟔 𝒂𝒙 − 𝟎. 𝟎𝟓𝟐 𝒂𝒚 − 𝟎. 𝟗𝟒𝟕 𝒂𝒛

(c) |𝐺| = 100 =

(2𝑥 𝑦) + 2(𝑧 − 𝑥)

+ (3𝑥𝑦𝑧)

∴ 𝟏𝟎𝟎𝟎𝟎 = 𝟒𝒙𝟒 𝒚𝟐 + 𝟒 (𝒛 − 𝒙)𝟐 + 𝟗 𝒙𝟐 𝒚𝟐 𝒛𝟐 ⋯ ⋯ ⋯ ⋯ 𝒆𝒒𝒖(𝟏) (d) @ 𝑄(−3, 𝑦, 5)

∴𝑦=±

( )

10000 = 324 𝑦 + 256 + 2025 𝑦

9744 = 2349 𝑦

9744 = ±𝟐. 𝟎𝟑𝟔𝟕 2349

(e) ∵ 𝑟 ⃑ = 𝑄⃑ − 𝑃⃑ = −3 𝑎 ± 2.0367 𝑎 + 5 𝑎 = −5 𝑎 + (3 ± 2.0367) 𝑎 + 𝑎 ∴ 𝑟 ⃑ =

5 + (3 ± 2.0367) + 1 =

− 2𝑎 −3𝑎 +4𝑎

5 + 5.0367 + 1 5 + 0.9633 + 1

=

=

𝟕. 𝟏𝟔𝟕 5.189

Problem 6

Sheet 1

𝐴 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑖𝑒𝑙𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑎𝑠 𝐺 = 2𝑥 𝑦 𝑎 − 2(𝑧 − 𝑥) 𝑎 + 3𝑥𝑦𝑧 𝑎 , 𝑃𝑟𝑒𝑝𝑎𝑟𝑒 𝑠𝑘𝑒𝑡𝑐ℎ𝑒𝑠 𝑜𝑓 𝐺 , 𝐺 , 𝐺 , 𝑎𝑛𝑑 |𝐺|, 𝑎𝑙𝑙 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒𝑑 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑥 = 2, 𝑦 = −1; Solution 𝐺 (2, −1, 𝑧) = 2𝑥 𝑦 = 2(4)(−1) = −8 𝐺 (2, −1, 𝑧) = −2(𝑧 − 𝑥) = −2(𝑧 − 2) = −2 𝑧 + 4 𝐺 (2, −1, 𝑧) = 3𝑥𝑦𝑧 = 3(2)(−1)𝑧 = −6 𝑧 |𝐺| =

64 + (−2 𝑧 + 4) + (−6𝑧)

=

64 + 4 𝑧 − 16𝑧 + 16 + 36 𝑧

=

80 + 40 𝑧 − 16𝑧

= 4 2.5 𝑧 − 𝑧 + 5

0 ≤ 𝑧 ≤ 10

Problem 7

Sheet 1

𝐴 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑖𝑒𝑙𝑑 𝑖𝑠 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑎𝑠 𝐹 = 2(𝑥 + 𝑦) sin 𝜋𝑧 𝑎 − (𝑥 + 𝑦) 𝑎 + 𝑎𝑙𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑡

𝑎) 𝐹 = 0

10 𝑎 , 𝑥 +𝑦

𝑠𝑝𝑒𝑐𝑖𝑓𝑦 𝑡ℎ𝑒 𝑙𝑜𝑐𝑢𝑠 𝑜𝑓 𝑐) |𝐹 | = 1

𝑏) 𝐹 = 0

Solution - Correct 𝐴𝑛𝑠.: (𝑎) 𝑦 = −𝑥 (𝑝𝑙𝑎𝑛𝑒), 𝑧 = 0, ±1, ±2, … (𝑝𝑙𝑎𝑛𝑒𝑠); (𝑏) 𝑦 = −𝑥 (𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟); (𝑐) 𝑥 + 𝑦 = 10 (𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟)

𝑎𝑡 𝐹 = 0 2(𝑥 + 𝑦) sin 𝜋𝑧 = 0 ∴ 2(𝑥 + 𝑦) = 0 𝑜𝑟 sin 𝜋𝑧 = 0 ∴ 𝑃𝑙𝑎𝑛𝑒𝑠 𝑎𝑡 𝒚 = −𝒙 , 𝒛 = 𝟎, ±𝟏, ±𝟐, ±𝟑 ⋯ (𝑥 + 𝑦) = 0

𝑎𝑡 𝐹 = 0

∴ 𝐿𝑜𝑐𝑢𝑠 𝑜𝑓 𝑝𝑎𝑟𝑏𝑜𝑙𝑖𝑐 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝒚 = −𝒙𝟐 10 =1 𝑥 +𝑦

𝑎𝑡 |𝐹 | = 1 ∴ 𝑥 + 𝑦 = 10

∴ 𝐿𝑜𝑐𝑢𝑠 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝒙𝟐 + 𝒚𝟐 = √𝟏𝟎

Problem 8

𝟐

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑓𝑖𝑒𝑙𝑑𝑠 𝐺 = 5 (𝑥 + 𝑦) 𝑎 + 10 𝑎 , 𝑎) |𝐺 |

𝑏) |𝐺 |

𝑎𝑛𝑑 𝐺 = 5 𝑎 + 2𝑥𝑦 𝑎 𝑓𝑖𝑛𝑑 𝑎𝑡 𝑃(3,2,0) 𝑐) 𝑎 , 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺 𝑒) 𝑎 , 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺 ⃑ + 𝐺 ⃑ .

𝑑) 𝑎 , 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺 Solution 𝑎𝑡 𝑃(3,2,0) ∴ |𝐺 | =

𝐺 = 25 𝑎 + 10 𝑎 , 25 + 10 = √725 = 𝟐𝟔. 𝟗𝟐𝟔

∴ 𝑎⃑ =

𝐺⃑ 1 = 25 𝑎 + 10 𝑎 𝐺⃑ √725

∴ 𝑎⃑ =

𝐺⃑ 1 = 5 𝑎 + 12 𝑎 ⃑ 13 𝐺

𝐺 ⃑ + 𝐺 ⃑ = 25 𝑎 + 10 𝑎 ∴ 𝑎 ⃑ = 𝒙𝒙

𝐺 = 5 𝑎 + 12 𝑎 𝑎𝑛𝑑

|𝐺 | =

5 + 12 = √169 = 𝟏𝟑

= 𝟎. 𝟗𝟐𝟖𝟓 𝒂𝒙 + 𝟎. 𝟑𝟕𝟏𝟒 𝒂𝒚

= 𝟎. 𝟑𝟖𝟓 𝒂𝒙 + 𝟎. 𝟗𝟐𝟑 𝒂𝒚

+ 5 𝑎 + 12 𝑎

𝐺⃑ + 𝐺 ⃑ 1 = 𝐺⃑ + 𝐺 ⃑ √30 + 22

= 30 𝑎 + 22 𝑎

30 𝑎 + 22 𝑎

= 𝟎. 𝟖𝟎𝟔𝟒𝟎𝟓 𝒂𝒙 + 𝟎. 𝟓𝟗𝟏𝟑𝟔𝟒 𝒂𝒚

Problem 9

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑓𝑖𝑒𝑙𝑑, 𝐺⃑ =

2𝑥 𝑎 + (𝑦 + 𝑧 + 1) 𝑎 + (5𝑥 − 𝑧 ) 𝑎 , 𝑓𝑖𝑛𝑑: 1+𝑦

(𝑎) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺 𝑎𝑡 𝑃(1, 2, −3); (𝑏) 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐺 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑦 = 0 𝑝𝑙𝑎𝑛𝑒 𝑎𝑡 𝑄(2, 0,4); (𝑐) 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓

𝐺 𝑑𝑥 𝑑𝑦 𝑎 𝑎𝑡 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑧 = 1.

Solution - Correct 𝐴𝑛𝑠.: (𝑎) 0.0995 𝑎 − 0.995 𝑎 ; (𝑏) 34.7 ; (𝑐) 36

𝐺 ⃑ = 0.4 𝑎 + 0 𝑎 − 4 𝑎

𝑎𝑡 𝑃(1, 2, −3) ∴𝑎 ⃑=

𝐺⃑ = 𝐺⃑

1 (0.4) + 4

[0.4 𝑎 − 4 𝑎 ] = 𝟎. 𝟎𝟗𝟗𝟓 𝒂𝒙 − 𝟎. 𝟗𝟓𝟓 𝒂𝒛

(b) 𝐺⃑ = 4𝑎 +5𝑎 −6𝑎

𝑎𝑡 𝑄(2, 0,4) ∵ 𝐺 ⃑. 𝑎 ⃑ = 𝐺 ⃑ cos 𝛼 ∵5=

4 + 5 + 6 cos 𝛼

∴ cos 𝛼 =

5 √77

𝛼 = 55.26

𝟗𝟎 − 𝜶 = 𝟑𝟒. 𝟕𝟑

(c) 𝐺 𝑑𝑥 𝑑𝑦 𝑎 =

(5𝑥 − 𝑧 ) 𝑑𝑥 𝑑𝑦 𝑎

𝑎𝑡 𝑧 = 1 𝑝𝑙𝑎𝑛𝑒 ∴𝐼=

(5𝑥 − 1) 𝑑𝑥 𝑑𝑦 =

5𝑥 −𝑥 2

(𝑦) = 𝟑𝟔

Problem 10

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑝𝑜𝑖𝑛𝑡𝑠 𝑃(2, 5, −1), 𝑎) 𝑡ℎ𝑒 𝑣𝑒𝑐𝑜𝑟 𝑅 ⃑

𝑄(−1, −4,1),

𝑎𝑛𝑑 𝑇(5, 0,2),

𝑓𝑖𝑛𝑑

𝑏) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅 ⃑

𝑐) 𝑇ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑒𝑟𝑚𝑖𝑡𝑒𝑟 𝑜𝑓 ∆ 𝑃𝑄𝑇

𝑒) 𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅 ⃑ 𝑜𝑛 𝑅 ⃑

𝑑) 𝑇ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒 𝑎𝑡 𝑄

𝑓) 𝑇ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑡ℎ𝑎𝑡 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑓𝑟𝑜𝑚 𝑄 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟𝑦 𝑡𝑜 𝑡ℎ𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑜𝑟 𝑖𝑡𝑠 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛. Solution (𝒂)

𝑟 ⃑ = 𝑄⃑ − 𝑃⃑ = − 𝑎 − 4 𝑎 + 𝑎

(𝒃)

