4.4 Lines and Planes in Space

Lines in Three-Dimensional Space

A direction vector of a straight line is a vector parallel to the line.

In three dimensions geometry we can determine the equation of a line using its direction and any fixed point on the line.

Suppose a line passes through a fixed point A with $$\overrightarrow{OA} = \vec{a}$$ $$\small{\text{and that the line is parallel to a vector}} \,\,\vec{b} .$$ Let R be any point on the line with $$\overrightarrow{OR} = \vec{r}\, .$$

fig 4.4-1

Then
$$\overrightarrow{OR} = \overrightarrow{OA} + \overrightarrow{AR} \, .$$ $$\text{Since} \,\, \overrightarrow{AR} \parallel \vec{b}\, ,$$ $$\overrightarrow{AR} = t \, \vec{b} \,\,\, \text{for some} \,\, t \in \mathbb{R}\, ,$$ $$\vec{r} = \vec{a} + t \, \vec{b} \, .$$ $$\text{So} \,\,\, \vec{r} = \vec{a} + t\vec{b}, \quad t \in \mathbb{R}$$ is the vector equation of the line.

In three dimensions,
$$\small{\text{Let} \quad \overrightarrow{OR} = \vec{r} = \begin{pmatrix}x\\y\\z\end{pmatrix} \, ,}$$ $$\small{ \vec{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}\, .}$$ $$ \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} + t \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$$ is the vector equation of the line where R(x, y, z) is any point on the line, B(a1, a2, a3) is the known (fixed) point on the line and
$$\vec{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$$ is the direction vector of the line.

$$\text{Since} \,\, \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}a_1 + tb_1\\ a_2 + tb_2\\ a_3 + tb_3\end{pmatrix} \, ,$$ we can write the parametric equation of the line as
x   =   a   +   tb 1 1 y   =   a   +   tb 2 2 where t ∈   ℝ z   =   a   +   tb 3 3 is the parameter.
Each point on the line corresponds to exactly one value of t.

By equating t values, we obtain the Cartesian equation of the line
x - a1 ⁄b1 = y - a2 ⁄b2 = z - a3 ⁄b3

Example 16.
Find the Cartesian equation of the line with vector equation
$$\small{\vec{r} = \begin{pmatrix}1\\4\\-1\end{pmatrix} + t \begin{pmatrix}3\\2\\5\end{pmatrix}\, .}$$ Solution
The vector equation is
$$\small{\vec{r} = \vec{a} + t \vec{b}\, ,\, \, \text{therefore} }$$ $$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\4\\-1\end{pmatrix} + t \begin{pmatrix}3\\2\\5\end{pmatrix}\, , \,\, t \isin \mathbb{R}\,. }$$ The parametric equations are:
x = 1 + 3t,         t = x - 1 ⁄3 ,
y = 4 + 2t,         t = y - 4 ⁄2 ,
z = -1 + 5t,         t = z + 1 ⁄5 .
The Cartesian equation of the line is
x - 1 ⁄3 = x - 4 ⁄2 = x + 1 ⁄5 .

Example 17.
Does the point A(3, -2, 2) lie on the line with equation
x + 1 ⁄2 = 4 - y ⁄3 = 2z ⁄3 ?
Solution
Consider the given Cartesian equation passing through the point A(3, -2, 2).
x + 1 ⁄2 = 3 + 1 ⁄2 = 2
4 - y ⁄3 = 4 + 2 ⁄3 = 2
2x ⁄3 = 2 × 2 ⁄3 = 4 ⁄3
and so
2 = 2 ≠ 4 ⁄3 .
The coordinates do not satisfy the Cartesian equation. Therefore the point does not lie on the line.

Planes in Three Dimensions

To determine the vector equation of the plane, we requires extension of the ideas of the equation of line. Think of a very simple example, the xy-plane. The position vector of any point in the xy-plane is a sum of scalar multiples of î and ĵ, so î and ĵ can be considered direction vectors for the xy-plane.

More generally, for any plane through the origin, if we fix two nonparallel vectors in that plane, the position vector of any point in the plane is a sum of scalar multiples of those two vectors.
fig 4.4-2


If plane does not pass through the origin, we have to make a further turn: we need to shift the plane and its vectors by adding the position vector of some point on the plane.

