Exercise 1

Calculate the head of the vector \vec { AB } knowing that its components are (3, -1) and its tail is A = (-2, 4).

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Exercise 2

Given points A = (0, a) and B = (1, 2), calculate the value of a if the magnitude of the vector \vec { AB } is one.

Exercise 3

Normalize the vectors: \vec { u } = (1, \sqrt { 2 } ), \vec { v } = (-4, 3) and \vec { w } = (6, -8).

Exercise 4

Determine the unit vector, \vec { u }, which is in the same direction as the vector \vec { v } = 8 \vec { i } - 6 \vec { j }.

Exercise 5

Calculate the coordinates of D so that the quadrilateral formed by the vertices: A = (-1, -2), B = (4, -1), C = (5, 2) and D; is a parallelogram.

Exercise 6

The vectors \vec { u } = (1, 4) and \vec { v } = (1, 3) form a basis. Express this in basis the vector \vec { w } = (-1, -1).

Exercise 7

Find the value of k so that the angle that forms between \vec { u } = (3, k) and \vec { v } = (2, -1) is:

1 { 90 }^{ \circ }

2 { 0 }^{ \circ }

3 { 45 }^{ \circ }

Exercise 8

Calculate the value of a so that the vectors \vec { u } = 3 \vec { i } + 4 \vec { j } and \vec { v } = a \vec { i } - 2 \vec { j } form an angle of { 45 }^{ \circ }.

Exercise 9

If \left \{ \vec { u }, \vec { u } \right \} is an orthonormal basis, calculate:

1 \vec { u } . \vec { u }

2 \vec { u } . \vec { v }

3 \vec { v } . \vec { u }

4 \vec { v } . \vec { v }

 

Solution of exercise 1

Calculate the head of the vector \vec { AB } knowing that its components are (3, -1) and its tail is A = (-2, 4).

3 = { x }_{ B } - (-2) \qquad { x }_{ B } = 1

-1 = { y }_{ B } - 4 \qquad { y }_{ B } = 3

B (1, 3)

 

Solution of exercise 2

Given points A = (0, a) and B = (1, 2), calculate the value of a if the magnitude of the vector \vec { AB } is one.

d(A, B) = 1 \qquad \sqrt { { (1 - 0) }^{ 2 } + { (2 - a) }^{ 2 } } = 1

1 + 4 - 4a + { a }^{ 2 } = 1 \qquad a = 2

 

Solution of exercise 3

Normalize the vectors: \vec { u } = (1, \sqrt { 2 } ), \vec { v } = (-4, 3) and \vec { w } = (6, -8).

\left | \vec { u } \right | = \sqrt { 1 + 2 } = \sqrt { 3 } \qquad \frac { \vec { u } }{ \left | \vec { u } \right | } = (\frac { 1 }{ \sqrt { 3 } }, \frac { \sqrt { 2 } }{ \sqrt { 3 } } )

\left | \vec { v } \right | = \sqrt { 16 + 9 } = 5 \qquad \frac { \vec { v } }{ \left | \vec { v } \right | } = (- \frac { 4 }{ 5 }, \frac { 3 }{ 5 } )

\left | \vec { w } \right | = \sqrt { 36 + 64 } = 10 \qquad \frac { \vec { w } }{ \left | \vec { w } \right | } = (\frac { 6 }{ 10 }, - \frac { 8 }{ 10 } )

 

Solution of exercise 4

Determine the unit vector, \vec { u }, which is in the same direction as the vector \vec { v } = 8 \vec { i } - 6 \vec { j }.

\vec { u } = \frac { \vec { v } }{ \sqrt { v } }

\vec { u } = \frac { 8 \vec { i } - 6 \vec { j } }{ \sqrt { { 8 }^{ 2 } + { 6 }^{ 2 } } } = \frac { 8 \vec { i } - 6 \vec { j } }{ 10 } = \frac { 4 }{ 5 } \vec { i } - \frac { 3 }{ 5 } \vec { j }

 

Solution of exercise 5

Calculate the coordinates of D so that the quadrilateral formed by the vertices: A = (-1, -2), B = (4, -1), C = (5, 2) and D; is a parallelogram.