𝑎 ⃑=

𝑟 ⃑ 𝑟 ⃑

=

1 √3 + 9 + 2

− 2𝑎 +5𝑎 − 𝑎

−3 𝑎 − 9 𝑎 + 2 𝑎

=

= −𝟑 𝒂𝒙 − 𝟗 𝒂𝒚 + 𝟐 𝒂𝒛 𝟏 √𝟗𝟒

−𝟑 𝒂𝒙 − 𝟗 𝒂𝒚 + 𝟐 𝒂𝒛

∵ 𝑟 ⃑ = 𝑇⃑ − 𝑄⃑ = 5 𝑎 + 0 𝑎 + 2 𝑎 − − 𝑎 − 4 𝑎 + 𝑎 = 6 𝑎 + 4 𝑎 + 𝑎 ∵ 𝑟 ⃑ = 𝑃⃑ − 𝑇⃑ = 2 𝑎 + 5 𝑎 − 𝑎 − 5 𝑎 + 0 𝑎 + 2 𝑎 = −3 𝑎 + 5 𝑎 − 3𝑎 (𝒄)

𝑟 ⃑ = √53 |𝑟 ⃑ | = √43

𝑝𝑒𝑟𝑚𝑖𝑡𝑒𝑟 𝑜𝑓 ∆ 𝑃𝑄𝑇 = 𝑟 ⃑ + 𝑟 ⃑ + |𝑟 ⃑ | = √94 + √53 + √43 = 𝟐𝟑. 𝟓𝟑𝟑

(d) ∴ 𝑅 ⃑. 𝑅 ⃑ = 𝑅 ⃑ . 𝑅 ⃑ cos 𝛼 ∴ cos 𝛼 =

6𝑎 +4𝑎 +𝑎 . 3𝑎 +9𝑎 −2𝑎 𝑅 ⃑. 𝑅 ⃑ = 𝑅 ⃑. 𝑅 ⃑ √53. √94

=

(18 + 36 − 2) √4982

=

52 √4982

∴ 𝜶 = 𝟒𝟐. 𝟓𝟒𝟕𝒐 (e) 𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟 ⃑ 𝑜𝑛 𝑅 ⃑ =

𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟 ⃑ 𝑜𝑛 𝑅 ⃑ =

− 𝑎 − 4 𝑎 + 𝑎 . 3 𝑎 − 5 𝑎 + 3𝑎 𝑟 ⃑. 𝑅 ⃑ = √9 + 25 + 9 |𝑅 ⃑| −3 + 20 + 3 √43

=

20 √43

(f) 𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑉𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑟 ⃑ 𝑜𝑛 𝑅 ⃑ =

𝑟 ⃑. 𝑅 ⃑ 𝑅 ⃑ 20 3 𝑎 − 5 𝑎 + 3𝑎 = √43 |𝑅 ⃑| |𝑅 ⃑| √43

=

20 3 𝑎 − 5 𝑎 + 3𝑎 43

Problem 11

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑝𝑜𝑖𝑛𝑡𝑠 𝐸(2, 5, 1),

𝐹(−1, 4, −2),

𝑎𝑛𝑑 𝐺(3, −2,4),

𝑎) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝐸 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝐹; (𝑖. 𝑒. 𝑎 𝑏) 𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑅

&𝑅

𝑓𝑖𝑛𝑑

⃑)

;

𝑐) 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑒𝑟𝑚𝑖𝑡𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐸𝐹𝐺; 𝑑) 𝑇ℎ𝑒 𝑠𝑐𝑎𝑙𝑒𝑟 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅 ⃑ 𝑜𝑛 𝑅 ⃑; Solution - Correct 𝐴𝑛𝑠.: (𝑎) − 0.688 𝑎 − 0.229 𝑎 − 0.688 𝑎 ; (𝑏) 98.6 ; (𝑐) 21.4; (𝑑) − 0.651

∵ 𝑅 ⃑ = 𝐹⃑ − 𝐸⃑ = − 𝑎 + 4 𝑎 − 2 𝑎 ∴𝑎

⃑=

− 2𝑎 +5𝑎 + 𝑎

𝑅 ⃑ 1 = −3 𝑎 − 𝑎 − 3 𝑎 𝑅 ⃑ √3 + 1 + 3

=

𝟏 √𝟏𝟗

= −3 𝑎 − 𝑎 − 3 𝑎 −𝟑 𝒂𝒙 − 𝒂𝒚 − 𝟑 𝒂𝒛

∴ 𝒂𝑹𝑬𝑭⃑ = −𝟎. 𝟔𝟖𝟖 𝒂𝒙 − 𝟎. 𝟐𝟐𝟗𝟒 𝒂𝒚 − 𝟎. 𝟔𝟖𝟖 𝒂𝒛 (b) ∵ 𝑅 ⃑ = 𝐺⃑ − 𝐸⃑ = 3 𝑎 − 2 𝑎 + 4 𝑎 ∵ 𝑅 ⃑. 𝑅 ⃑ = 𝑅 ⃑ . 𝑅 ⃑ cos 𝛼

− 2𝑎 +5𝑎 + 𝑎

= 𝑎 −7𝑎 +3𝑎

cos 𝛼 =

𝑅 ⃑ = √59

𝑅 ⃑. 𝑅 ⃑ −3 + 7 − 9 −5 = = 𝑅 ⃑. 𝑅 ⃑ √1121 √19 √59

∴ 𝜶 = 𝟗𝟖. 𝟓𝟖𝟖𝟓𝒐 (𝒄) ∵ 𝑅 ⃑ = 𝐹⃑ − 𝐺⃑ = − 𝑎 + 4𝑎 − 2𝑎

− 3 𝑎 − 2𝑎 + 4𝑎

= −4 𝑎 + 6 𝑎 − 6 𝑎

∴ 𝑝𝑒𝑟𝑚𝑖𝑡𝑒𝑟 𝑜𝑓 ∆ 𝐸𝐹𝐺 = 𝑅 ⃑ + 𝑅 ⃑ + 𝑅 ⃑ = √19 + √59 + √88 = 𝟐𝟏. 𝟒𝟐 (𝑑) 𝑇ℎ𝑒 𝑠𝑐𝑎𝑙𝑒𝑟 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅 ⃑ 𝑜𝑛 𝑅 ⃑ = 𝑅 ⃑. 𝑎 ⃑ = 𝑅 ⃑.

𝑅 ⃑ = 𝑅 ⃑ cos 𝛼 𝑅 ⃑

= √19 cos 98.5885 = −𝟎. 𝟔𝟓𝟎𝟗𝟓

𝑅 ⃑ = √88

Problem 12

Sheet 1

𝐸𝑥𝑝𝑟𝑒𝑠𝑠 𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠: (𝑎) 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝐺 𝑒𝑥𝑡𝑒𝑛𝑑𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑚𝑖𝑑 − 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑗𝑜𝑖𝑛𝑖𝑛𝑔 𝐴(2, −3,5) 𝑡𝑜 𝐵(6, −5,5) (𝑏) 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝐷 𝑒𝑥𝑡𝑒𝑛𝑑𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝐶(−2,7,3) 𝑡𝑜 𝑡ℎ𝑒 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑗𝑜𝑖𝑛𝑖𝑛𝑔 𝐴 𝑡𝑜 𝐵 (𝑐) 𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑅

𝑡ℎ𝑎𝑡 𝑖𝑠 𝑖𝑛 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅

𝑑) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅 ⃑. Solution (a) 𝑀 = 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡

2 + 6 −3 − 5 5 + 5 , , = (4, −4, 5 ) 2 2 2

∴ 𝑮⃑ = 𝟒 𝒂𝒙 − 𝟒 𝒂𝒚 + 𝟓 𝒂𝒛 (b) ∵ 𝑅 ⃑ = 𝑀⃑ − 𝐶⃑ = 4 𝑎 − 4 𝑎 + 5 𝑎

− −2 𝑎 + 7 𝑎 + 3 𝑎

= 𝟔 𝒂𝒙 − 𝟏𝟏 𝒂𝒚 + 𝟐 𝒂𝒛

(c) ∵ 𝑅 ⃑ = 𝐵⃑ − 𝐴⃑ = 6 𝑎 − 5 𝑎 + 5 𝑎 − 2 𝑎 − 3 𝑎 + 5 𝑎 = 4 𝑎 − 2 𝑎 + 0 𝑎 ∵ 𝑅 ⃑ = 𝐶⃑ − 𝐴⃑ = −2 𝑎 + 7 𝑎 + 3 𝑎 − 2 𝑎 − 3 𝑎 + 5 𝑎 = 0 𝑎 + 10 𝑎 − 2 𝑎 ∴ 𝑅 ⃑. 𝑅 ⃑ = 4 𝑎 − 2 𝑎 + 0 𝑎 . 0 𝑎 + 10 𝑎 − 2 𝑎 ∴ 𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑅

𝑡ℎ𝑎𝑡 𝑖𝑠 𝑖𝑛 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅

= −𝟐𝟎

=𝑅 ⃑

𝑅 ⃑ −20 −20 = = = −𝟏. 𝟗𝟔 𝑅 ⃑ √0 + 10 + 2 √104

(d) ∵ 𝑅 ⃑ = 𝐶⃑ − 𝐵⃑ = −2 𝑎 + 7 𝑎 + 3 𝑎 ∴𝑎

⃑=

∴ 𝒂𝑹𝑩𝑪⃑ =

− 6𝑎 −5𝑎 +5𝑎

𝑅 ⃑ 1 = −8 𝑎 + 12 𝑎 − 2 𝑎 𝑅 ⃑ √8 + 12 + 2 𝟏 √𝟓𝟑

−𝟒 𝒂𝒙 + 𝟔 𝒂𝒚 − 𝒂𝒛

=

= −8 𝑎 + 12 𝑎 − 2 𝑎 1

2√53

−8 𝑎 + 12 𝑎 − 2 𝑎

Problem 13

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝐴 = −4 𝑎 + 2 𝑎 + 3 𝑎 , 𝑎𝑛𝑑 𝐵 = 3 𝑎 + 4 𝑎 − 𝑎 , 𝑓𝑖𝑛𝑑: 𝑎) 𝑇ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 5 𝐴⃑ − 2 𝐵⃑, 𝑖. 𝑒. 5 𝐴⃑ − 2 𝐵⃑ 𝑏) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓

5 𝐴⃑ − 2 𝐵⃑ ; 𝐴⃑

𝑐) 𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐴 𝑖𝑠 𝑝𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝐵 ; (𝑖. 𝑒. 𝐴

=? 𝑤ℎ𝑒𝑟𝑒 𝐴 ⃑ 𝐵⃑)

𝑒) 𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝐵 (𝑖. 𝑒. 𝐴 ⃑) Solution - Correct 𝐴𝑛𝑠.: (𝑎) 31.1; (𝑏) − 0.835 𝑎 + 0.0642 𝑎 + 0.546 𝑎 ; (𝑐) − 0.808 𝑎 − 1.077 𝑎 + 0.269 𝑎 ; (𝑑) − 3.19 𝑎 + 3.08 𝑎 + 2.73 𝑎 ;