Consider A, B and C be three noncolinear points on the plane with
$$\small{\overrightarrow{OA} = \vec{a} \, . \,\,\, \text{Let} \,\, \vec{d_1}\,\, \text{and} \,\, \vec{d_2}}$$ be the two nonparallel vectors on the plane.
fig 4.4-3


If the plane passing through A and having two nonparallel vectors
$$\small{\vec{d_1} \,\, \text{and} \,\, \vec{d_2}, \,\, \text{the position vector} \,\, \vec{r} } $$ of any point R on the plane is given by following equation.

$$\vec{r} = \vec{a} + t_1 \vec{d}_1 + t_2 \vec{d_2} $$

Any vector that is perpendicular to a plane is called a normal vector or simply normal to the plane. We can find the normal to a plane by finding the cross product of the two nonparallel vectors of the plane. Normal is perpendicular to every line on the plane.

To find the different vector form of the equation of a plane,
$$ \small{\text{we can use the position vector} \,\, \vec{a} }$$ $$\small{ \text{of one point and the normal vector} \,\, \vec{n} }$$ is perpendicular to the plane.

fig 4.4-4

Thus we consider a plane passing through a point A with position vector
$$\small{\vec{a} \,\, \text{and} \,\, \vec{n} = a\hat{i} + b\hat{j} + c\hat{k} }$$ is perpendicular to the given plane.
$$\small{ \text{Let} \,\, \vec{r} \,\, \text{be the position vector of an arbitrary point} }$$ R(x, y, z) on the plane. Since AR is a line in the plane, it follows that AR is at right angle
$$\small{ \text{ to the normal vector}\,\, \vec{n}\,.}$$ $$\small{\vec{n} \cdot \overrightarrow{AR} = 0.}$$ $$\small{ \text{The vector} \,\, \overrightarrow{AR} \,\, \text{is given by} \,\, \overrightarrow{AR} = \vec{r} - \vec{a} \,\, \text{and so} }$$ $$\small{\left(\vec{r} - \vec{a}\right) \cdot \vec{n} = 0 \,.}$$ This is the vector equation of the plane. We can write this in an alternative form as
$$\small{ \vec{r} \cdot \vec{n} - \vec{a} \cdot \vec{n} = 0 }$$ $$\small{ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} }$$ $$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}a\\b\\c\end{pmatrix} = \vec{a} \cdot \vec{n} }$$
ax + by + cz = d     where
$$d = \vec{a} \cdot \vec{n}. $$
Note that d is a constant.


The Cartesian equation of a plane has the form
ax + by + cz = d where $$\small{ \begin{pmatrix}a\\b\\c\end{pmatrix} \,\, \text{is the normal vector of the plane}. }$$


Example 18.
Find the vector equation of the plane containing points M(2, 2, -2), N(1, -1, 3) and P(4, 0, 2).
Solution
$$\small{ \text{Let} \,\, \vec{a} = \overrightarrow{OM} = \begin{pmatrix}2\\2\\-2\end{pmatrix}\, .} $$ We can find that
$$\small{ \vec{d_1} = \overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} }$$ $$\small{ = \begin{pmatrix}1\\-1\\3\end{pmatrix} - \begin{pmatrix}2\\2\\-2\end{pmatrix} = \begin{pmatrix}-1\\-1\\5\end{pmatrix} ,} $$ $$\small{ \vec{d_2} = \overrightarrow{MP} = \overrightarrow{OP} - \overrightarrow{OM} }$$ $$\small{ = \begin{pmatrix}4\\0\\2\end{pmatrix} - \begin{pmatrix}2\\2\\-2\end{pmatrix} = \begin{pmatrix}2\\-2\\4\end{pmatrix} .} $$ Therefore the vector equation of the plane is $$\small{\vec{r} = \vec{a} + t_1\vec{d_1} + t_2\vec{d_2} }$$ $$\small{ = \begin{pmatrix}2\\2\\-2\end{pmatrix} + t_1 \begin{pmatrix}-1\\-3\\5\end{pmatrix} + t_2 \begin{pmatrix}2\\-2\\4\end{pmatrix} .} $$
fig 4.4-5