\vec { AB } = \vec { DC }

(4 + 1, - 1 + 2) = (5 - { x }_{ D }, 2 - { y }_{ D } )

5 = 5 - { x }_{ D } \qquad { x }_{ D } = 0

1 = 2 - { y }_{ D } \qquad { y }_{ D } = 1

Hence, D = (0, 1)

Solution of exercise 6

The vectors \vec { u } = (1, 4) and \vec { v } = (1, 3) form a basis. Express this in basis the vector \vec { w } = (-1, -1).

(-1, -1) = a(1, 4) + b (1, 3)

-1 = a + b \qquad -1 = 4a + 3b

a = -1 - b \qquad -1 = 4a + 3b

Replacing the value of a in the second equation:

-1 = 4(-1 - b) + 3b

-1 = -4 - 4b + 3b

3 = -b

b = -3

Plugging the value of b in the first equation:

-1 = a + b

-1 = a - 3

a = 2

 

\vec { w } = 2 \vec { u } - 3 \vec { v }

 

Solution of exercise 7

Find the value of k so that the angle that forms between \vec { u } = (3, k) and \vec { v } = (2, -1) is:

1 { 90 }^{ \circ }

\widehat { uv } = { 90 }^{ \circ } \qquad \vec { u } . \vec { v } = 0

3 . 2 + k (-1) = 0 \qquad k = 6

2 { 0 }^{ \circ }

\widehat { uv } = { 0 }^{ \circ } \qquad \frac { { u }_{ 1 } }{ { u }_{ 2 } } = \frac { { v }_{ 1 } }{ { v }_{ 2 }}

\frac { 3 }{ k } = \frac { 2 }{ -1 } \qquad k = - \frac { 3 }{ 2 }

 

3 { 45 }^{ \circ }

\widehat { uv } = { 45 }^{ \circ } \qquad \cos { { 45 }^{ \circ } } = \frac { \sqrt { 2 } }{ 2 }

\frac { \sqrt { 2 } }{ 2 } = \frac { 3 . 2 + k (-1) }{ \sqrt { 9 + { k }^{ 2 } } . \sqrt { 4 + 1 } }

6 { k }^{ 2 } + 48k - 54 = 0

After solving the above equation:

k = -9 \qquad k = 1

 

Solution of exercise 8

Calculate the value of a so that the vectors \vec { u } = 3 \vec { i } + 4 \vec { j } and \vec { v } = a \vec { i } - 2 \vec { j } form an angle of { 45 }^{ \circ }.

\cos { { 45 }^{ \circ } } = \frac { 3a - 8 }{ \sqrt { { 3 }^{ 2 } + { 4 }^{ 2 } } . \sqrt { { a }^{ 2 } + { (-2) }^{ 2 }  } } \qquad \frac { \sqrt { 2 } }{ 2 } = \frac { 3a - 8 }{ 5 }

\sqrt { 2 . ( { a }^{ 2 } + 4 ) } = \frac { 6a - 16 }{ 5 } \qquad 2 . ( { a }^{ 2 } + 4 ) = \frac { 36 { a }^{ 2 } - 192a + 256 }{ 25 }

7 { a }^{ 2 } + 96a - 28 = 0

{ a }_{ 1 } = \frac { 2 }{ 7 } \qquad { a }_{ 2 } = -14

 

Solution of exercise 9

If \left \{ \vec { u }, \vec { u } \right \} is an orthonormal basis, calculate:

1 \vec { u } . \vec { u } = 1 . 1 . \cos { { 0 }^{ \circ } } = 1

2 \vec { u } . \vec { v } = 1 . 1 . \cos { { 90 }^{ \circ } } = 0

3 \vec { v } . \vec { u } = 1 . 1 . \cos { { 90 }^{ \circ } } = 0

4 \vec { v } . \vec { v } = 1 . 1 . \cos { { 0 }^{ \circ } } = 1

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.