5 𝐴⃑ − 2 𝐵⃑ = 5 −4 𝑎 + 2 𝑎 + 3 𝑎 ∴ 5 𝐴⃑ − 2 𝐵⃑ = (b) 𝐸⃑ =

−2 3𝑎 +4𝑎 − 𝑎

= −26 𝑎 + 2 𝑎 + 17 𝑎

26 + 2 + 17 = √𝟗𝟔𝟗 = 𝟑𝟏. 𝟏

−26 𝑎 + 2 𝑎 + 17 𝑎 5 𝐴⃑ − 2 𝐵⃑ −26 𝑎 + 2 𝑎 + 17 𝑎 = = 𝐴⃑ √4 + 2 + 3 √29

∴ 𝑎⃑ =

𝐸⃑ = 𝐸⃑

1

−26 𝑎 + 2 𝑎 + 17 𝑎

26 + 2 + 17 29

√29

=

𝟏 √𝟗𝟔𝟗

−𝟐𝟔 𝒂𝒙 + 𝟐 𝒂𝒚 + 𝟏𝟕 𝒂𝒛

∴ 𝒂𝑬⃑ = −𝟎. 𝟖𝟑𝟓𝟐 𝒂𝒙 + 𝟎. 𝟎𝟔𝟒 𝒂𝒚 + 𝟎. 𝟓𝟒𝟔 𝒂𝒛 (c) ∵ 𝐴 ⃑ = 𝐴 ⃑ .𝑎 ⃑ ∴ 𝐴 ⃑ = 𝐴 cos 𝛼 . 𝑎 ⃑ ∴𝐴 ⃑=

𝐴⃑. 𝐵⃑ −12 + 8 − 3 3 𝑎 + 4 𝑎 − 𝑎 cos 𝛼 . 𝑎 ⃑ = . |𝐵| √9 + 16 + 1 √9 + 16 + 1

=

−𝟕 𝟑 𝒂𝒙 + 𝟒 𝒂𝒚 − 𝒂𝒛 𝟐𝟔

∴ 𝐴 ⃑ = −𝟎. 𝟖𝟎𝟕 𝒂𝒙 − 𝟏. 𝟎𝟕𝟕 𝒂𝒚 + 𝟎. 𝟐𝟔𝟗𝟐 𝒂𝒛 (d) ∵ 𝐴⃑ = 𝐴 ⃑ + 𝐴 ⃑ ∴ 𝐴 ⃑ = 𝐴⃑ − 𝐴 ⃑ = −4 𝑎 + 2 𝑎 + 3 𝑎

+

7 3𝑎 +4𝑎 − 𝑎 26

= −𝟑. 𝟏𝟗 𝒂𝒙 + 𝟑. 𝟎𝟕𝟕𝒂𝒚 + 𝟐. 𝟕𝟑 𝒂𝒛

Problem 14

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑡ℎ𝑟𝑒𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝐴(2, −1, 2), 𝑎) 𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑅

𝐵(−4, 1,4),

𝑎𝑛𝑑 𝐶(4, 3, −1),

𝑓𝑖𝑛𝑑

&𝑅

𝑏) 𝑇ℎ𝑒 (𝑠𝑐𝑎𝑙𝑎𝑟) 𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶 𝑐) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 ∆ 𝐴𝐵𝐶 Solution ∵ 𝑅 ⃑ = 𝐵⃑ − 𝐴⃑ = −4 𝑎 + 𝑎 + 4 𝑎

− 2𝑎 − 𝑎 +2𝑎

= −6 𝑎 + 2 𝑎 + 2 𝑎 𝑅 ⃑ =

∵ 𝑅 ⃑ = 𝐶⃑ − 𝐴⃑ = 4 𝑎 + 3 𝑎 − 𝑎

− 2𝑎 − 𝑎 +2𝑎

=2𝑎 +4𝑎 −3𝑎 𝑅 ⃑ =

∴ 𝛼 = cos

𝑅 ⃑. 𝑅 ⃑ = cos 𝑅 ⃑. 𝑅 ⃑

−12 + 8 − 6 √44 √29

= cos

6 + 2 + 2 = √44

−10 √1276

2 + 4 + 3 = √29

= 𝟏𝟎𝟔. 𝟐𝟓𝟕𝒐

(b) Area of the triangle ABC

𝑎 𝑎 𝑎 ∵ 𝑅 ⃑ × 𝑅 ⃑ = −6 2 2 = (−6 − 8) 𝑎 − (18 − 4) 𝑎 + (−24 − 4) 𝑎 = −14 𝑎 − 14 𝑎 − 28 𝑎 2 4 −3 1 ∴ 𝒂𝒓𝒆𝒂 = 𝑅 ⃑ × 𝑅 ⃑ = −7 𝑎 − 7 𝑎 − 14 𝑎 = 7 + 7 + 14 = √𝟐𝟗𝟒 = 𝟏𝟕. 𝟏𝟒𝟔 𝒔𝒒𝒖. 𝒖𝒏𝒊𝒕𝒔 2 (c)

Normal to the triangle ABC

∴𝑎⃑=

−𝟏𝟒 𝒂𝒙 − 𝟏𝟒 𝒂𝒚 − 𝟐𝟖 𝒂𝒛 −14 𝑎 − 14 𝑎 − 28 𝑎 𝑅 ⃑×𝑅 ⃑ = = 𝑅 ⃑×𝑅 ⃑ √14 + 14 + 28 √𝟏𝟏𝟕𝟔

Problem 15

Sheet 1

𝐹𝑖𝑛𝑑 𝑖𝑛 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠: 𝑎) 𝑖𝑛 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸 𝑎𝑡 𝑃(2,3, −4) 𝑖𝑓 𝐸⃑ = (𝑥 + 𝑦 + 𝑧 )

𝑥 𝑦 +𝑧

𝑎 +

𝑦 √𝑥 + 𝑧

𝑎 +

𝑧 𝑥 +𝑦

𝑎 ;

𝑏) 𝑝𝑟𝑒𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑀(1, −5,5), 𝑁(−2,4,0), 𝑎𝑛𝑑 𝑄(2,3,4) & ℎ𝑎𝑣𝑖𝑛𝑔 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑥 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡; (𝑖. 𝑒. 𝑎 ⃑ ⊥ 𝑀𝑁𝑄) 𝑐) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑅 ⃑ 𝑎𝑛𝑑 𝑅 ⃑. Solution - Correct 𝐴𝑛𝑠.: (𝑎) 0.295 𝑎 + 0.494 𝑎 − 0.818 𝑎 ; (𝑏) 0.674 𝑎 − 0.1740 𝑎 − 0.718 𝑎 ; (𝑐) 31.9 ;

∵ 𝐸⃑ = 29 0.4 𝑎 +

∴ 𝑎⃑ =

𝐸⃑ = 𝐸⃑

3 √20

𝑎 −

4 √13

1 3 4 (0.4) + + 20 13

𝑎

0.4 𝑎 +

3 √20

𝑎 −

4 √13

𝑎

= 𝟎. 𝟕𝟑𝟕 𝟎. 𝟒 𝒂𝒙 +

𝟑 √𝟐𝟎

𝒂𝒚 −

𝟒 √𝟏𝟑

𝒂𝒛

∴ 𝒂𝑬⃑ = 𝟎. 𝟐𝟗𝟒𝟖 𝒂𝒙 + 𝟎. 𝟒𝟗𝟒𝟒 𝒂𝒚 − 𝟎. 𝟖𝟏𝟖 𝒂𝒛 (b) ∵ 𝑅 ⃑ = 𝑁⃑ − 𝑀⃑ = −2 𝑎 + 4 𝑎 + 0 𝑎

− 𝑎 −5𝑎 +5𝑎

= −3 𝑎 + 9 𝑎 − 5 𝑎 𝑅 ⃑ =

∵ 𝑅 ⃑ = 𝑄⃑ − 𝑀⃑ = 2 𝑎 + 3 𝑎 + 4 𝑎

− 𝑎 −5𝑎 +5𝑎

=𝑎 +8𝑎 − 𝑎 𝑅 ⃑ =

Normal to the triangle MNQ

𝑎 𝑎 ⃑ ⃑ ∵ 𝑅 × 𝑅 = −3 9 1 8 ∴𝑎⃑=

3 + 9 + 5 = √115 1 + 8 + 1 = √66

𝑎 −5 = (−9 + 40) 𝑎 − (3 + 5) 𝑎 + (−24 − 9) 𝑎 = 31 𝑎 − 8 𝑎 − 33 𝑎 −1

𝟑𝟏 𝒂𝒙 − 𝟖 𝒂𝒚 − 𝟑𝟑 𝒂𝒛 31 𝑎 − 8 𝑎 − 33 𝑎 𝑅 ⃑×𝑅 ⃑ = = = 𝟎. 𝟔𝟕𝟒 𝒂𝒙 − 𝟎. 𝟏𝟕𝟒 𝒂𝒚 − 𝟎. 𝟕𝟏𝟕𝟕 𝒂𝒛 𝑅 ⃑×𝑅 ⃑ √31 + 8 + 33 √𝟐𝟏𝟏𝟒

(c) Angle between MN & MQ

∴ 𝛼 = cos

𝑅 ⃑. 𝑅 ⃑ = cos 𝑅 ⃑. 𝑅 ⃑

−3 + 72 + 5 √115 √66

= cos

74 √7590

= 𝟑𝟏. 𝟖𝟓𝟑𝟖𝒐

Problem 16

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛 𝑟⃑ = 3 𝑎 + 8 𝑎 + 𝑎 ,

𝑟 ⃑ = 𝑎 − 6 𝑎 + 5 𝑎 , 𝑎𝑛𝑑 𝑟 ⃑ = 7 𝑎 − 4 𝑎 , 𝑓𝑖𝑛𝑑:

𝑎) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑟𝑒𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑟⃑ 𝑎𝑛𝑑 𝑟 ⃑ (𝑖. 𝑒. 𝑓𝑖𝑛𝑑 𝑎 𝑁𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝑟⃑, 𝑟 ⃑) 𝑏) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑟𝑒𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑟⃑ − 𝑟 ⃑, 𝑟 ⃑ − 𝑟 ⃑, 𝑎𝑛𝑑 𝑟 ⃑ − 𝑟⃑; 𝑐) 𝑇ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 ℎ𝑒𝑎𝑑𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑟⃑, 𝑟 ⃑, 𝑎𝑛𝑑 𝑟 ⃑; Solution 𝑎 ∵ 𝑟⃑ × 𝑟 ⃑ = 3 1 ∴𝑎⃑=

𝑎 𝑎 8 1 = 46 𝑎 − 14 𝑎 − 26 𝑎 −6 5

𝟒𝟔 𝒂𝒙 − 𝟏𝟒 𝒂𝒚 − 𝟐𝟔 𝒂𝒛 46 𝑎 − 14 𝑎 − 26 𝑎 𝑟⃑ × 𝑟 ⃑ = = |𝑟⃑ × 𝑟 ⃑| √46 + 14 + 26 √𝟐𝟗𝟖𝟖