Example 19.
Find the Cartesian equation of the plane containing points M(2, 2, -2), N(1, -1, 3) and P(4, 0, 2).
Solution
The equation of the plane containing M, N and P is
$$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\2\\-2\end{pmatrix} + t_1 \begin{pmatrix}-1\\-3\\5\end{pmatrix} + t_2 \begin{pmatrix}2\\-2\\4\end{pmatrix} . \,\, \text{(See Example 18)} }$$ x = 2 - t1 + 2t2           ______(1)
y = 2 - 3t1 - 2t2           ______(2)
z = -2 + 5t1 + 4t2           ______(3)
Add equation (1) and (2),
x + y = 4 - 4t1           ______(4)
Multiply equation (2) by 2 and then add (3),
2y + z = 2 - t1           ______(5)
From equation (4) and (5), we obtain
x + y - 4 ⁄-4 = 2y + z - 2 ⁄-1
-x - y + 4 = -8y - 4z + 8
x - 7y - 4z = -4 .

Example 20.
Determine whether points A(3, -1, 4) , B(2, 1, 1), C(4, 3, 1) and D(-3, 1, 4) lie in the same plane.
Solution
The equation of the plane containing A, B and C is
$$\small{ \vec{r} = \overrightarrow{OA} + t_1 \overrightarrow{AB} + t_2 \overrightarrow{AC} }$$ $$\small{ = \begin{pmatrix}3\\-1\\4\end{pmatrix} + t_1 \begin{pmatrix}-1\\2\\-3\end{pmatrix} + t_2 \begin{pmatrix}1\\4\\-3\end{pmatrix} . }$$ If D is on the plane, then
$$\small{\vec{r} = \overrightarrow{OD} ,}$$ 3 - t1 + t2 = -3           ______(1)
-1 + 2t1 + 4t2 = 1           ______(2)
4 - 3t1 - 3t2 = 4           ______(3)
Multiply equation (1) by 2, we get
-2t1 + 2t2 = -12 .           ______(4)
From (2) and (4), we obtain
t1 = 13 ⁄3 ,   t2 = - 5 ⁄3 .
Substitute the values of t1 and t2 in left hand side of equation (3), we get
4 - 3( 13 ⁄3) - 3 ( - 5 ⁄3) = -4 ≠ 4 .
So, D does not lie on the same plane as A, B and C .

Example 21.
Find a vector equation of the plane containing the line $$\small{ \vec{r} = \begin{pmatrix}-2\\ 1\\ 2\end{pmatrix} + t \begin{pmatrix}-1\\1\\1\end{pmatrix} }$$ and point A(3, -1, 2).
Solution
$$\small{ \text{Let} \,\, \overrightarrow{OA} = \begin{pmatrix}3\\-1\\2\end{pmatrix} .}$$ $$\small{ \text{The vector} \,\, \vec{d_1} = \begin{pmatrix}-1\\1\\1\end{pmatrix} }$$ is on the plane.
B(-2, 1, 2) is on the plane and so
$$\small{ \vec{d_2} = \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} }$$ $$\small{ = \begin{pmatrix}-2\\1\\2\end{pmatrix} - \begin{pmatrix}3\\-1\\2\end{pmatrix} = \begin{pmatrix}-5 \\2\\0\end{pmatrix} . }$$ $$\small{ \text{We see that} \,\, \vec{d_2} \,\, \text{is also on the plane.} }$$ Therefore the vector equation of the plane is $$\small{ \vec{r} = \begin{pmatrix}3\\-1\\2\end{pmatrix} + t_1 \begin{pmatrix}-1\\1\\1\end{pmatrix} + t_2 \begin{pmatrix}-5\\2\\0\end{pmatrix} . }$$
fig 4.4-6



Example 22.
$$\small{ \text{Vector} \,\, \vec{n} = \begin{pmatrix}2 \\ 4\\-2\end{pmatrix} \,\, \text{is perpendicular} }$$ to the plane which contains point A(1, -5, 2) .
(a) Write an equation of the plane in the form
$$\small{ \vec{r} \cdot \vec{n} = d .}$$ (b) Find the Cartesian equation of the plane.

