Exercise 1

Find two unit vectors for (2, −2, 3) and (3, −3, 2) and determine the orthogonal vector for the two.

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Exercise 2

Find a unit vector that is perpendicular to \overrightarrow { u } = (2, 3, 4) and \overrightarrow { v } = (-1, 3, -5).

Exercise 3

Given the vectors \overrightarrow { u } = 3i - j + k and \overrightarrow { v } = 2i - 3j + k, find the product \overrightarrow { u } \times \overrightarrow { v } and verify that this vector is orthogonal to \overrightarrow { u } and \overrightarrow { v }. Also, find the vector \overrightarrow { v } \times \overrightarrow { u } and compare it with \overrightarrow { u } \times \overrightarrow { v }.

Exercise 4

Consider the following figure:

Determine:

1 The coordinates of D if ABCD is a parallelogram.

2 The area of the parallelogram.

Exercise 5

Given the points A = (1, 0, 1), B = (1, 1, 1) and C = (1, 6, a), determine:

1 What values of a are collinear.

2 Determine if values exist for a so that A, B, and C are three vertices of a parallelogram of area 3. If values do exist, determine the coordinates of C:

Exercise 6

A = (−3, 4, 0), B = (3, 6, 3) and C = (−1, 2, 1) are the three vertices of a triangle.

1. Calculate the cosine of each of the three angles in the triangle.

2. Calculate the area of the triangle.

 

 

Solution of exercise 1

Find two unit vectors for (2, −2, 3) and (3, −3, 2) and determine the orthogonal vector for the two.

\overrightarrow { w } = \begin{vmatrix} i & j & k \\ 2 & -2 & 3 \\ 3 & -3 & 2 \end{vmatrix} = i \begin{vmatrix} -2 & 3 \\ -3 & 2 \end{vmatrix} - j \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} + k \begin{vmatrix} 2 & -2 \\ 3 & -3 \end{vmatrix}

\overrightarrow { w } = i(-4 + 9) - j(4 - 9) + k(-6 + 6)

\overrightarrow { w } = 5i + 5j

 

\left| \overrightarrow { w } \right| = \sqrt { { 5 }^{ 2 } + { 5 }^{ 2 } } = 5 \sqrt { 2 }

\hat { u } = (\frac { 5 }{ 5 \sqrt { 2 } }, \frac { 5 }{ 5 \sqrt { 2 } }, 0 ) = (\frac { 1 }{ \sqrt { 2 } }, \frac { 1 }{ \sqrt { 2 }}, 0)

\hat { v } = (\frac { -5 }{ 5 \sqrt { 2 } }, \frac { -5 }{ 5 \sqrt { 2 } }, 0 ) = (\frac { -1 }{ \sqrt { 2 } }, \frac { -1 }{ \sqrt { 2 }}, 0)

 

Solution of exercise 2

Find a unit vector that is perpendicular to \overrightarrow { u } = (2, 3, 4) and \overrightarrow { v } = (-1, 3, -5).

\overrightarrow { u } \times \overrightarrow { v } = \begin{vmatrix} i & j & k \\ 2 & 3 & 4 \\ -1 & 3 & -5 \end{vmatrix} = i \begin{vmatrix} 3 & 4 \\ 3 & -5 \end{vmatrix} - j \begin{vmatrix} 2 & 4 \\ -1 & -5 \end{vmatrix} + k \begin{vmatrix} 2 & 3 \\ -1 & 3 \end{vmatrix} = -27i + 6j + 9k

\left| \overrightarrow { u } \times \overrightarrow { v } \right| = \sqrt { { (-27) }^{ 2 }+{ 6 }^{ 2 }+{ 9 }^{ 2 } } = \sqrt { 846 }

\overrightarrow { w } = ( \frac { -27 }{ \sqrt { 846 } }, \frac { 6 }{ \sqrt { 846 } }, \frac { 9 }{ \sqrt { 846 }} )

 

Solution of exercise 3

Given the vectors \overrightarrow { u } = 3i - j + k and \overrightarrow { v } = 2i - 3j + k, find the product \overrightarrow { u } \times \overrightarrow { v } and verify that this vector is orthogonal to \overrightarrow { u } and \overrightarrow { v }. Also, find the vector \overrightarrow { v } \times \overrightarrow { u } and compare it with \overrightarrow { u } \times \overrightarrow { v }.