(b) 𝑅 ⃑ = 𝑟⃑ − 𝑟 ⃑ = 3 𝑎 + 8 𝑎 + 𝑎

− 𝑎 −6𝑎 +5𝑎

𝑅 ⃑ = 𝑟 ⃑ − 𝑟⃑ = 𝑎 − 6 𝑎 + 5 𝑎

− 7𝑎 +0𝑎 −4𝑎

𝑅 ⃑ = 𝑟 ⃑ − 𝑟⃑ = 7 𝑎 + 0 𝑎 − 4 𝑎

− 3𝑎 +8𝑎 + 𝑎

= 2 𝑎 + 14 𝑎 − 4 𝑎 = −6 𝑎 − 6 𝑎 + 9 𝑎 =4𝑎 −8𝑎 −5𝑎

𝑎 𝑎 𝑎 ⃑ ⃑ ∵ 𝑅 × 𝑅 = 2 14 −4 = 102 𝑎 + 6 𝑎 + 72 𝑎 −6 −6 9 ∴𝑎 ⃑=

102 𝑎 + 6 𝑎 + 72 𝑎 102 𝑎 + 6 𝑎 + 72 𝑎 𝑅 ⃑×𝑅 ⃑ = = = 𝟎. 𝟖𝟏𝟔 𝒂𝒙 + 𝟎. 𝟎𝟒𝟖 𝒂𝒚 + 𝟎. 𝟓𝟕𝟔 𝒂𝒛 𝑅 ⃑×𝑅 ⃑ √102 + 6 + 72 √15624

𝑎 𝑎 𝑎 ⃑ ⃑ 𝑪𝒉𝒆𝒄𝒌: 𝑅 × 𝑅 = −6 −6 9 = 102 𝑎 + 6 𝑎 + 72 𝑎 4 −8 −5 (c) Area of the triangle ABC

∵ 𝑅 ⃑ × 𝑅 ⃑ = 102 𝑎 + 6 𝑎 + 72 𝑎 (𝑎𝑠 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑎𝑏𝑜𝑣𝑒) 1 ∴ 𝒂𝒓𝒆𝒂 = 𝑅 ⃑ × 𝑅 ⃑ = 51 𝑎 + 3 𝑎 + 36 𝑎 = 51 + 3 + 36 = √𝟑𝟗𝟎𝟔 = 𝟔𝟐. 𝟓 𝒔𝒒𝒖. 𝒖𝒏𝒊𝒕𝒔 2

Problem 17

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑡ℎ𝑟𝑒𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑀(6,2, −3), 𝑁(−2,3,0), 𝑎𝑛𝑑 𝑃(−4,6,5),

𝑓𝑖𝑛𝑑:

𝑎) 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑; (𝑖. 𝑒. 𝑎𝑟𝑒𝑎 𝑜𝑓 ∆𝑀𝑁𝑃) 𝑏) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑡ℎ𝑖𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒; 𝑐) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑏𝑖𝑠𝑒𝑐𝑡𝑖𝑛𝑔 𝑡ℎ 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑎𝑡 𝑀. Solution - Correct 𝐴𝑛𝑠.: (𝑎) 20.3; (𝑏) ± 0.0983 𝑎 − 0.836 𝑎 + 0.541 𝑎 ; (𝑐) − 0.851 𝑎 + 0.211 𝑎 + 0.480 𝑎 ;

(a) ∵ 𝑅 ⃑ = 𝑁⃑ − 𝑀⃑ = −2 𝑎 + 3 𝑎 + 0 𝑎

− 6𝑎 +2𝑎 −3𝑎

= −8 𝑎 + 𝑎 + 3 𝑎 𝑅 ⃑ =

∵ 𝑅 ⃑ = 𝑃⃑ − 𝑀⃑ = −4 𝑎 + 6 𝑎 + 5 𝑎

− 6𝑎 +2𝑎 −3𝑎

8 + 1 + 3 = √74

= −10 𝑎 + 4 𝑎 + 8 𝑎 𝑅 ⃑ =

10 + 4 + 8 = √180 𝑎 𝑎 𝑎 ∵ 𝑅 ⃑ × 𝑅 ⃑ = −8 1 3 = −4 𝑎 + 34 𝑎 − 22 𝑎 −10 4 8 1 ∴ 𝑨𝒓𝒆𝒂 = 𝑅 ⃑ × 𝑅 ⃑ = −2 𝑎 + 17 𝑎 − 11 𝑎 = 2 + 17 + 11 = √𝟒𝟏𝟒 = 𝟐𝟎. 𝟑𝟒𝟕 𝒔𝒒𝒖. 𝒖𝒏𝒊𝒕𝒔 2 (b) Unit vector normal to MNP

∴𝑎⃑=±

−𝟒 𝒂𝒙 + 𝟑𝟒 𝒂𝒚 − 𝟐𝟐 𝒂𝒛 −4 𝑎 + 34 𝑎 − 22 𝑎 𝑅 ⃑×𝑅 ⃑ =± =± 𝑅 ⃑×𝑅 ⃑ √4 + 34 + 22 √𝟏𝟔𝟓𝟔

∴ 𝒂𝑵⃑ = ± −𝟎. 𝟎𝟗𝟖𝟑 𝒂𝒙 + 𝟎. 𝟖𝟑𝟓𝟓 𝒂𝒚 − 𝟎. 𝟓𝟒𝟎𝟔 𝒂𝒛

(c) Bisecting unit vector

∴ 𝑅⃑ = 𝑎

=

∴ 𝑎⃑ =

⃑+𝑎 −8 √74

−

⃑= 10

√180

−8 𝑎 + 𝑎 + 3 𝑎 −10 𝑎 + 4 𝑎 + 8 𝑎 𝑅 ⃑ 𝑅 ⃑ + = + 𝑅 ⃑ 𝑅 ⃑ √74 √180 𝑎 +

1 √74

+

4 √180

𝑎 +

3 √74

𝑅⃑ = −𝟎. 𝟖𝟓𝟏𝟒𝟓 𝒂𝒙 + 𝟎. 𝟐𝟏 𝒂𝒚 + 𝟎. 𝟒𝟖 𝒂𝒛 𝑅⃑

+

8 √180

𝑎

Problem 18

Sheet 1

(𝑎) 𝐸𝑥𝑝𝑟𝑒𝑠𝑠 𝑡ℎ𝑒 𝑓𝑖𝑒𝑙𝑑 𝐹 = 2𝑥𝑦𝑧 𝑎 ± 5(𝑥 + 𝑦 + 𝑧) 𝑎 𝑖𝑛 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 (𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 & 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠). (𝑏) 𝐹𝑖𝑛𝑑 |𝐹| 𝑎𝑡 𝑃(𝜌 = 2, 𝜑 = 60, 𝑧 = 3). Solution 𝐹⃑ = 2𝑥𝑦𝑧 𝑎 ± 5(𝑥 + 𝑦 + 𝑧) 𝑎 = 𝐹 𝑎 + 𝐹 𝑎 + 𝐹 𝑎 𝐹 = 𝐹⃑ 𝑎 = 2𝑥𝑦𝑧 𝑎 𝑎 ± 5(𝑥 + 𝑦 + 𝑧) 𝑎 𝑎 = 2𝑥𝑦𝑧 cos 𝜑 = 2(𝜌 cos 𝜑)(𝜌 sin 𝜑)𝑧 cos 𝜑 𝐹 = 2 𝜌 𝑧 (cos 𝜑)(sin 𝜑) 𝐹 = 𝐹⃑ 𝑎 = 2𝑥𝑦𝑧 𝑎 𝑎 ± 5(𝑥 + 𝑦 + 𝑧) 𝑎 𝑎 = 2𝑥𝑦𝑧(− sin 𝜑) = −2(𝜌 cos 𝜑)(𝜌 sin 𝜑)𝑧 sin 𝜑 𝐹 = −2 𝜌 𝑧 (sin 𝜑)(cos 𝜑) 𝐹 = 𝐹⃑ 𝑎 = 2𝑥𝑦𝑧 𝑎 𝑎 ± 5(𝑥 + 𝑦 + 𝑧) 𝑎 𝑎 = ±5(𝑥 + 𝑦 + 𝑧) = ±5(𝜌 cos 𝜑 + 𝜌 sin 𝜑 + 𝑧) ∴ 𝑭⃑ = 𝟐 𝝆𝟐 𝒛 𝐜𝐨𝐬 𝟐 𝝋 (𝐬𝐢𝐧 𝝋) 𝒂𝝆 − 𝟐 𝝆𝟐 𝒛 𝐬𝐢𝐧𝟐 𝝋 (𝐜𝐨𝐬 𝝋) 𝒂𝝋 ± 𝟓(𝝆 𝐜𝐨𝐬 𝝋 + 𝝆 𝐬𝐢𝐧 𝝋 + 𝒛) 𝒂𝒛 (b) 𝑎𝑡 𝑃(𝜌 = 2, 𝜑 = 60, 𝑧 = 3) 𝐹⃑ = 2 (2 )(3)

1 4

3 √3 𝑎 − 2(2 )(3) 2 4

1 1 √3 𝑎 ±5 2 +2 +3 𝑎 2 2 2

𝐹⃑ = 3√3 𝑎 − 9 𝑎 ± 20 + 5√3 𝑎 𝐹⃑ =

3√3

𝑭⃑ = 𝟏𝟏. 𝟖𝟔𝟖𝟑

+ 9 + 4 + √3

=

27 + 81 + 4 + √3

Problem 19

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑃(𝜌 = 5, 𝜑 = 60, 𝑧 = 2), 𝑎𝑛𝑑 𝑄(𝜌 = 2, 𝜑 = 110, 𝑧 = −1): (𝑎) 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑅 ⃑ 𝑏) 𝑔𝑖𝑣𝑒 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑃 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑄. (𝑖. 𝑒. 𝑎 𝑐) 𝑔𝑖𝑣𝑒 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑃 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑄 (𝑖. 𝑒. 𝑎

⃑) ⃑)

Solution 𝐴𝑛𝑠.: (𝑎) 5.01; (𝑏) − 0.635 𝑎 − 0.489 𝑎 − 0.598 𝑎 ; (𝑐) − 0.714 𝑎 + 0.306 𝑎 − 0.598 𝑎 ;