Solution
(a) The vector equation of the plane is
$$\small{ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}. }$$ $$\small{ \vec{r} \cdot \vec{n} = \begin{pmatrix}1\\-5\\2\end{pmatrix} \cdot \begin{pmatrix}2\\4\\-2\end{pmatrix} }$$
= 2 - 20 - 4 = -22

$$\small{ \text{(b)} \qquad \vec{r} \cdot \vec{n} = d }$$ $$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}2\\4\\-2\end{pmatrix} = -22 }$$
2x + 4y - 2z = -22
fig 4.4-7



Exercise 4.4
1. Find the vector equation of the line:
$$\small{ \text{(a) \,\,\, parallel to} \,\, \begin{pmatrix}2\\1\\3\end{pmatrix} }$$ and through the point (1, 3, -7).
(b) through (0, 1, 2) and with direction vector î + ĵ - 2k̂ .
(c) parallel to the x-axis and through the point (-2, 2, 2).
Solution
$$\small{ \text{(a) \quad Let} \,\, A= (1, 3, -7), \qquad \vec{b} = \begin{pmatrix}2\\1\\3\end{pmatrix}}$$ $$\small{ \vec{a} = \overrightarrow{OA} = \begin{pmatrix}1\\3\\-7\end{pmatrix} }$$ The vector equation is
$$\small{ \vec{r} = \vec{a} + t \vec{b} , \quad t \isin \mathbb{R} }$$ $$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\3\\-7\end{pmatrix} + t \begin{pmatrix}2\\1\\3\end{pmatrix} }$$

$$\small{ \text{(b) \quad Let} \quad A = (0,1,2) \,\, \text{and} \,\, \vec{b} = \hat{i} + \hat{j} - 2\hat{k} }$$ $$\small{ \vec{a} = \overrightarrow{OA} = \begin{pmatrix}0\\1\\2\end{pmatrix}, \,\,\, \vec{b} = \begin{pmatrix}1\\1\\-2\end{pmatrix} }$$ The vector equation of the line is
$$\small{ \vec{r} = \vec{a} + t \vec{b} , \quad t \isin \mathbb{R} }$$ $$\small{ \vec{r} = \begin{pmatrix}0\\1\\2\end{pmatrix} + t \begin{pmatrix}1\\1\\-2\end{pmatrix} }$$

$$\small{ \text{(c) \quad Let} \quad \vec{a} = \begin{pmatrix}-2\\2\\2\end{pmatrix}\, , \,\,\, \vec{b} = \begin{pmatrix}1\\0\\0\end{pmatrix} }$$ The vector equation of the line is
$$\small{\vec{r} = \vec{a} + t \vec{b} \,,\, \quad t \isin \mathbb{R} } $$ $$\small{ \vec{r} = \begin{pmatrix}-2\\2\\2\end{pmatrix} + t \begin{pmatrix}1\\0\\0\end{pmatrix} }$$

2. (a) Find the Cartesian equation of the line with parametric equation
x = 3t + 1, y = 4 - 2t, z = 3t - 1.
(b) Find the unit vector in the direction of the line.
Solution
(a)
x = 3t + 1  ,     y = 4 - 2t   ,     z = 3t - 1
x - 1 = 3t ,     y - 4 = -2t ,     z + 1 = 3t
x - 1⁄3 = t ,     y - 4⁄-2 = t ,     z + 1⁄3 = t

The Cartesian equation of the line is
x - 1⁄3 = y - 4⁄2 = x + 1⁄3

(b) The direction value of the line is (3, -2, 3).
$$\small{ \text{Let} \quad \vec{a} = 3\hat{i} - 2\hat{j} + 3\hat{k} } $$ $$ \small{ \left|\vec{a}\right| = \sqrt{3^2 + (-2)^2 + 3^2} = \sqrt{22}}$$ The unit vector in the direction of the line is
$$ \small{\hat{a} = \dfrac{\vec{a}}{\left|\vec{a}\right|} }$$ $$\small{= \dfrac{3\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{22}} }$$

3. Find the equation of the plane:
$$\small{ \text{(a) \,\, with normal vector}\,\, \begin{pmatrix}2\\-1\\3\end{pmatrix} }$$ and which passes through (-1, 2, 4).
(b) perpendicular to the line joining points A(2, 3, 1) and B(5, 7, 2) and which passes through A.
(c) containing A(3, 2, 1) and the line x = 1 + t , y = 2 - t, z = 3 + 2t.

