PST 3 - problem solving task 3 PDF

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PST 3 Q1. A) Find the Vectors  AB∧  BC .

 AB:

 0A

=

⟨ 2 ,−1,0 ⟩

 0 B = ⟨ 1,3 ,−1 ⟩  0A AB=  0 B−  AB= ⟨ 1,3 ,−1 ⟩ −⟨ 2,−1,0 ⟩  AB= ⟨ −1,4 , −1⟩  AB=−1 i+4 j−1 k  BC :

 0 B = ⟨ 1,3 ,−1⟩  0 C = ⟨−1,0.5,0 ⟩

 BC = 0 C− 0B  BC=⟨ −1,0.5,0⟩− ⟨ 1,3 ,−1 ⟩  BC=⟨ −2 , −2.5,1⟩  BC=−2i−2.5 j+1 k

B) Using your answer to part (a), determine the area of the parallelogram which has adjacent sides  AB∧  BC .

 AB=−1 i+4 j−1 k  BC=−2i−2.5 j+1 k  BC=(−1i+ 4 j−1 k)×(−2 i−2.5 j+1 k ) AB×   AB×  BC= (−2∗2 ) i∗i+ (−1∗−2.5 ) i∗ j+( −1∗1 ) i∗k + ( 4∗−2 ) j∗i+ ( 4∗−2.5 ) j∗j+ ( 4∗1 ) j∗k + (−1∗2 ) k∗i+ (−  BC =2.5 k +1 j+8 k +4 i−2 j−2.5 i AB×   BC =1.5 i−1 j+10.5 k AB× 

| AB×  BC |= √ 1.52 + 12+ 10.52=10.65 units C) Are the vectors v

and

w

parallel, perpendicular, or neither? Justify why.

v =−1i+2 j−1k , w=3 i−3 j+2 k Parallel test:

3 x= =3 1 3 y= =1.5 2

not same ratio, therefore, not parallel.

2 z= =2 1 Perpendicular test:

v ∙ w=( −1∗3 ) +(2∗−3 )+ ( 2∗2 )

¿−3−6+ 4 ¿−5≠ 0

not equal to zero, therefore, not perpendicular.

These vectors aren’t parallel as they don’t have the same ratios when like values are divided by each other. They also aren’t perpendicular as the dot product of the vectors doesn’t equal 0.

D) Find the vector projection of u

in the direction of w .

u=−1i+0 j+2 k , w =3 i−3 j+2 k Projw ( u )= Where,

( u‖w∙ w‖ )‖ww‖

u ∙ w= (− 1 )( 3 ) +(0 )( −3)+ ( 2 )( 2)

¿−3+4=1

‖w ‖= √3 2+3 2+22 = √22 Projw ( u )= ¿

( √122 ) 3i−3√ 22j+2 k

3 i−3 j+ 2 k 3 i 3 j 2 k = − + =0.14 i−0.14 j+ 0.09 k 22 22 22 22

Q2. A) Sketch the vector addition of the three forces if the resultant force on the bracket is 150N to the right.

Fr=150 i−153.2 j+ 150 i−153.18 j+ 150 i−172 j ¿ 450 i−478.38 j B) Find the magnitude and direction of F2 if the resultant force on the bracket is 150N directed to the right.

|F 2|=√ 1502+ 153.18 2=214.39 N (magnitude ) θ=tan

O A

θ=tan−1

( 153.18 150 )

θ=45.6 °(direction)

Q3. A) Encode the word “HELLO” into a string of numbers for transmission to your friend. Show the matrix multiplication in full. A=

|−21 −13 |

H, E, L, L, O = 24,9,5,5,2 E = AM E=

E=

E=

|−21 −13 ||249 55 02| + 3∗9 −2∗5+3∗5 |−2∗24 1∗24−1∗9 1∗5−1∗5 |1915 50 −24|

|

−2∗2+3∗0 1∗2−1∗0

E = 19,15,5,0,-4,2 B) Solve for the decoding matrix

A

−1

E=M

A−1=

|

|

1 d −b det( A) −c a

Where det(A) = ad-bc = 2-3 = -1

A−1=

|

|

1 −1 −3 −1 −1 −2

| | M=|1 3||24 5 2| 1 2 9 5 0 1∗24 + 3∗9 1∗5+3∗5 M=| 1∗24 + 2∗9 1∗5+2∗5 M =|24 5 2| 9 5 0 −1 A =1 3 1 2

|

1∗2+3∗0 1∗2−1∗2

=HELLO

C) Your friend sends you a reply of “26,-3,20,-1,15,-3”. Use your answer to decode it.

|11 32||−263 −201 −315 | 1∗26+3∗−3 1∗20 + 3∗−1 ¿| 1∗26+2∗− 3 1∗20 + 2∗−1 ¿|17 17 6 | 20 18 9 ¿

= It is me.

|

1∗15 + 3∗−3 1∗15+ 2∗−3...