(a) 𝐶𝑦𝑙𝑖𝑛𝑑𝑒𝑟𝑖𝑐𝑎𝑙

𝐶𝑎𝑟𝑡𝑖𝑧𝑖𝑎𝑛

𝑥 = 𝜌 cos 𝜑 𝑦 = 𝜌 sin 𝜑 𝑧=𝑧 𝑃(𝜌 = 5, 𝜑 = 60, 𝑧 = 2) 𝑃 2.5, 2.5 √3, 2 𝑄(𝜌 = 2, 𝜑 = 110, 𝑧 = −1) 𝑄(2 cos 110 , 2 sin 110 , −1) ∴ 𝑅 ⃑ = 𝑄⃑ − 𝑃⃑ = 2 cos 110 𝑎 + 2 sin 110 𝑎 − 𝑎 − 2.5 𝑎 + 2.5 √3 𝑎 + 2 𝑎 ∴ 𝑅 ⃑ = −3.184 𝑎 − 2.450742 𝑎 − 3 𝑎 𝑅 ⃑ =

3.184 + 2.450742 + 3 = 𝟓. 𝟎𝟏𝟒𝟑𝟕𝟗

(b) ∴𝑎

⃑=

−3.184 𝑎 − 2.450742 𝑎 − 3 𝑎 𝑅 ⃑ = = −𝟎. 𝟔𝟑𝟓 𝒂𝒙 − 𝟎. 𝟒𝟖𝟖𝟖 𝒂𝒚 − 𝟎. 𝟓𝟗𝟖𝟑 𝒂𝒛 5.014379 𝑅 ⃑

(c) 𝜑 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 = 𝜑 = 60 𝐴 = 𝐴 cos 𝜑 + 𝐴 sin 𝜑 = −0.635 cos 60 − 0.4888 sin 60 = −0.7408 𝐴 = −𝐴 sin 𝜑 + 𝐴 cos 𝜑 = +0.635 sin 60 − 0.4888 cos 60 = 0.3055 𝑟𝑎𝑑 𝑧 = 𝑧 = −0.5983 ∴ 𝒂𝑹𝑷𝑸⃑ = −𝟎. 𝟕𝟒𝟎𝟖 𝒂𝝆 + 𝟎. 𝟑𝟎𝟓𝟓 𝒂𝝋 − 𝟎. 𝟓𝟗𝟖𝟑 𝒂𝒛

Problem 20

Sheet 1

𝐹𝑖𝑛𝑑 𝑖𝑛 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠: − 𝑎) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑎𝑡 𝑃(𝜌 = 5, 𝜑 = 53.13, 𝑧 = −2), 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐹 = 𝜌𝑧 cos 𝜑 𝑎 − 𝜌𝑧 sin 𝜑 𝑎 + 𝜌𝑎 𝑎 𝑏) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑎𝑡 𝑃 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑎 𝑐) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑎𝑡 𝑄(𝜌 = 5, 𝜑 = −36.87, 𝑧 = −2) 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑎 𝑑) 𝐺 = 2 𝑎 − 4 𝑎 + 4 𝑎 𝑎𝑡 𝑃. Solution

Problem 21

Sheet 1

(𝑎) 𝐺𝑖𝑣𝑒 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑡ℎ𝑎𝑡 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑓𝑟𝑜𝑚 𝑃(𝜌 = 4, 𝜑 = 10 , 𝑧 = 1) 𝑡𝑜 𝑄(𝜌 = 7, 𝜑 = 75 , 𝑧 = 4). (𝑏) 𝐺𝑖𝑣𝑒 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑀(𝑥 = 5, 𝑦 = 1, 𝑧 = 2) 𝑡ℎ𝑎𝑡 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑡𝑜 𝑁(2,4,6). (𝑐) 𝐻𝑜𝑤 𝑓𝑎𝑟 𝑖𝑠 𝑖𝑡 𝑓𝑟𝑜𝑚 𝐴(110, 60 , 1) 𝑡𝑜 𝐵(30, −125 , 10)? 𝑖. 𝑒. 𝑅 ⃑ Solution - Correct 𝐴𝑛𝑠.: (𝑎) − 2.13 𝑎 + 6.07 𝑎 + 3 𝑎 ; (𝑏) − 2.35 𝑎 + 3.53 𝑎 + 4 𝑎 ; (𝑐) 105.4;

(a) 𝑃(4,

10 ,

1) = 𝑃(4 cos 10 ,

4 sin 10 ,

1)

𝑄(7,

75 ,

4) = 𝑄(7 cos 75 ,

7 sin 75 ,

4)

∴ 𝑅 ⃑ = 𝑄⃑ − 𝑃⃑ = 7 cos 75 𝑎 + 7 sin 75 𝑎 + 4 𝑎

− 4 cos 10 𝑎 + 4 sin 10 𝑎 + 𝑎

∴ 𝑹𝑷𝑸⃑ = −𝟐. 𝟏𝟐𝟕𝟓 𝒂𝒙 + 𝟔. 𝟎𝟔𝟔𝟖 𝒂𝒚 + 𝟑 𝒂𝒛 (b) ∴ 𝑅 ⃑ = 𝑁⃑ − 𝑀⃑ = 2 𝑎 + 4 𝑎 + 6 𝑎

− 5𝑎 + 𝑎 +2𝑎

= −3 𝑎 + 3 𝑎 + 4 𝑎

Conversion ∴ 𝐴 ⃑ = 𝑅 ⃑. 𝑎 = −3 𝑎 + 3 𝑎 + 4 𝑎 . 𝑎 = −3 𝑎 𝑎 + 3 𝑎 𝑎 + 0 ∴ 𝐴 ⃑ = −3 cos 𝜑 + 3 sin 𝜑 𝜑 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑀 = 𝜑 = tan ∴ 𝐴 ⃑ = −3

5

+3

1

=

𝑦 𝑥

= tan

1 = 11.31 5

−12

= −2.35 √26 √26 √26 ∴ 𝐴 ⃑ = 𝑅 ⃑. 𝑎 = −3 𝑎 + 3 𝑎 + 4 𝑎 . 𝑎 = −3 𝑎 𝑎 + 3 𝑎 𝑎 + 0 −1 5 18 ∴ 𝐴 ⃑ = −3 sin 𝜑 + 3 cos 𝜑 = −3 +3 = √26 √26 √26 ∴ 𝑹𝑴𝑵⃑ = −𝟐. 𝟑𝟓 𝒂𝝆 + 𝟑. 𝟓𝟑 𝒂𝝋 + 𝟒 𝒂𝒛 (c) ∴ 𝑅 ⃑ = 𝐵⃑ − 𝐴⃑ = (30 cos 125 − 110 cos 60)𝟐 + (30 sin 125 − 110 sin 60)𝟐 + (10 + 20)𝟐 = 𝟏𝟎𝟓. 𝟒𝟎𝟕𝟒

Problem 22 𝐿𝑒𝑡 𝐸 =

Sheet 1

10 + 8 sin 𝜑 𝑎 + 8 cos 𝜑 𝑎 − 5 𝑎 . 𝑆𝑘𝑒𝑡𝑐ℎ |𝐸| 𝜌

(𝑎) 𝑣𝑠. 𝜑 𝑓𝑜𝑟 𝜌 = 2, 𝑧 = 3 (𝑏) 𝑣𝑠. 𝜌 𝑓𝑜𝑟 𝜑 = 0, 𝑧 = 3

𝑐) 𝑣𝑠. 𝜌 𝑓𝑜𝑟 𝜑 = 90, 𝑧 = 3 .

Solution (𝒂) 𝑺𝒌𝒆𝒕𝒄𝒉 |𝑬| 𝒗𝒔. 𝝋 𝒇𝒐𝒓 𝝆 = 𝟐, 𝒛 = 𝟑 |𝐸| =

(5 + 8 sin 𝜑) + (8 cos 𝜑) + (5) =

|𝐸| =

114 + 80 sin 𝜑

𝑬 = (𝟓 + 𝟖 𝐬𝐢𝐧 𝝋)𝒂𝝆 + 𝟖 𝐜𝐨𝐬 𝝋 𝒂𝝋 − 𝟓 𝒂𝒛 (25 + 80 sin 𝜑 + (8 sin 𝜑) ) + (8 cos 𝜑) + (5)

(𝒃) 𝑺𝒌𝒆𝒕𝒄𝒉 |𝑬| 𝒗𝒔. 𝝆 𝒇𝒐𝒓 𝝋 = 𝟎, 𝒛 = 𝟑

|𝐸| =

10 𝜌

+ (8) + (5) =

(a)

10 +8 𝜌

+ (0) + (5) = (b)

𝟏𝟎 𝒂𝝆 + 𝟖 𝒂𝝋 − 𝟓 𝒂𝒛 𝝆

100 + 89 𝜌

(𝒄) 𝑺𝒌𝒆𝒕𝒄𝒉 |𝑬| 𝒗𝒔. 𝝆 𝒇𝒐𝒓 𝝋 = 𝟗𝟎, 𝒛 = 𝟑

|𝐸| =

𝑬=

100 160 + + 64 + 25 = 𝜌 𝜌

𝑬=

𝟏𝟎 + 𝟖 𝒂𝝆 + 𝟎 𝒂𝝋 − 𝟓 𝒂𝒛 𝝆

100 160 + + 89 𝜌 𝜌 (c)

Problem 23

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑝𝑜𝑖𝑛𝑡𝑠 𝐴(𝑥 = 2, 𝑦 = 3, 𝑧 = −1), 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠:

𝑎𝑛𝑑 𝐵(𝜌 = 4, 𝜑 = −50 , 𝑧 = 2), 𝑓𝑖𝑛𝑑 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟𝑖𝑐𝑎𝑙

(𝑎) 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝐵 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑡𝑜𝑤𝑎𝑟𝑑 𝑝𝑜𝑖𝑛𝑡 𝐴

(𝑏) 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝐴 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑡𝑜𝑤𝑎𝑟𝑑 𝑝𝑜𝑖𝑛𝑡 𝐵.

