Exercise 1

Find the area of the triangle whose vertices are the points A = (1, 1, 1), B = (3, 2, 1) and C = (−1, 3, 2).

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Exercise 2

Find the volume of the tetrahedron whose vertices are the points A = (0,0,0), B = (2, 1, 3), C = (−1, 3, 1) and D = (4, 2, 1).

Exercise 3

Given the line r = \frac {x - 1}{2} = \frac {y + 1}{1} = \frac {z}{1}  and the plane \pi = x + y + z - 4 = 0 , find the equation of the line, s, which is the orthogonal projection of r on π.

Exercise 4

Calculate the distance between the following lines:

r = \frac {x - 2}{3} = \frac {y - 2}{-1} = \frac {z + 1}{4}

s = \begin{cases} x = 5 + \lambda \\ y = -1 \\ z = 8 + 2 \lambda \end{cases}

Exercise 5

Find the symmetric point of Point A = (3, 2, 1) to the plane x + y + z + 21 = 0 .

Exercise 6

Calculate the area of the triangle whose vertices are the points of intersection of the plane 6x + 3y + 2z = 6  with the coordinate axes.

Exercise 7

Given the plane \pi = x + 2y + 3z = 1  and the point A = (1, 1, 1), calculate the coordinates of the base (endpoint) of the perpendicular from A to the plane.

Exercise 8

Determine the equation of the plane π that is \sqrt{6}  distant from the origin and is parallel to the plane 2x + y - z = 3 .

Exercise 9

Find the distance between the point A = (3, 2, 7) and the line of the first octant (+,+,+).

Exercise 10

Calculate the area of the square whose sides are on the lines:

r = \frac {x - 1}{1} = \frac {y}{2} = \frac {z - 2}{1}

s = \begin{cases} x - y + z = 2 \\3x - y - z = -4\end{cases}

 

Solution of exercise 1

Find the area of the triangle whose vertices are the points A = (1, 1, 1), B = (3, 2, 1) and C = (−1, 3, 2).

\overrightarrow{AB} = (3 - 1, 2 - 1, 1 - 1) = (2, 1, 0)

\overrightarrow {AC} = (-1 - 1, 3 - 1, 2 - 1) = (-2, 2, 1)

\rightarow {AB} \times \overrightarrow {AC} =\begin {bmatrix}  \overrightarrow {i}& \overrightarrow{j} & \overrightarrow{k} \\ 2 & 1 & 0 \\ -2 & 2 & 1 \\ \end {bmatrix} = \overrightarrow {i} - 2 \overrightarrow {j} + 6 \overrightarrow {k}

|\overrightarrow {AB} \times \overrightarrow {AC}| = \sqrt {1^2 + (-2)^2 + 6^2} = \sqrt{41}

A_r = \frac {1}{2} \sqrt{41}u^2

 

Solution of exercise 2

Find the volume of the tetrahedron whose vertices are the points A = (0,0,0), B = (2, 1, 3), C = (−1, 3, 1) and D = (4, 2, 1).

\overrightarrow {AB} = (2 - 0, 1 - 0, 3 - 0) = (2, 1, 3)

\overrightarrow {AC} =(-1 - 0, 3 - 0, 1 - 0) = (-1, 3, 1)

\overrightarrow {AD} = (4 - 0, 2 - 0, 1 - 0) = (4, 2, 1)

V = \frac {1}{6} = \begin {bmatrix} 2 & 1 & 3  \\ -1 & 3 & 1 \\ 4 & 2 & 1 \\ \end {bmatrix} = \frac {35}{6} u^3

 

Solution of exercise 3

Given the line r = \frac {x - 1}{2} = \frac {y + 1}{1} = \frac {z}{1}  and the plane \pi = x + y + z - 4 = 0 , find the equation of the line, s, which is the orthogonal projection of r on π.

The line, s, is the intersection of the plane, π, with the plane, πp, that contains the line, r, and it is perpendicular to π.

The plane, πp, is determined by the point A = (2, −1, 0), the vector (2, 1, 1) and the normal vector, (1, 1, 1), of the perpendicular plane π.

Solution of Exercise 3

\begin {bmatrix} x - 1 & 2 & 1  \\ y + 1 & 1 & 1 \\ z & 1 & 1\\ \end {bmatrix} = 0

-y + z - 1 = 0

s = \begin{cases} x + y + z - 4 = 0 \\ -y + z - 1 = 0 \end{cases}

 

Solution of exercise 4

Calculate the distance between the following lines:

Calculate the distance between the following lines:

r = \frac {x - 2}{3} = \frac {y - 2}{-1} = \frac {z + 1}{4}

s = \begin{cases} x = 5 + \lambda \\ y = -1\\ z = 8 + 2 \lambda \end{cases}

A = (2, 2, -1)             \overrightarrow {u} = (3, -1, 4)

B = (5, -1, 8)             \overrightarrow {V} = (1, 0, 2)

\overrightarrow {AB} = (5 - 2, 2 + 1, 8 + 1) = (3, -3, 9)

|[\overrightarrow {AB}, \overrightarrow {u}, \overrightarrow{v}]| = \begin {vmatrix} 3 & -1 & 4  \\ 1 & 0 & 2 \\ 3 & -3 & 9 \\ \end {vmatrix} = 9

\overrightarrow {u} \times \overrightarrow {v} = \sqrt {(-2)^2 + (-2)^2 + 1^2} = 3

d (r, s) = \frac {9}{3} = 3

 

Solution of exercise 5

Find the symmetric point of Point A = (3, 2, 1) to the plane x + y + z + 21 = 0.