\overrightarrow { u } \times \overrightarrow { v } = \begin{vmatrix} i & j & k \\ 3 & -1 & 1 \\ 2 & -3 & 1 \end{vmatrix} = i \begin{vmatrix} -1 & 1 \\ -3 & 1 \end{vmatrix} - j \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} + k \begin{vmatrix} 3 & -1 \\ 2 & -3 \end{vmatrix}

\overrightarrow { u } \times \overrightarrow { v } = i(-1 + 3) - j(3 - 2) + k(-9 + 2) = 2i - j - 7k

(\overrightarrow { u } \times \overrightarrow { v }) \bot \overrightarrow { u } \qquad (\overrightarrow { u } \times \overrightarrow { v }) . \overrightarrow { u } = 0

(2, -1, -7) . (3, -1, 1) = 6 + 1 - 7 = 0

 

(\overrightarrow { u } \times \overrightarrow { v }) \bot \overrightarrow { v } \qquad (\overrightarrow { u } \times \overrightarrow { v }) . \overrightarrow { v } = 0

(2, -1, -7) . (2, -3, 1) = 4 + 3 - 7 = 0

\overrightarrow { v } \times \overrightarrow { u } = \begin{vmatrix} i & j & k \\ 2 & -3 & 1 \\ 3 & -1 & 1 \end{vmatrix} = i \begin{vmatrix} -3 & 1 \\ -1 & 1 \end{vmatrix} - j \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + k \begin{vmatrix} 2 & -3 \\ 3 & -1 \end{vmatrix}

\overrightarrow { v } \times \overrightarrow { u } = \begin{vmatrix} i & j & k \\ 2 & -3 & 1 \\ 3 & -1 & 1 \end{vmatrix} = i \begin{vmatrix} -3 & 1 \\ -1 & 1 \end{vmatrix} - j \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + k \begin{vmatrix} 2 & -3 \\ 3 & -1 \end{vmatrix}

\overrightarrow { v } \times \overrightarrow { u } = i(-3 + 1) - j(2 - 3) + k(-2 + 9) = -2i + j + 7k

\overrightarrow { v } \times \overrightarrow { u } = - (\overrightarrow { u } \times \overrightarrow { v })

 

Solution of exercise 4

Consider the following figure:

Determine:

1 The coordinates of D if ABCD is a parallelogram.

(x - 1, y - 1, z) = (2+1, 2+1, 1)

 

x - 1 = 3 \qquad x = 4

y - 1 = 3 \qquad y = 4

z = 1

 

D = (4, 4, 1)

2 The area of the parallelogram.

A = \left| \overrightarrow { BC } \times \overrightarrow { BA }  \right|

\overrightarrow { BC } = (3, 3, 1) \qquad \overrightarrow { BA } = (2, 2, 1)

\overrightarrow { BC } \times \overrightarrow { BA } = \begin{vmatrix} i & j & k \\ 3 & 3 & 1 \\ 2 & 2 & 1 \end{vmatrix} = i \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} - j \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} + k \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix}

\overrightarrow { BC } \times \overrightarrow { BA } = i (3 - 2) - j (3 - 2) + k (6 - 6) = i - j

A = \left| \overrightarrow { BC } \times \overrightarrow { BA }  \right| = \sqrt { { 1 }^{ 2 } +  { (-1) }^{ 2 } } = \sqrt { 2 } { u }^{ 2 }

 

Solution of exercise 5

Given the points A = (1, 0, 1), B = (1, 1, 1) and C = (1, 6, a), determine:

1 What values of a are collinear.

If A, B, and C are collinear, the vectors \overrightarrow { AB } and \overrightarrow { AC } are linearly dependent and have proportional components.

