Exercise 1

Determine the equations of the coordinate axes and the coordinate planes.

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Exercise 2

Determine the equation of the plane that contains the lines:

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

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

Exercise 3

Determine the equation of the plane that contains the point A = (2, 5, 1) and the line:

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

Exercise 4

Find the intersecting point between the plane x + 2y − z − 2 = 0, the line determined by the point (1, −3, 2) and the vector \overrightarrow {u} = (2, 4, 1) .

Exercise 5

Determine, in intercept form, the equation of the plane that passes through the points A = (2, 0, 0), B = (0, 4, 0) and C = (0, 0, 7).

Exercise 6

π is a plane that passes through P = (1, 2, 1) and intersects the positive coordinate semi-axes at points A, B and C. If ABC is an equilateral triangle, determine the equations of π.

Exercise 7

Find the equation of the plane that passes through the point P = (1, 1, 1) and is parallel to:

\left\{\begin{matrix} x = 1 + 2\lambda - 3 \mu \\ y = 3 + 2\mu\\ z = -1 - \mu \end{matrix}\right

 

Exercise 8

Determine the equation of the plane that contains the line \frac {x - 2}{1} = \frac{y - 2}{-2} = \frac {z - 4}{3} and is parallel to the line  \left\{\begin{matrix} x = 1 + 3\lambda \\y = 1 + 2 \lambda\\ z = \lambda \end{matrix}\right.

Exercise 9

Calculate the equation of the plane that passes through the point (1, 1, 2) and is parallel to the following lines:

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

s = \left\{\begin{matrix} 2x - y + z = -2 \\ -x + y + 3z = 1 \end{matrix}\right

 

Solution of exercise 1

Determine the equations of the coordinate axes and the coordinate planes.

x - axis           O = (0, 0, 0)             \overrightarrow {i} = (1, 0, 0)

\frac {x}{1} = \frac {y}{0}  = \frac{z}{0}

\left\{\begin{matrix} y = 0 \\ z = 0 \end{matrix}\right

y - axis           O = (0, 0, 0)            \overrightarrow {j} (0, 1, 0)

\frac {x}{0} = \frac {y}{1} = \frac {z}{0}

\left\{\begin{matrix} x = 0 \\ z = 0 \end{matrix}\right

z - axis          O (0, 0, 0)             \overrightarrow{k} = (0, 0, 1)

\frac{x}{0} = \frac{y}{0} = \frac{z}{1}

\left\{\begin{matrix} x = 0 \\ y = 0 \end{matrix}\right

XOY   O = (0, 0, 0)       \overrightarrow {i} = (1, 0, 0)      \overrightarrow {j} = (0, 1, 0)

A = \begin {bmatrix} x & 1 & 0  \\ y & 0 & 1 \\ z & 0 & 0 \\ \end {bmatrix}

z = 0

XOZ     0 = (0, 0, 0)     \overrightarrow {i} = (1, 0, 0)            \overrightarrow {k} = (0, 0, 1)

A = \begin {vmatrix} x & 1 & 0  \\ y & 0 & 0 \\ z & 0 & 1 \\ \end {vmatrix}

y = 0

YOZ       O = (0, 0, 0)        \overrightarrow {j} = (0, 1, 0)            \overrightarrow {k} = (0, 0, 1)

A = \begin {vmatrix} x & 0 & 0  \\ y & 1 & 0 \\ z & 0 & 1 \\ \end {vmatrix}

x = 0

 

Solution of exercise 2

Determine the equation of the plane that contains the lines:

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

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

A = (-2, 1, -3)

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

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

A = \begin {vmatrix} x + 2 & 1 & -1  \\ y - 1 & 3 & 4 \\ z + 3 & -1 & -2 \\ \end {vmatrix} = 0

-2x + 3y + 7z + 14 = 0

 

Solution of exercise 3

Determine the equation of the plane that contains the point A = (2, 5, 1) and the line:

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

B = (2, 1, -1)

\overrightarrow {AB} = (2 - 2, 1 - 5, -1 -1) = (0, -4, -2)

A = (2, 5, 1)

\overrightarrow {u} = (1, 3, 1)

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

\begin {vmatrix} x - 2 & 1 & 0  \\ y - 5 & 3 & -4\\ z - 1 & 1 & -2 \\ \end {vmatrix} = 0

x - y + 2z + 1 =0

 

Solution of exercise 4

Find the intersecting point between the plane x + 2y − z − 2 = 0, the line determined by the point (1, −3, 2) and the vector \overrightarrow {u} = (2, 4, 1) .

\left\{\begin{matrix} x = 1 + 2\lambda \\ y = -3 + 4 \lambda \\ z = 2 + \lambda \end{matrix}\right

(1 + 2\lambda) + 2(-3 + 4 \lambda) - (2 + \lambda) - 2 = 0

\lambda = 1

B = (3, 1, 3)

 

Solution of exercise 5

Determine, in intercept form, the equation of the plane that passes through the points A = (2, 0, 0), B = (0, 4, 0) and C = (0, 0, 7).

\frac {x}{2} + \frac {y}{4} + \frac {z}{7} = 1

 

Solution of exercise 6

π is a plane that passes through P = (1, 2, 1) and intersects the positive coordinate semi-axes at points A, B and C. If ABC is an equilateral triangle, determine the equations of π.

A = (a, 0, 0)          B = (0, b, 0)              C = (0, 0, c)

\frac {x}{a} + \frac {y}{a} + \frac {z}{c} = 1

a = b = c

As the triangle is equilateral, the three line segments are equal.

\frac {x}{a} + \frac {y}{a} + \frac {z}{a} = 1

x + y + z = a

1 + 2 + 1 = a

a = 4

x + y + z = 4

 

Solution of exercise 7

Find the equation of the plane that passes through the point P = (1, 1, 1) and is parallel to:

\left\{\begin{matrix} x = 1 + 2 \lambda - 3 \mu \\ y = 3 + 2\mu\\ z = -1 - \mu \end{matrix}\right

P = (1, 1, 1)

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

\overightarrow {v} = (-3, 2, -1)

\begin {vmatrix} x -  & 2 & -3  \\ y - 1 & 0 & 2\\ z - 1 & 0 & -1 \\ \end {vmatrix} = 0

-2 \begin {vmatrix} y - 1 & 2  \\ z - 1 & -1\\ \end {vmatrix} = 0

-2 (- y + 1 - 2z + 2) =0

y + 2z - 3 = 0

 

Solution of exercise 8

Determine the equation of the plane that contains the line \frac {x - 2}{1} = \frac{y - 2}{-2} = \frac {z - 4}{3} and is parallel to the line  \left\{\begin{matrix} x = 1 + 3\lambda \\y = 1 + 2 \lambda\\ z = \lambda \end{matrix}\right.

 

The point A = (2, 2, 4) and the vector \overrightarrow {u} = (1, -2, 3)  belong to the plane because the line is in the plane.

The vector \overrightarrow {v} = (3, 2, 1)  is a vector in the plane because it is parallel to the line.

A = (2, 2, 4)

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

\overrightarrow {v} = (3, 2, 1)

2 \begin {vmatrix} x - 2 & 1 & 3  \\ y - 2 & -2 & 2\\ z - 4 & 3 & 1 \end {vmatrix} = 0

x - y - z + 4 = 0

 

Solution of exercise 9

Calculate the equation of the plane that passes through the point (1, 1, 2) and is parallel to the following lines:

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

s = \left\{\begin{matrix} 2x - y + z = -2 \\ -x + y + 3z = 1 \end{matrix}\right

\left\{\begin{matrix} 2x - y + z = -2 \\ -x + y + 3z = 1 \end{matrix}\right

 

\left\{\begin{matrix} 2x - y = -2 - z\\ -x + y = 1 - 3z \end{matrix}\right

x = \frac { \begin {vmatrix}-2 - z& -1  \\ 1 - 3z & 1\\ \end {vmatrix}} { \begin {vmatrix}2& -1  \\ 1 & 1\\ \end {vmatrix}}} = 1 - 4z

y = \frac { \begin {vmatrix}2& -2 - z  \\ -1 & 1 - 3z\\ \end {vmatrix}} { \begin {vmatrix}2 & -1 \\ 1 & 1\\ \end {vmatrix}}} = =-7z

s = \left\{\begin{matrix} x = 1 - 4\lambda \\ y = - 7\lambda\\ z = \lambda \end{matrix}\right

A = (1, 1, 2)

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

\overrightarrow {v} = (-4, -7, 1)

 

\begin {vmatrix} x - 1 & -1 & - 4 \\ y - 1 & 1 & -7\\ z - 2 & 2 & 1 \end {vmatrix} = 0

15x - 7y + 11z - 30 = 0

 

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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.