Solution of Exercise 5

First, compute r, which is the line that passes through Point A and is perpendicular to π.

r = \frac {x - 3}{1} = \frca {y - 2}{1} = \frac {z - 1}{1}

Then, find the point of intersection of the line r and the plane π.

\begin{cases} x - y - 1 = 0 \\ x - z - 2 = 0 \\ x + y + z + 21 = 0 \end{cases}

M = (-6, -7, -8)

Given the coordinates of the midpoint of the line segment, the endpoint A' can be found.

-6 = \frac {3 + x}{2}

-7 = \frac {2 + y}{2}

-8 = \frac {1 + z}{2}

A' = (-15, -16, -17)

 

Solution of exercise 6

Calculate the area of the triangle whose vertices are the points of intersection of the plane 6x + 3y + 2z = 6  with the coordinate axes.

y = 0, z = 0 \rightarrow x = 1                  A = (1, 0, 0)

x = 0, y = 0, z = 0 \rightarrow y = 2                  B = (0, 2, 0)

x = 0, y = 0 \rightarrow z = 3                  C = (0, 0, 3)

\overrightarrow {AB} = (0 - 1, 2 - 0, 0 - 0) = (-1, 2, 0)

\overrightarrow {AC} = (0 - 1, 0 - 0, 3 - 0) = (-1, 0, 3)

\overrightarrow {AB} \times \overrightarrow {AC} = \begin {vmatrix} \overrightarrow {i} & \overrightarrow{j} & \overrightarrow{k}  \\ -1 & 2 & 0\\ -1 & 0 & 3 \\ \end {bmatrix} = 6\overrightarrow {i} + 3 \overrightarrow {j} + 2 \overrightarrow {k}

A _r = \frac {1}{2} sqrt{6^2 + 3^2 + 2^2} = \frac {7}{2} u^2

 

Solution of exercise 7

Given the plane \pi = x + 2y + 3z = 1 and the point A = (1, 1, 1), calculate the coordinates of the base (endpoint) of the perpendicular from A to the plane.

Solution of Exercise 7

A = (1, 1, 1)           \overrightarrow {u_r} = \overrightarrow{n} = (1, 2, 3)

 

r = \begin{cases} 1 + \lambda \\ 1 + 2 \lambda \\ 1 + 3 \lambda \end{cases}

The foot of the perpendicular is the point of intersection between the plane and the line.

(1 + \lambda) + 2 (1 + 2\lambda) + 3(1 + 3\lambda) = 1

\lambda = - \frac {5}{14}

A' (\frac {9}{14}, \frac {4}{14}, -\frac {1}{14})

 

Solution of exercise 8

Determine the equation of the plane π that is \sqrt{6}  distant from the origin and is parallel to the plane 2x - y - z = 3.

2x + y - z + k = 0

d(O, \pi) = \frac {|0 + 0 - 0 + k|} {\sqrt {2^2 + 1^2 + 1^2}} = \frac {|k|}{\sqrt{6}} = \sqrt{6}

k = 6           k = -6

\pi_1 = 2x + y - z + 6 = 0

\pi_2 = 2x + y - z - 6 = 0

 

Solution of exercise 9

Find the distance between the point A = (3, 2, 7) and the line of the first octant (+,+,+).

\begin{cases} x = \lambda \\ y = \lambda \\ z = \lambda \end{cases}

B = (0, 0, 0)

\overrightarrow {u_r} = (1, 1, 1)

\overrightarrow {AB} = (0 - 3, 0 - 2, 0 - 7) = (-3, -2, -7)

\overrightarrow{AB} \times \overrightarrow{u_r} = \begin {bmatrix}  \overrightarrow {i}& \overrightarrow{j} & \overrightarrow{k} \\ -3 & - 2 & -7 \\ 1 & 1 & 1 \\ \end {bmatrix} = 5\overrightarrow {i} - 4 \overrightarrow {j} -  \overrightarrow {k}

d(A, r) = \frac {\sqrt{5^2 + (-4)^2 + (-1)^2}} {\sqrt{1^2} + 1^2 + 1^2}} = \frac {\sqrt{42}} {\sqrt{3}} = \sqrt{14}

 

Solution of exercise 10

Calculate the area of the square whose sides are on the lines:

r = \frac {x - 1}{1} = \frac {y}{2} = \frac {z - 2}{1}

s = \begin{cases} x - y + z = 2 \\3x - y - z = -4\end{cases}

Line r.

A(1, 0, 2)                   \overrightarrow{u} = (1, 2, 1)

Line s.

\overrightarrow{v} = = \begin {bmatrix}  \overrightarrow {i}& \overrightarrow{j} & \overrightarrow{k} \\ 1 & - 1 & 1 \\ 3 & -1 & - 1 \\ \end {bmatrix} = 2\overrightarrow {i} + 4 \overrightarrow {j} + 2  \overrightarrow {k}

s = \begin{cases} x - y + z = 2 \\3x - y - z = -4\end{cases}

-2x + 2z = 6

x = 0, z = 3, y = 1

B = (0, 1, 3)

\overrightarrow{v} = (2, 4, 2)

\overrightarrow{AB} = (0 - 1, 2 - 0, 3 - 2) = (-1, 1, 1)

The distance of r to s is equal to the distance of the point B to the line r.

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

d(B, r) = \frac {\sqrt{(-1)^2 + 2^2 + 3^2}} {\sqrt{1^2 + 2^2 + 1^2}} = \frac {\sqrt{14}}{\sqrt{6}} = \frac {\sqrt{7}}{\sqrt{3}}

The side of the square is equal to the distance between the lines r and s.

A_o = \frac {7}{3} u^2

 

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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.