Solution
$$\small{\text{(a)} \quad \vec{n} = \begin{pmatrix}2\\-1\\3\end{pmatrix} \, , \,\,\, \vec{r} = \begin{pmatrix}-1\\2\\4\end{pmatrix}}$$ The vector equation of the plane is
$$\small{ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} }$$ $$\small{\begin{pmatrix}-1\\2\\4\end{pmatrix} \cdot \begin{pmatrix}2\\-1\\3\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}2\\-1\\3\end{pmatrix}}$$ -2 - 2 + 12 = 2x - y + 3z
8 = 2x - y + 3z
The equation of the plane is 2x - y + 3z = 8

(b)     A(2, 3, 1),       B(5, 7, 2)
$$\small{ \vec{a} = \overrightarrow{OA} = \begin{pmatrix}2\\3\\1\end{pmatrix}\, , } $$ $$\small{ \vec{b} = \overrightarrow{OB} = \begin{pmatrix}5\\7\\2\end{pmatrix}} $$ $$\small{ \text{Normal vector} \,\, \, \vec{n} = \overrightarrow{AB} }$$ $$\small{ = \overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{OB} - \overrightarrow{OA}}$$ $$ \small{ = \begin{pmatrix}5\\7\\2\end{pmatrix} - \begin{pmatrix}2\\3\\1\end{pmatrix} }$$ $$ \small{ \vec{n} = \begin{pmatrix}3\\4\\1\end{pmatrix}}$$ The equation of the plane is
$$\small{ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}} $$ $$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}3\\4\\1\end{pmatrix} = \begin{pmatrix}2\\3\\1\end{pmatrix} \cdot \begin{pmatrix}3\\4\\1 \end{pmatrix}} $$ 3x + 4y + z = 6 + 12 + 1
3x + 4y + z = 19
The equation of the plane is
3x + 4y + z = 19

fig 4.4-8

(c)     x = 1 + t ,     y = 2 - t ,     z = 3 + 2t
x - 1 = t , y - 2 = -t , z - 3 = 2t
t = x - 1 , - (y - 2) = t ,     z - 3⁄2 = t

The Cartesian equation of the line is
x - 1⁄1 = y - 2⁄-1 = z - 3⁄2
The direction value of the line is (1, -1, 2)
The coordinates B(1, 2, 3) satisfy the Cartesian equation.
∴ B(1, 2, 3) lies on the plane.
A(3, 2, 1)
$$\small{ \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{OB} - \overrightarrow{OA} } $$ $$\small{ = \begin{pmatrix}1\\2\\3\end{pmatrix} - \begin{pmatrix}3\\2\\1\end{pmatrix} = \begin{pmatrix}-2\\0\\2\end{pmatrix} }$$ Directed value of AB = (-2, 0, 2)
Let (a, b, c) be directed value of the plane.
$$\small{ \begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}-2\\0\\2\end{pmatrix} \times \begin{pmatrix}1\\-1\\2\end{pmatrix} }$$ -2           0           2           -2           0 1           -1           2           1           -1
$$\small{ = \begin{pmatrix}0\times 2 \quad - \quad (-1)\times 2\\ 2\times 1 \quad - \quad 2\times (-2)\\ -2\times (-1) \quad - \quad 1\times 0 \end{pmatrix}} $$ $$\small{ = \begin{pmatrix}0+2\\2+4\\2-0\end{pmatrix} }$$ $$\small{ \begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}2\\6\\2\end{pmatrix} }$$ a = 2,       b = 6 ,       c = 2
ax + by + cz = d
2x + 6y + 2z = d

A(3, 2, 1)
2(3) + 6(2) + 2(1) = d
6 + 12 + 2 = d
20 = d

2x + 6y + 2z = d
2 (x + 3y + z) = 20
x + 3y + z = 10
The equation of the plane is   x + 3y + z = 10

fig 4.4-9

4. Find the equation of the plane through A(-1, 2, 1), B(4, 1, 1) and C(2, 0, 3):
(a) in vector form           (b) in Cartesian form.

Solution
A(-1, 2, 1), B(4, 1, 1), C(2, 0, 3):
$$\small{ \text{(a) \quad Let} \,\,\, \vec{a} = \overrightarrow{OA} = \begin{pmatrix}-1\\2\\1\end{pmatrix} } $$ $$\small{\vec{d_1} = \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} }$$ $$\small{ = \overrightarrow{OB} - \overrightarrow{OA}} $$ $$\small{ = \begin{pmatrix}4\\1\\1\end{pmatrix} - \begin{pmatrix}-1\\2\\1\end{pmatrix} = \begin{pmatrix} 5\\-1\\0\end{pmatrix} }$$ $$\small{ \vec{d_2} = \overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} }$$ $$\small{ = \overrightarrow{OC} - \overrightarrow{OA}}$$ $$\small{ = \begin{pmatrix}2\\0\\3\end{pmatrix} - \begin{pmatrix}-1\\2\\1\end{pmatrix} = \begin{pmatrix}3\\-2\\2\end{pmatrix} }$$ $$\small{ \vec{r} = \vec{a} + t_1 \vec{d_1} + t_2 \vec{d_2}} $$ $$\small{\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-1\\2\\1\end{pmatrix} + t_1 \begin{pmatrix}5\\-1\\0\end{pmatrix} + t_2 \begin{pmatrix}3\\-2\\2\end{pmatrix} }$$ $$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-1 + 5t_1 + 3t_2\\ 2 - t_1 - 2t_2\\ 1 + 2t_2\end{pmatrix}}$$ x = -1 + 5t1 + 3t2     ______(1)
y = 2 - t1 - 2t2     ______(2)
z = 1 + 2t2
z - 1 = 2t2
z - 1⁄2 = t2     ______(3)

equation (1)   +   equation (2) × 5,
    x = -1 + 5t1 + 3t2
  5y = 10 - 5t1 - 10t2

x + 5y = 9 - 7t2
x + 5y - 9⁄-7 = t2     ______(4)
From equation (3) and (4), we obtain
z - 1⁄2 = x + 5y - 9⁄-7
-7z + 7 = 2x + 10y - 18
7 + 18 = 2x + 10y + 7z
2x + 10y + 7z = 25

5. Find the Cartesian equation of the plane with vector equation
$$\small{ \vec{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t_1 \begin{pmatrix}1\\1\\2\end{pmatrix} + t_2 \begin{pmatrix}2\\-1\\5\end{pmatrix} .} $$ Solution
$$\small{ \vec{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t_1 \begin{pmatrix}1\\1\\2\end{pmatrix} + t_2 \begin{pmatrix}2\\-1\\5\end{pmatrix}} $$ $$\small{ \begin{pmatrix}x\\y\\x\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \begin{pmatrix}t_1\\t_1\\2t_1\end{pmatrix} + \begin{pmatrix}2t_2\\-t_2\\5t_2\end{pmatrix}} $$ $$\small{ \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1 + t_1 + 2t_2\\ 2 + t_1 - t_2\\ 3 + 2t_1 + 5t_2\end{pmatrix} }$$ x = 1 + t1 + 2t2     ______(1)
y = 2 + t1 - t2     ______(2)
z = 3 + 2t1 + 5t2     ______(3)

Equation (1) - equation (2)
    x = 1 + t1 + 2t2
- (y = 2 + t1 - t2)

x - y = -1         + 3t2
x - y + 1⁄3 = t2     ______(4)

Equation (2)× 2   -   equation (3)
    2y = 4 + 2t1 - 2t2
- (z = 3 + 2t1 + 5t2)

2y - z = 1         - 7t2
2y - z - 1⁄-7 = t2     ______(5)
From equation (4) and (5), we obtain the Cartesian equation
x - y + 1⁄3 = 2y - z - 1⁄-7
-7x + 7y - 7 = 6y - 3z - 3
-7 + 3 = 7x - 7y + 6y - 3z
-4 = 7x - y - 3z
7x - y - 3z = -4