Solution 𝐴𝑛𝑠.: (𝑎) − 0.738 𝑎 + 0.510 𝑎 + 0.442 𝑎 ; (𝑏) − 0.696 𝑎 − 0.565 𝑎 + 0.442 𝑎 ;

𝐵(𝜌 = 4, 𝜑 = −50 , 𝑧 = 2) ≡ (𝑥 = 4 cos −50 , 𝑦 = 4 sin −50 , 𝑧 = 2) ≡ (𝑥 = 2.57, 𝑦 = −3.064, 𝑧 = 2) (a) ∴ 𝑅 ⃑ = 𝐴⃑ − 𝐵⃑ = 2 𝑎 + 3 𝑎 − 𝑎 𝐴 ∵ 𝐴 𝐴

cos 𝜑 = − sin 𝜑 0

sin 𝜑 cos 𝜑 0

0 𝐴 0 𝐴 1 𝐴

− 2.57 𝑎 − 3.064 𝑎 + 2 𝑎 cos 𝜑 = − sin 𝜑 0

sin 𝜑 cos 𝜑 0

= −0.57 𝑎 + 6.064 𝑎 − 3 𝑎

0 −0.57 0 6.064 −3 1

∴ 𝑅 ⃑ = (−0.57 cos 𝜑 + 6.064 sin 𝜑) 𝑎 + (0.57 sin 𝜑 + 6.064 cos 𝜑) 𝑎 − 3 𝑎 𝜑 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝐵 = 𝜑 = tan

𝑦 = tan 𝑥

−3.064 = −50 2.57

∴ 𝑅 ⃑ = −5.01168 𝑎 + 3.46122 𝑎 − 3 𝑎 ∴

−5.01168 𝑎 + 3.46122 𝑎 − 3 𝑎 𝑅 ⃑ = = −𝟎. 𝟕𝟑𝟖 𝒂𝝆 + 𝟎. 𝟓𝟏 𝒂𝝋 − 𝟎. 𝟒𝟒𝟏𝟖 𝒂𝒛 6.7895 𝑅 ⃑

(b) ∴ 𝑅 ⃑ = 𝐵⃑ − 𝐴⃑ = 2.57 𝑎 − 3.064 𝑎 + 2 𝑎 𝐴 ∵ 𝐴 𝐴

cos 𝜑 = − sin 𝜑 0

sin 𝜑 cos 𝜑 0

0 𝐴 0 𝐴 1 𝐴

− 2𝑎 +3𝑎 − 𝑎

cos 𝜑 = − sin 𝜑 0

sin 𝜑 cos 𝜑 0

= +0.57 𝑎 − 6.064 𝑎 + 3 𝑎

0 +0.57 0 −6.064 3 1

∴ 𝑅 ⃑ = (0.57 cos 𝜑 − 6.064 sin 𝜑) 𝑎 + (−0.57 sin 𝜑 − 6.064 cos 𝜑) 𝑎 + 3 𝑎 𝜑 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝐴 = 𝜑 = tan

𝑦 = tan 𝑥

3 = 56.3 → 2

∴ 𝑅 ⃑ = −4.73 𝑎 − 3.838 𝑎 + 3 𝑎 ∴

−4.73 𝑎 − 3.838 𝑎 + 3 𝑎 𝑅 ⃑ = = −𝟎. 𝟔𝟗𝟔𝟔 𝒂𝝆 − 𝟎. 𝟓𝟔𝟓 𝒂𝝋 + 𝟎. 𝟒𝟒𝟏𝟖 𝒂𝒛 6.7895 𝑅 ⃑

cos 𝜑 = 2⁄√13 sin 𝜑 = 3⁄√13

Problem 24

Sheet 1

(𝑎) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 4 < 𝜌 < 6, 30 < 𝜑 < 60 , 2 < 𝑧 < 5 (𝑏) 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑛𝑔𝑒𝑠𝑡 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑡ℎ𝑎𝑡 𝑙𝑖𝑒𝑠 𝑒𝑛𝑡𝑖𝑟𝑒𝑙𝑦 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒? (𝑐) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒. Solution ∵ 𝑑𝑣 = 𝜌 𝑑𝑟 (𝑑𝜑) (𝑑𝑧) ∴𝑣=

𝜌 𝑑𝜌 𝑑𝜑 𝑑𝑧 =

𝜌 2

𝜑 1

𝑧 1

=

6 −4 2

𝜋 𝜋 𝜋 [3] = 𝟓𝝅 − [5 − 2] = [10] 3 6 6

(b) (𝜌 = 6, 𝜑 = 60 , 𝑧 = 5 ) ≡ (𝑥 = 6 cos 60 , 𝑦 = 6 sin 60 , 𝑧 = 5 ) ≡ 𝑥 = 3,

𝑦 = 3√3,

𝑧=5

(𝜌 = 4, 𝜑 = 30 , 𝑧 = 2 ) ≡ (𝑥 = 6 cos 60 , 𝑦 = 6 sin 60 , 𝑧 = 2 ) ≡ 𝑥 = 2,

𝑦 = 2√3,

𝑧=2

𝑙𝑜𝑛𝑔𝑒𝑠𝑡 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑙𝑒𝑛𝑔𝑡ℎ =

3 𝑎 + 3√3 𝑎 + 5 𝑎

− 2 𝑎 + 2√3 𝑎 + 2 𝑎

= 𝑎 + √3 𝑎 + 3 𝑎

𝑙𝑜𝑛𝑔𝑒𝑠𝑡 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑙𝑒𝑛𝑔𝑡ℎ = √1 + 3 + 9 = √13 = 𝟑. 𝟔 (c) ∵ 𝑑𝑠 = ± 𝜌 𝑑𝜑 𝑑𝑧 𝑎

∴𝑠=𝜌

± 𝑑𝜌 𝑑𝑧 𝑎

𝑑𝜑 𝑑𝑧 𝑎 +

∴ 𝑠 = 𝜌(5 − 2)

± 𝜌 𝑑𝜌 𝑑𝜑 𝑎

𝑑𝜌 𝑑𝑧 𝑎 +

𝜋 𝜋 𝜋 𝜋 − 𝑎 + (5 − 2)(6 − 4) 𝑎 + − 3 6 3 6

∴𝑠

= 𝑠 (𝑎𝑡 𝜌 = 4) + 𝑠 (𝑎𝑡 𝜌 = 6) + 2 𝑠 + 2 𝑠 = 4

∴𝑠

= 𝟏𝟐 +

𝟐𝟓𝝅 𝟑

𝜌 𝑑𝜌 𝑑𝜑 𝑎

𝜌 2

𝑎 =𝜌

𝜋 5𝜋 𝑎 +6𝑎 + 𝑎 2 3

𝜋 𝜋 5𝜋 10𝜋 +6 + 2(6) + 2 = 2𝜋 + 3𝜋 + 12 + 2 2 3 3

Problem 25

Sheet 1

𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑠𝑦𝑠𝑡𝑒𝑚 𝑛𝑎𝑚𝑒𝑑, 𝑔𝑖𝑣𝑒 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝐴(2, −1, −3) 𝑡ℎ𝑎𝑡 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑡𝑜 𝐵(1,3,4) (𝑎) 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛

(𝑏) 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟𝑖𝑐𝑎𝑙

(𝑐) 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙.

Solution - Correct 𝐴𝑛𝑠.: (𝑎) − 𝑎 + 4 𝑎 + 7 𝑎 ; (𝑏) − 2.68 𝑎 + 3.13 𝑎 + 7 𝑎 ; (𝑐) − 7.22 𝑎 − 2.03 𝑎 + 3.13 𝑎 ;

∴ 𝑅 ⃑ = 𝐵⃑ − 𝐴⃑ = 1 𝑎 + 3 𝑎 + 4 𝑎 − 2 𝑎 − 𝑎 − 3 𝑎 = − 𝒂𝒙 + 𝟒 𝒂𝒚 + 𝟕 𝒂𝒛 (b) 𝐴 cos 𝜑 sin 𝜑 0 𝐴 cos 𝜑 sin 𝜑 0 −1 ∵ 𝐴 = − sin 𝜑 cos 𝜑 0 𝐴 = − sin 𝜑 cos 𝜑 0 4 0 0 1 𝐴 0 0 1 7 𝐴

𝑹𝑨𝑩⃑ = 𝟖. 𝟏𝟐𝟒

∴ 𝑅 ⃑ = (− cos 𝜑 + 4 sin 𝜑) 𝑎 + (sin 𝜑 + 4 cos 𝜑) 𝑎 + 7 𝑎 𝜑 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝐴 = 𝜑 = tan

∴𝑅 ⃑=

−6 √5

𝑎 +

7 √5

𝑦 = tan 𝑥

−1 cos 𝜑 = 2⁄√5 = −26.565 → 2 sin 𝜑 = −1⁄√5 𝑹𝑨𝑩⃑ = 𝟖. 𝟏𝟐𝟐𝟕

𝑎 + 7 𝑎 = −𝟐. 𝟔𝟖 𝒂𝝆 + 𝟑. 𝟏𝟑 𝒂𝝋 + 𝟕 𝒂𝒛

(c) 𝐴 ∵ 𝐴 𝐴

sin 𝜃 cos 𝜑 = cos 𝜃 cos 𝜑 − sin 𝜑

sin 𝜃 sin 𝜑 cos 𝜃 sin 𝜑 cos 𝜑

𝐴 𝐴 𝐴

cos 𝜃 − sin 𝜃 0

sin 𝜃 cos 𝜑 = cos 𝜃 cos 𝜑 − sin 𝜑

sin 𝜃 sin 𝜑 cos 𝜃 sin 𝜑 cos 𝜑

cos 𝜃 −1 − sin 𝜃 4 0 7

∴ 𝑅 ⃑ = (− sin 𝜃 cos 𝜑 + 4 sin 𝜃 sin 𝜑 + 7 cos 𝜃) 𝑎 + (− cos 𝜃 cos 𝜑 + 4 cos 𝜃 sin 𝜑 − 7 sin 𝜃) 𝑎 + (sin 𝜑 + 4 cos 𝜑) 𝑎

𝜃 & 𝜑 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝐴 𝜃 = tan

𝑥 +𝑦 𝑧

𝜑 = tan

𝑦 = tan 𝑥

∴𝑅 ⃑= −

∴𝑅 ⃑=

√5 2 √14 √5

−27 √14

= tan

−1 = −26.565 2

+4

𝑎 +

√2 + 1 = tan −3

√5 −1 √14 √5

−6 √70

+

+7

12 √70

−3 √14

−7

√5 √14

√5 = 36.7 −3

cos 𝜃 = −3⁄√14 sin 𝜃 = √5⁄√14

cos 𝜑 = 2⁄√5 sin 𝜑 = −1⁄√5 𝑎 + −

𝑎 +

∴ 𝑹𝑨𝑩⃑ = −𝟕. 𝟐𝟏𝟔 𝒂𝒓 − 𝟐. 𝟎𝟑𝟐 𝒂𝜽 + 𝟑. 𝟏𝟑 𝒂𝝋

−3 2 √14 √5 7

√5

+4

𝑎 =

−3 −1 √14 √5

−27 √14

−7

𝑎 +

√5 √14

−17 √70

𝑹𝑨𝑩⃑ = 𝟖. 𝟏𝟐𝟑

𝑎 +

𝑎 +

−1 √5 7

√5

+4

𝑎

2 √5

𝑎

Problem 26

Sheet 1

𝑎𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑎𝑠 𝐸 =

100 cos 𝜃 50 cos 𝜃 𝑎 + 𝑎 . 𝐴𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑜𝑠𝑒 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑟 𝑟

𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑟 = 2, 𝜃 = 60, 𝜑 = 20, 𝑓𝑖𝑛𝑑 (𝑎) |𝐸| (𝑏) 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 (𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠) 𝑖𝑛 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸. Solution 100 cos 𝜃 𝑟

|𝐸| =

+

50 cos 𝜃 𝑟 50 2

|𝐸| =

𝑎𝑡 𝑟 = 2, 𝜃 = 60, 𝜑 = 20

+

25 2

= 6.9877

(b) 𝐴 𝐴 ∵ 𝐴

sin 𝜃 cos 𝜑 = sin 𝜃 sin 𝜑 cos 𝜃

∴ 𝐸⃑ = sin 𝜃 cos 𝜑

cos 𝜃 cos 𝜑 cos 𝜃 sin 𝜑 − sin 𝜃

100 cos 𝜃 𝑟3

+ cos 𝜃 ∴ 𝐸⃑ =

50 𝑟3

− sin 𝜑 𝐴 cos 𝜃 𝐴 𝐴 0

+ cos 𝜃 cos 𝜑

100 cos 𝜃 𝑟3

− sin 𝜃

sin 𝜃 cos 𝜑 = sin 𝜃 sin 𝜑 cos 𝜃

50 cos 𝜃 𝑟3

50 cos 𝜃 𝑟3

cos 𝜃 cos 𝜑 cos 𝜃 sin 𝜑 − sin 𝜃

𝑎 + sin 𝜃 sin 𝜑

100 cos 𝜃 𝑟3

100 cos 𝜃 ⎡ ⎤ − sin 𝜑 𝑟 ⎢ ⎥ cos 𝜃 ⎢ 50 cos 𝜃 ⎥ 0 ⎢ 𝑟 ⎥ ⎣ ⎦ 0

+ cos 𝜃 sin 𝜑

50 cos 𝜃 𝑟3

𝑎

𝑎

(sin 2𝜃 cos 𝜑 + cos2 𝜃 cos 𝜑)𝑎 + (sin 2𝜃 sin 𝜑 + cos2 𝜃 sin 𝜑) 𝑎 + (2 cos2 𝜃 − 0.5 sin 2𝜃) 𝑎

𝑎𝑡 𝑟 = 2, 𝜃 = 60, 𝜑 = 20

𝐸⃑ =

50 1.048 𝑎𝑥 + 0.3817 𝑎𝑦 + 0.067 𝑎𝑧 8

𝑬⃑ = 𝟔. 𝟓𝟓 𝒂𝒙 + 𝟐. 𝟑𝟖𝟓𝟔𝟐𝟓 𝒂𝒚 + 𝟎. 𝟒𝟏𝟖𝟕𝟓 𝒂𝒛 ∴ 𝑎⃑ =

6.55 𝑎 + 2.385625 𝑎 + 0.41875 𝑎 𝐸⃑ = 𝐸⃑ √48.769

∴ 𝒂𝑬⃑ =

𝑬⃑ = 𝟎. 𝟗𝟑𝟖 𝒂𝒙 + 𝟎. 𝟑𝟒𝟏𝟔 𝒂𝒚 + 𝟎. 𝟎𝟔 𝒂𝒛 𝑬⃑

Problem 27

Sheet 1

(𝑎) 𝐹𝑖𝑛𝑑 𝑎 𝑖𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑡 𝑃(3, −4,5).

(𝑏) 𝐹𝑖𝑛𝑑 𝑎 𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑡 𝑃.

Solution - Correct 𝐴𝑛𝑠.: (𝑎) 0.424 𝑎 + 0.424 𝑎 + 0.8 𝑎 ; (𝑏) 0.424 𝑎 − 0.566 𝑎 − 0.707 𝑎 ;

(a) ∴ 𝑟⃑ = 3 𝑎 − 4 𝑎 + 5 𝑎 𝐴 sin 𝜃 cos 𝜑 sin 𝜃 sin 𝜑 ∵ 𝐴 = cos 𝜃 cos 𝜑 cos 𝜃 sin 𝜑 𝐴 − sin 𝜑 cos 𝜑

𝑎⃑ = 3 cos 𝜃 𝐴 − sin 𝜃 𝐴 0 𝐴

sin 𝜃 cos 𝜑 = cos 𝜃 cos 𝜑 − sin 𝜑

sin 𝜃 sin 𝜑 cos 𝜃 sin 𝜑 cos 𝜑

cos 𝜃 3 − sin 𝜃 0 0 0

∴ 𝑎 ⃑ = (3 sin 𝜃 cos 𝜑) 𝑎 + (3 cos 𝜃 cos 𝜑) 𝑎 + (−3 sin 𝜑) 𝑎 𝜃 & 𝜑 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝐴 𝜃 = tan

𝑥 +𝑦 𝑧

𝜑 = tan

𝑦 = tan 𝑥

∴ 𝑎⃑ = 3

= tan

√3 + 4 = tan 5

−4 = −53.13 3

cos 𝜑 = 3⁄5 sin 𝜑 = −4⁄5

1 3 1 3 −4 9 9 12 𝑎 + 3 𝑎 + −3 𝑎 = 𝑎 + 𝑎 + 𝑎 5 5 5√2 5√2 √2 5 √2 5

∴ 𝑎 ⃑ = 1.2728 𝑎 + 1.2728 𝑎 + 2.4 𝑎 ∴

cos 𝜃 = 1⁄√2 sin 𝜃 = 1⁄√2

1 = 45

|𝑎 ⃑| = 3

𝒂𝒙⃑ = 𝟎. 𝟒𝟐𝟒𝟐 𝒂𝒓 + 𝟎. 𝟒𝟐𝟒𝟐 𝒂𝜽 + 𝟎. 𝟖 𝒂𝝋 |𝒂𝒙⃑|

(b) 𝐴 ∵ 𝐴 𝐴 ∴ 𝑎⃑ =

∴ 𝑎⃑ =

∴

sin 𝜃 cos 𝜑 = sin 𝜃 sin 𝜑 cos 𝜃 3 5√2

cos 𝜃 cos 𝜑 cos 𝜃 sin 𝜑 − sin 𝜃

− sin 𝜑 𝐴 cos 𝜃 𝐴 𝐴 0

sin 𝜃 cos 𝜑 = sin 𝜃 sin 𝜑 cos 𝜃

cos 𝜃 cos 𝜑 𝑎 + cos 𝜃 sin 𝜑 𝑎 − sin 𝜃 𝑎

3 3 4 𝑎 − 𝑎 − 𝑎 10 5 5

=

= 0.18 𝑎 − 0.24 𝑎 − 0.3 𝑎

𝒂𝜽⃑ = 𝟎. 𝟒𝟐𝟒 𝒂𝒙 − 𝟎. 𝟓𝟔𝟓 𝒂𝒚 − 𝟎. 𝟕𝟎𝟕 𝒂𝒛 |𝒂𝜽⃑|

cos 𝜃 cos 𝜑 cos 𝜃 sin 𝜑 − sin 𝜃

− sin 𝜑 cos 𝜃 0

3

0 3 5√2 0

1 3 1 −4 1 𝑎 + 𝑎 − 𝑎 5√2 √2 5 √2 5 √2 |𝑎 ⃑| = √0.18

Problem 28

Sheet 1

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑀(𝑟 = 5, 𝜃 = 20 , 𝜑 = 120 ),

𝑎𝑛𝑑 𝑁(𝑟 = 5, 𝜃 = 80 , 𝜑 = 30 ), 𝑓𝑖𝑛𝑑

𝑎) 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑀 𝑡𝑜 𝑁 (𝑖. 𝑒. 𝑀𝑁⃑) 𝑏) 𝑔𝑖𝑣𝑒 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑀 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑡𝑜𝑤𝑎𝑟𝑑 𝑁. 𝑐) 𝑔𝑖𝑣𝑒 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑀 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑑𝑖𝑟𝑒𝑐𝑡𝑒𝑑 𝑡𝑜𝑤𝑎𝑟𝑑 𝑁. Solution 𝑃(𝑥,

𝑦,

𝑧) = 𝑃(𝑟 sin 𝜃 cos 𝜑 ,

𝑀(5,

20 ,

120 ) = 𝑀(−0.855,

𝑁(5,

80 ,

30 ) = 𝑁(1.7057,

𝑟 sin 𝜃 sin 𝜑 , 1.481, 0.9848,

𝑟 cos 𝜃) 4.6985)

0.3473)

∴ 𝑅 ⃑ = 𝑁⃑ − 𝑀⃑ = 1.7057 𝑎 + 0.9848 𝑎 + 0.3473 𝑎

− −0.855 𝑎 + 1.481 𝑎 + 4.6985 𝑎

∴ 𝑹𝑴𝑵⃑ = 𝟐. 𝟓𝟎𝟕 𝒂𝒙 − 𝟎. 𝟒𝟗𝟔𝟐 𝒂𝒚 − 𝟒. 𝟑𝟓𝟏𝟐 𝒂𝒛 (b) ∴ 𝑎⃑ =

𝑅 ⃑ = 𝟎. 𝟓𝟎𝟓 𝒂𝒙 − 𝟎. 𝟎𝟗𝟕𝟖𝟏 𝒂𝒚 − 𝟎. 𝟖𝟓𝟕𝟕 𝒂𝒛 𝑅 ⃑

(c) ∴ 𝐴 ⃑ = 𝐴⃑. 𝑎 = 𝐴 (𝑎 . 𝑎 ) + 𝐴 𝑎 . 𝑎 + 𝐴 (𝑎 . 𝑎 ) = 𝐴 sin 𝜃 cos 𝜑 + 𝐴 sin 𝜃 sin 𝜑 + 𝐴 cos 𝜑 ∴ 𝐴 ⃑ = 𝐴⃑. 𝑎 = 𝐴 (𝑎 . 𝑎 ) + 𝐴 𝑎 . 𝑎 + 𝐴 (𝑎 . 𝑎 ) = 𝐴 cos 𝜃 cos 𝜑 + 𝐴 cos 𝜃 sin 𝜑 − 𝐴 sin 𝜃 ∴ 𝐴 ⃑ = 𝐴⃑. 𝑎 = 𝐴 𝑎 . 𝑎 + 𝐴 𝑎 . 𝑎 + 𝐴 𝑎 . 𝑎 = 𝐴 cos 𝜑 + 𝐴 sin 𝜑

Using the angles of the starting point “M” ∴ 𝐴 ⃑ = 0.505 sin 20 cos 120 − 0.09781 sin 20 sin 120 − 0.8577 cos 120 = −0.9213 ∴ 𝐴 ⃑ = 0.505 cos 20 cos 120 − 0.09781 cos 20 sin 120 + 0.8577 sin 20 = −0.02352 ∴ 𝐴 ⃑ = 0.505 cos 120 − 0.09781 sin 120 = −0.3372

∴ 𝑎⃑ = −0.9213 𝑎 − 0.02352 𝑎 − 0.3372 𝑎

.