\overrightarrow { AB } = (1 - 1, 1 - 0, 1 - 1) = (0, 1, 0)

\overrightarrow { AC } = (1 - 1, 6 - 0, a - 1) = (0, 6, a - 1)

(0, 6, a - 1) = k (0, 1, 0) \qquad a - 1 =0 \qquad a = 1

 

2 Determine if values exist for a so that A, B, and C are three vertices of a parallelogram of area 3. If values do exist, determine the coordinates of C:

(\overrightarrow { AB } \times \overrightarrow { AC } ) = \begin{vmatrix} i & j & k \\ 0 & 1 & 0 \\ 0 & 6 & a - 1 \end{vmatrix} = i \begin{vmatrix} 1 & 0 \\ 6 & a-1 \end{vmatrix} - j \begin{vmatrix} 0 & 0 \\ 0 & a-1 \end{vmatrix} + k \begin{vmatrix} 0 & 1 \\ 0 & 6 \end{vmatrix}

(\overrightarrow { AB } \times \overrightarrow { AC } ) = i (a - 1) - j (0 - 0) + k (0 - 0) = i(a - 1)

A = \left| \overrightarrow { AB } \times \overrightarrow { AC } \right| = 3

3 = \sqrt { { (a - 1) }^{ 2 } + { 0 }^{ 2 } + { 0 }^{ 2 } }

a - 1 = 3 \qquad or \qquad a -1 = -3

a = 4 \qquad or \qquad a = -2

C (1, 6, 4) \qquad or \qquad C (1, 6, -2)

 

Solution of exercise 6

A = (−3, 4, 0), B = (3, 6, 3) and C = (−1, 2, 1) are the three vertices of a triangle.

1. Calculate the cosine of each of the three angles in the triangle.

\overrightarrow { AB } = (3 + 3, 6 - 4, 3 - 0) = (6, 2, 3) \qquad \qquad \overrightarrow { BA } = (-6, -2, -3)

\overrightarrow { AC } = (-1 + 3, 2 - 4, 1 - 0) = (2, -2, 1) \qquad \qquad \overrightarrow { CA } = (-2, 2, -1)

\overrightarrow { BC } = (-1 - 3, 2 - 6, 1 - 3) = (-4, -4, -2) \qquad \qquad \overrightarrow { CB } = (4, 4, 2)

 

\cos { \overrightarrow { AB }, \overrightarrow { AC } } = \frac { 12 - 4 +3 }{ \sqrt { 36 + 4 + 9 } \sqrt { 4 + 4 + 1}} = \frac { 11 }{ 21 }

\cos { \overrightarrow { BA }, \overrightarrow { BC } } = \frac { 24 + 8 + 6 }{ \sqrt { 36 + 4 + 9 } \sqrt { 16 + 16 + 4 }} = \frac { 38 }{ 21 }

\cos { \overrightarrow { CA }, \overrightarrow { CB } } = \frac { -8 + 8 - 2 }{ \sqrt { 4 + 4 + 1 } \sqrt { 16 + 16 + 4 }} = \frac { -1 }{ 9 }

2. Calculate the area of the triangle.

A = \frac { 1 }{ 2 } \left| \overrightarrow { AB } \times \overrightarrow { AC } \right|

\overrightarrow { AB } \times \overrightarrow { AC } = \begin{vmatrix} i & j & k \\ 6 & 2 & 3 \\ 2 & -2 & 1 \end{vmatrix} = i \begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} - j \begin{vmatrix} 6 & 3 \\ 2 & 1 \end{vmatrix} + k \begin{vmatrix} 6 & 2 \\ 2 & -2 \end{vmatrix} = 8i - 16k

\overrightarrow { AB } \times \overrightarrow { AC } = \sqrt { { 8 }^{ 2 } + { 16 }^{ 2 } } = \sqrt { 64 + 256 } = 8 \sqrt { 5 }

 

A = \frac { 1 }{ 2 } \left| \overrightarrow { AB } \times \overrightarrow { AC } \right| = \frac { 1 }{ 2 } (8 \sqrt { 5 }) = 4 \sqrt { 5 }

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.