Precalculus

Analytic Geometry

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Cartesian Coordinates in Space

Position of a Point in Space

Three mutually perpendicular axes in space with a common origin O and the same scale unit form a Cartesian rectangular coordinate system in space. One of the indicated axes is called the x-axis, the other is the y-axis, and the third is the z-axis. The coordinates of any point in space are determined by three real numbers: x, y, z. They are also known as the abscissa, ordinate and applicate, respectively.

Three-dimensional coordinate system
Figure 1.

The fact that point M has coordinates x, y and z is denoted as \({M\left({x,y,z}\right)}.\) The origin O has coordinates \({\left({0,0,0}\right).}\)

Octants

The xy-, yz- and xz-planes divide the space into eight octants. They are numbered as shown in the figure.

Eight octants in the coordinate space
Figure 2.

Distance Between Two Points in Space

The distance between two points \(M_1\left({x_1,y_1,z_1}\right)\) and \(M_2\left({x_2,y_2,z_2}\right)\) in space is determined by the formula

\[d\left({M_1,M_2}\right) = \sqrt{\left({x_2 - x_1}\right)^2 + \left({y_2 - y_1}\right)^2 + \left({z_2 - z_1}\right)^2 }.\]
Distance between two points in space
Figure 3.

Area of a Triangle

The area of a triangle with vertices \(A\left({x_1,y_1,z_1}\right),\) \(B\left({x_2,y_2,z_2}\right)\) and \(C\left({x_3,y_3,z_3}\right)\) is found by the formula

\[A = \frac{1}{2}\sqrt{ \left| {\begin{array}{*{20}{c}} {y_1}&{z_1}&1\\ {y_2}&{z_2}&1\\ {y_3}&{z_3}&1 \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} {z_1}&{x_1}&1\\ {z_2}&{x_2}&1\\ {z_3}&{x_3}&1 \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|^2}\]
Area of a triangle with 3 vertices in space
Figure 4.

Volume of a Pyramid

The volume of a pyramid whose vertices have the coordinates \(A\left({x_1,y_1,z_1}\right),\) \(B\left({x_2,y_2,z_2}\right),\) \(C\left({x_3,y_3,z_3}\right)\) and \(D\left({x_4,y_4,z_4}\right)\) is determined by the expression

\[V = \left({\pm}\right)\frac{1}{6}\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&{z_1}&1\\ {x_2}&{y_2}&{z_2}&1\\ {x_3}&{y_3}&{z_3}&1\\ {x_4}&{y_4}&{z_4}&1 \end{array}} \right|\]

or

\[V = \left({\pm}\right)\frac{1}{6}\left| {\begin{array}{*{20}{c}} {x_1}-{x_4}&{y_1}-{y_4}&{z_1}-{z_4}\\ {x_2}-{x_4}&{y_2}-{y_4}&{z_2}-{z_4}\\ {x_3}-{x_4}&{y_3}-{y_4}&{z_3}-{z_4} \end{array}} \right|\]

We pick the sign on the right side of the formulas such that the volume is positive.

Volume of a pyramid
Figure 5.

Solved Problems

Example 1.

Find the images the points \(A\left({1,2,3}\right),\) \(B\left({0,1,1}\right),\) and \(C\left({a,b,c}\right)\) symmetrical with respect to

  1. \(xy-\)plane
  2. \(yz-\)plane
  3. \(xz-\)plane
  4. \(x-\)axis
  5. \(y-\)axis
  6. \(z-\)axis
  7. origin

Solution.

  1. \(xy-\)plane:
    \[A\left({1,2,3}\right) \mapsto A_1\left({1,2,-3}\right),\; B\left({0,1,1}\right) \mapsto B_1\left({0,1,-1}\right),\;C\left({a,b,c}\right) \mapsto C_1\left({a,b,-c}\right);\]
  2. \(yz-\)plane:
    \[A\left({1,2,3}\right) \mapsto A_2\left({-1,2,3}\right),\; B\left({0,1,1}\right) \mapsto B_2\left({0,1,1}\right),\;C\left({a,b,c}\right) \mapsto C_2\left({-a,b,c}\right);\]
  3. \(xz-\)plane:
    \[A\left({1,2,3}\right) \mapsto A_3\left({1,-2,3}\right),\; B\left({0,1,1}\right) \mapsto B_3\left({0,-1,1}\right),\;C\left({a,b,c}\right) \mapsto C_3\left({a,-b,c}\right);\]
  4. \(x-\)axis:
    \[A\left({1,2,3}\right) \mapsto A_4\left({1,-2,-3}\right),\; B\left({0,1,1}\right) \mapsto B_4\left({0,-1,-1}\right),\;C\left({a,b,c}\right) \mapsto C_4\left({a,-b,-c}\right);\]
  5. \(y-\)axis:
    \[A\left({1,2,3}\right) \mapsto A_5\left({-1,2,-3}\right),\; B\left({0,1,1}\right) \mapsto B_5\left({0,1,-1}\right),\;C\left({a,b,c}\right) \mapsto C_5\left({-a,b,-c}\right);\]
  6. \(z-\)axis:
    \[A\left({1,2,3}\right) \mapsto A_6\left({-1,-2,3}\right),\; B\left({0,1,1}\right) \mapsto B_6\left({0,-1,1}\right),\;C\left({a,b,c}\right) \mapsto C_6\left({-a,-b,c}\right);\]
  7. origin:
    \[A\left({1,2,3}\right) \mapsto A_7\left({-1,-2,-3}\right),\; B\left({0,1,1}\right) \mapsto B_7\left({0,-1,-1}\right),\;C\left({a,b,c}\right) \mapsto C_7\left({-a,-b,-c}\right).\]

Example 2.

Name the octants in which the points can be located if their coordinates satisfy the equation

  1. \(x + y = 0\)
  2. \(x - z = 0\)
  3. \(xy \gt 0\)
  4. \(xyz \gt 0\)
  5. \(xyz \lt 0\)

Solution.

  1. If \(x + y = 0,\) then \(x\) and \(y\) must have different signs. It is possible in octants \(II,\) \(IV,\) \(VI,\) and \(VIII.\) See figure above.
  2. If \(x - z = 0,\) then \(x\) and \(z\) have the same sign. It is possible in octants \(I,\) \(IV,\) \(VI,\) and \(VII.\)
  3. If \(xy \gt 0,\) then \(x\) and \(y\) have the same sign. It is possible in octants \(I,\) \(III,\) \(V,\) and \(VII.\)
  4. If \(xyz \gt 0,\) then either all three coordinates are positive or two of them are negative. It is possible in octants \(I,\) \(III,\) \(VI,\) and \(VIII.\)
  5. The case \(xyz \lt 0\) is complementary to the previous one. Hence, the condition \(xyz \lt 0\) is satisfied in octants \(II,\) \(IV,\) \(V,\) and \(VII.\)

Example 3.

Find the coordinates of the point M on the \(x-\)axis which is equidistant from points \(A\left({1,1,2}\right)\) and \(B\left({-2,4,1}\right).\)

Solution.

Let \(M = M\left({x,0,0}\right).\) Then by the formula for the distance between points, we can write:

\[ d\left({M,A}\right) = d\left({M,B}\right),\]
\[\Rightarrow \sqrt{\left({1 - x}\right)^2 + \left({1 - 0}\right)^2 + \left({2 - 0}\right)^2} = \sqrt{\left({-2 - x}\right)^2 + \left({4 - 0}\right)^2 + \left({1 - 0}\right)^2},\]
\[\Rightarrow \sqrt{\left({1 - x}\right)^2 + 1 + 4} = \sqrt{\left({-2 - x}\right)^2 + 16 + 1},\]
\[\Rightarrow \left({1 - x}\right)^2 + 5 = \left({-2 - x}\right)^2 + 17,\]
\[\Rightarrow 1 - 2x + \cancel{x^2} + 5 = 4 + 4x + \cancel{x^2} +17, \Rightarrow 6x = -15, \Rightarrow x = -\frac{5}{2},\]

that is, the point M has coordinates \(\left({-\frac{5}{2},0,0}\right).\)

Example 4.

Prove that the triangle with vertices \(A\left({3,2,3}\right),\) \(B\left({6,-4,2}\right),\) \(C\left({3,-5,-1}\right)\) is a right triangle.

Solution.

This problem is easiest to solve using the scalar product, however we will use the formula for the distance between points. Find the lengths of the sides of the triangle:

\[AB = \sqrt{\left({6 - 3}\right)^2 + \left({-4 - 2}\right)^2 + \left({2 - 3}\right)^2} = \sqrt{3^2 + \left({-6}\right)^2 + \left({-1}\right)^2} = \sqrt{9 + 36 + 1} = \sqrt{46};\]
\[AC = \sqrt{\left({3 - 3}\right)^2 + \left({-5 - 2}\right)^2 + \left({-1 - 3}\right)^2} = \sqrt{0^2 + \left({-7}\right)^2 + \left({-4}\right)^2} = \sqrt{49 + 16} = \sqrt{65};\]
\[BC = \sqrt{\left({3 - 6}\right)^2 + \left({-5 + 4}\right)^2 + \left({-1 - 2}\right)^2} = \sqrt{\left({-3}\right)^2 + \left({-1}\right)^2 + \left({-3}\right)^2} = \sqrt{9 + 1 + 9} = \sqrt{19}.\]

Check that the Pythagorean theorem is fulfilled:

\[{AB}^2 + {BC}^2 = {AC}^2, \Rightarrow 46 + 19 = 65, \Rightarrow 65 = 65.\; \blacksquare\]

Example 5.

Find the area of a space triangle whose vertices are \(A\left({2,1,0}\right),\) \(B\left({1,0,1}\right)\) and \(C\left({2,-1,1}\right).\)

Solution.

We calculate the area of the triangle using the formula

\[A = \frac{1}{2}\sqrt{ \left| {\begin{array}{*{20}{c}} {y_1}&{z_1}&1\\ {y_2}&{z_2}&1\\ {y_3}&{z_3}&1 \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} {z_1}&{x_1}&1\\ {z_2}&{x_2}&1\\ {z_3}&{x_3}&1 \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|^2}.\]

Substituting the coordinates of the vertices we get

\[A = \frac{1}{2}\sqrt{ \left| {\begin{array}{*{20}{c}} 1&0&1\\ 0&1&1\\ -1&1&1 \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} 0&2&1\\ 1&1&1\\ 1&2&1 \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} 2&1&1\\ 1&0&1\\ 2&-1&1 \end{array}} \right|^2}.\]

Calculate each determinant separately:

\[\Delta_1 = \left| {\begin{array}{*{20}{c}} 1&0&1\\ 0&1&1\\ -1&1&1 \end{array}} \right| = 1\cdot \left| {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right| + 1 \cdot \left| {\begin{array}{*{20}{c}} 0&1\\ -1&1 \end{array}} \right| = 0+1 = 1;\]
\[\Delta_2 = \left| {\begin{array}{*{20}{c}} 0&2&1\\ 1&1&1\\ 1&2&1 \end{array}} \right| = -2\cdot \left| {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right| + 1\cdot \left| {\begin{array}{*{20}{c}} 1&1\\ 1&2 \end{array}} \right|= -2\cdot0 +1\cdot1 = 1;\]
\[\Delta_3 = \left| {\begin{array}{*{20}{c}} 2&1&1\\ 1&0&1\\ 2&-1&1 \end{array}} \right| = -1\cdot \left| {\begin{array}{*{20}{c}} 1&1\\ -1&1 \end{array}} \right| - 1\cdot \left| {\begin{array}{*{20}{c}} 2&1\\ 2&-1 \end{array}} \right|= -1\cdot2 -1\cdot\left({-4}\right) = 2.\]

From here we find the area of the space triangle:

\[A = \frac{1}{2}\sqrt{1^2 + 1^2 + 2^2} = \frac{\sqrt{6}}{2}.\]

Example 6.

Find the volume of a pyramid whose vertices are \(A\left({0,0,0}\right),\) \(B\left({0,2,1}\right),\) \(C\left({2,0,0}\right)\) and \(D\left({1,0,2}\right).\)

Solution.

To find the volume of the pyramid, we calculate the \(4\text{th}\) order determinant:

\[V = \left({\pm}\right)\frac{1}{6}\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&{z_1}&1\\ {x_2}&{y_2}&{z_2}&1\\ {x_3}&{y_3}&{z_3}&1\\ {x_4}&{y_4}&{z_4}&1 \end{array}} \right|.\]

Substitute the coordinates of the vertices and expand the determinant by the first row:

\[V = \left({\pm}\right)\frac{1}{6}\left| {\begin{array}{*{20}{c}} {0}&{0}&{0}&1\\ {0}&{2}&{1}&1\\ {2}&{0}&{0}&1\\ {1}&{0}&{2}&1 \end{array}} \right| = \left({\pm}\right)\frac{1}{6}\cdot\left({-1}\right)\cdot\left| {\begin{array}{*{20}{c}} {0}&{2}&{1}\\ {2}&{0}&{0}\\ {1}&{0}&{2} \end{array}} \right| = \left({\pm}\right)\frac{1}{6}\left| {\begin{array}{*{20}{c}} {0}&{2}&{1}\\ {2}&{0}&{0}\\ {1}&{0}&{2} \end{array}} \right|.\]

Now expand the \(3\text{rd}\) order determinant by the second row:

\[V = \left({\pm}\right)\frac{1}{6}\cdot\left({-2}\right)\cdot \left( {2\cdot 2 - 0\cdot 1} \right) = \left({\pm}\right)\frac{1}{3}\cdot4 = \left({\pm}\right)\frac{4}{3}.\]

We will choose the plus sign. Hence, \(V = \frac{4}{3}.\)