Problem 29

Sheet 1

𝐴 𝑐𝑙𝑜𝑠𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑖𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑏𝑦 3 < 𝑟 < 5, 0.1𝜋 < 𝜃 < 0.3𝜋, 1.2𝜋 < 𝜑 < 1.6𝜋 𝑎) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝑏) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑃 (𝑟 = 3, 𝜃 = 0.1𝜋, 𝜑 = 1.2𝜋) 𝑡𝑜 𝑃 (𝑟 = 5, 𝜃 = 0.3𝜋, 𝜑 = 1.6𝜋). 𝑐) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎. Solution – Correct 𝐴𝑛𝑠.: (𝑎) 14.91; (𝑏) 3.86; (𝑐) 36.8 ∵ 𝑑𝑣 = 𝑑𝑟 (𝑟 sin 𝜃 𝑑𝜑) (𝑟 𝑑𝜃) .

.

∴𝑣=

𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜑 = .

∴𝑣=

𝑑𝑣 = 𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜑

.

𝑟 3

[− cos 𝜃]

. .

. .

[𝜑]

=

5 −3 3

[cos 0.1𝜋 − cos 0.3𝜋][0.4𝜋]

98 [0.3633][0.4𝜋] = 𝟏𝟒. 𝟗𝟏𝟐𝟑 𝒄𝒖𝒃𝒊𝒄 𝒖𝒏𝒊𝒕 3

(b) ∵𝑑=

𝑟 + 𝑟 − 2𝑟 𝑟 cos 𝜃 cos 𝜃 − 2𝑟 𝑟 sin 𝜃 sin 𝜃 cos(𝜑 − 𝜑 )

∴𝑑=

5 + 3 − 2(5)(3) cos 54 cos 18 − 2(5)(3) sin 54 sin 18 cos(72) = √14.91 = 𝟑. 𝟖𝟔 𝒖𝒏𝒊𝒕 𝒍𝒆𝒏𝒈𝒕𝒉

(𝒄)

∵ 𝑑𝑠 = 𝑟 sin 𝜃 𝑑𝜃𝑑𝜑 𝑎 + 𝑟 sin 𝜃 𝑑𝑟𝑑𝜑 𝑎 + 𝑟 𝑑𝑟𝑑𝜃 𝑎 .

.

.

∴𝑠=𝑟

.

sin 𝜃 𝑑𝜃𝑑𝜑 𝑎 + sin 𝜃 .

𝑟𝑑𝑟𝑑𝜑 𝑎 +

.

∴ 𝑠 = 𝑟 [− cos 𝜃]

. .

[𝜑]

. .

𝑎 + sin 𝜃

𝑟 2

[𝜑]

. .

5 −3 2 ∴ 𝑠 = 0.4565 𝑟 𝑎 + 3.2𝜋 sin 𝜃 𝑎 + 1.6𝜋 𝑎 ∴ 𝑠 = 𝑟 [cos 18 − cos 54][0.4𝜋] 𝑎 + sin 𝜃

𝑠

+𝑠

𝑠

.

+𝑠

𝑠

.

+𝑠

∴𝑠

.

𝑎 +

= 3.2𝜋 sin 54 + 3.2𝜋 sin 18 = 11.2397 .

= 1.6𝜋 + 1.6𝜋 = 3.2𝜋 = 10.053

= 15.521 + 11.2397 + 10.053 = 𝟑𝟔. 𝟖𝟏

𝑟 2

[𝜃]

[0.4𝜋] 𝑎 +

= 0.4565(5 ) + 0.4565(3 ) = 15.521 .

𝑟 𝑑𝑟𝑑𝜃 𝑎

.

. .

𝑎

5 −3 2

[0.2𝜋]𝑎

Problem 30

Sheet 1

𝑎) 𝐺𝑖𝑣𝑒 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑡ℎ𝑎𝑡 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑓𝑟𝑜𝑚 𝑃(𝑟 = 4, 𝜃 = 20, 𝜑 = 10) 𝑡𝑜 𝑄(𝑟 = 7, 𝜃 = 120, 𝜑 = 75). 𝑏) 𝐺𝑖𝑣𝑒 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑡 𝑀(𝑥 = 5, 𝑦 = 1, 𝑧 = 2) 𝑡ℎ𝑎𝑡 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑡𝑜 𝑁(2,4,6). 𝑐) 𝐻𝑜𝑤 𝑓𝑎𝑟 𝑖𝑠 𝑖𝑡 𝑓𝑟𝑜𝑚 𝐴(𝑟 = 110, 𝜃 = 30, 𝜑 = 60) 𝑡𝑜 𝐵(𝑟 = 30, 𝜃 = 75, 𝜑 = 125). Solution 𝑃(𝑟,

𝜃,

𝜑) = 𝑃(𝑥 = 𝑟 sin 𝜃 cos 𝜑 ,

𝑃(4,

20 ,

𝑄(7,

120 ,

10 ) = 𝑃(1.3473,

𝑦 = 𝑟 sin 𝜃 sin 𝜑 ,

0.2375,

75 ) = 𝑄(1.569,

𝑧 = 𝑟 cos 𝜃)

3.7588)

5.8856, −3.5)

∴ 𝑅 ⃑ = 𝑄⃑ − 𝑃⃑ = 1.569 𝑎 + 5.8856 𝑎 − 3.5 𝑎

− 1.3473 𝑎 + 0.2375 𝑎 + 3.7588 𝑎

∴ 𝑹𝑷𝑸⃑ = 𝟎. 𝟐𝟐𝟏𝟕 𝒂𝒙 + 𝟓. 𝟔𝟒𝟖𝟏 𝒂𝒚 − 𝟕. 𝟐𝟓𝟖𝟖 𝒂𝒛 (b) ∴ 𝑅 ⃑ = 𝑁⃑ − 𝑀⃑ = 2 𝑎 + 4 𝑎 + 6 𝑎 − 5 𝑎 + 𝑎 + 2 𝑎 = −3 𝑎 + 3 𝑎 + 4 𝑎 𝐴 sin 𝜃 cos 𝜑 sin 𝜃 sin 𝜑 cos 𝜃 𝐴 sin 𝜃 cos 𝜑 sin 𝜃 sin 𝜑 cos 𝜃 −3 𝐴 𝐴 ∵ = cos 𝜃 cos 𝜑 cos 𝜃 sin 𝜑 − sin 𝜃 = cos 𝜃 cos 𝜑 cos 𝜃 sin 𝜑 − sin 𝜃 3 𝐴 − sin 𝜑 cos 𝜑 0 − sin 𝜑 cos 𝜑 0 𝐴 4 ∴ 𝑅 ⃑ = (−3 sin 𝜃 cos 𝜑 + 3 sin 𝜃 sin 𝜑 + 4 cos 𝜃) 𝑎 + (−3 cos 𝜃 cos 𝜑 + 3 cos 𝜃 sin 𝜑 − 4 sin 𝜃) 𝑎 + (3 sin 𝜑 + 3 cos 𝜑) 𝑎 𝜃 & 𝜑 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑀 𝜃 = tan

𝑥 +𝑦 𝑧

𝜑 = tan

𝑦 = tan 𝑥

∴ 𝑅 ⃑ = −3

∴ 𝑹𝑴𝑵⃑ =

= tan

√5 + 1 = tan 2

1 = 11.31 5

√26 5

+3

√𝟑𝟎

𝒂𝒓 −

𝟏𝟐𝟖

√26 1

√𝟕𝟖𝟎

𝒂𝜽 +

cos 𝜃 = 2⁄√30 sin 𝜃 = √26⁄√30

cos 𝜑 = 5⁄√26 sin 𝜑 = 1⁄√26 +4

√30 √26 √30 √26 1 5 + 3 +3 𝑎 √26 √26

−𝟒

√26 = 68.58 2

𝟏𝟖 √𝟐𝟔

2 √30

𝒂𝝋

𝑎 + −3

2

5

√30 √26

+3

2

1

√30 √26

−4

√26 √30

𝑎

(c) 𝐹𝑟𝑜𝑚 𝐴(𝑟 = 110, 𝜃 = 30, 𝜑 = 60) 𝑡𝑜 𝐵(𝑟 = 30, 𝜃 = 75, 𝜑 = 125) ∵𝑑=

𝑟 + 𝑟 − 2𝑟 𝑟 cos 𝜃 cos 𝜃 − 2𝑟 𝑟 sin 𝜃 sin 𝜃 cos(𝜑 − 𝜑 )

∴𝑑=

110 + 30 − 2(110)(30) cos(30) cos(75) − 2(110)(30) sin(30) sin(75) cos(125 − 60)

∴ 𝑑 = √10173.53 = 𝟏𝟎𝟎. 𝟖𝟔𝟒

Problem 31

Sheet 1

𝑃𝑜𝑖𝑛𝑡𝑠 𝐴(𝑟 = 100, 𝜃 = 90, 𝜑 = 0) & 𝐵(𝑟 = 100, 𝜃 = 90, 𝜑 = 5) 𝑎𝑟𝑒 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑎 100𝑚 𝑟𝑎𝑑𝑖𝑢𝑠. 𝑎) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒𝑖𝑟 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛, 𝑢𝑠𝑖𝑛𝑔 𝑎 𝑝𝑎𝑡ℎ 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒? 𝑏) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒𝑖𝑟 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛, 𝑢𝑠𝑖𝑛𝑔 𝑎 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 − 𝑙𝑖𝑛𝑒 𝑝𝑎𝑡ℎ? Solution - Correct 𝐴𝑛𝑠.: (𝑎) 8.72665; (𝑏) 8.72388.

∵ 𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑟 𝜑 = 100 × 5 ×

𝜋 = 𝟖. 𝟕𝟐𝟔𝟔𝟒𝟔 180

(b) ∵𝑑=

𝑟 + 𝑟 − 2𝑟 𝑟 cos 𝜃 cos 𝜃 − 2𝑟 𝑟 sin 𝜃 sin 𝜃 cos(𝜑 − 𝜑 ) =

𝑟 + 𝑟 − 2𝑟 𝑟 cos(𝜑 − 𝜑 )

∴𝑑=

100 + 100 − 2(100)(100) cos(5) = 100 1 + 1 − 2 cos(5) = 𝟖. 𝟕𝟐𝟑𝟖𝟖

Another solution for (b) 𝐴(𝑟 = 100, 𝜃 = 90, 𝜑 = 0) ≡ (𝑥 = 100, 𝑦 = 0, 𝑧 = 0) 𝐵(𝑟 = 100, 𝜃 = 90, 𝜑 = 5) ≡ (𝑥 = 100 cos 5 , 𝑦 = 100 sin 5 , 𝑧 = 0) ∴𝑑=

(100 − 100 cos 5) + (0 − 100 sin 5) + (0) = 100 (1 − cos 5) + (sin 5) = 𝟖. 𝟕𝟐𝟑𝟖𝟖

Problem 32

Sheet 1

(𝑎) 𝐹𝑖𝑛𝑑 𝑎 𝑖𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 & 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠. Solution

(𝑏) 𝐹𝑖𝑛𝑑 𝑎 𝑖𝑛 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 & 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠.