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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Further Vectors

1 Vector equations of a line
2 Vector equations of a plane
3 Vector product
4 Intersections
- 4.1 Point of intersection of a line and a plane
- 4.2 Line of intersection of 2 intersecting planes
5 Angles
6 Distances

Vector equations of a line

Lines can be written in several different ways:

There’s the c4 way: $\begin{bold}r \end{bold} = \begin{pmatrix}a \\b \\c \end{pmatrix}+\lambda\begin{pmatrix}d \\e \\f \end{pmatrix}$

The other c4 way: $\begin{bold}r \end{bold} = (a+ \lambda d)i+(b+ \lambda e)j+(c+ \lambda f)k$

Also the Cartesian form: $\begin{bold}r \end{bold} = \frac{x-a}{d}=\frac{y-b}{e}=\frac{z-c}{f}$

Vector equations of a plane

Similar to the c4 line equation: $\begin{bold}r\end{bold}=\begin{pmatrix}a \\b \\c \end{pmatrix}+\lambda\begin{pmatrix}d \\e \\f \end{pmatrix}+\mu\begin{pmatrix}g \\h \\i \end{pmatrix}$

Similar again: $\begin{bold}r=(1+ 4\lambda + 7\mu )i+(2+ 5\lambda + 8\mu )j+(3+ 6\lambda + 9\mu )k$ I used numbers for the sake of notation. 2 i’s won’t look pretty!

Here’s a new one: $\begin{bold}r.\begin{pmatrix}a \\b \\c \end{pmatrix}= \begin{pmatrix}d \\e \\f \end{pmatrix}\begin{bold}.\end{bold} \begin{pmatrix}a \\b \\c \end{pmatrix}$

Or: $\begin{bold}r.\end{bold}\begin{pmatrix}a \\b \\c \end{pmatrix}=k$

Or: ax+by+cz=k

Vector product

The definition of the vector product is $\begin{bold}a\end{bold}\times\begin{bold}b\end{bold} = \begin{bold}|a||b|sin\theta\begin{bold}n\end{bold}$ where $\begin{bold}n\end{bold}$ is the unit vector perpendicular to both a and b in the direction given by the right hand screw rule.

For vectors in component form, the vector product may be calculated using either of the following determinants, the second of which is given in the formula booklet:

$\begin{vmatrix} i & j & k \\x_1 & y_1 & z_1 \\x_2 & y_2 & z_2 \end{vmatrix}$ or $\begin{vmatrix} i & x_1 & x_2 \\j & y_1 & y_2 \\k & z_1 & z_2 \end{vmatrix}$

The vector product of any 2 parallel vectors is zero

The vector product is not commutative as $\begin{bold}a\end{bold}\times\begin{bold}b\end{bold}=-\begin{bold}b\end{bold}\times\begin{bold}a\end{bold}$

Area of a triangle is $\frac{1}{2}|a \times b|$ , of a parallelogram is $|a\times b|$ , $\begin{bold}a\end{bold}$ and $\begin{bold}b\end{bold}$ representing adjacent sides.

Intersections

Point of intersection of a line and a plane

Express the position vector of a general point on the line as a single vector, e.g.

$\begin{bold}r\end{bold}=\frac{x-1}{6}=\frac{y-2}{8}=\frac{z-3}{1}\Rightarrow \begin{bold}r\end{bold}=\begin{pmatrix} 1+6\lambda \\2+8\lambda \\3+\lambda\end{pmatrix}$

substitute this vector into the normal form of the equation of a plane and find $\lambda$ , e.g.

$\begin{bold}r.\end{bold}\begin{pmatrix}1\\ 2\\ -1\end{pmatrix}=3 \Rightarrow \begin{bold}r\end{bold}=\begin{pmatrix} 1+6\lambda \\2+8\lambda \\3+\lambda\end{pmatrix}.\begin{pmatrix}1\\ 2\\ -1\end{pmatrix}=3 \Rightarrow 1+6\lambda+2(2+8\lambda)-1(3+\lambda)=3$ etc.

Substitute your value of $\lambda$ back into the equation of the line.

Line of intersection of 2 intersecting planes

Find the direction vector of the line by using the fact that the line is perpendicular to both normal vectors, i.e. $\begin{bold}d\end{bold} = \begin{bold}n_1\end{bold} \times\begin{bold}n_2\end{bold}$

Find any point , common to both planes by using any value (usually 0) for x, y or z

Equation of line is $\begin{bold}r\end{bold}=a+\lambda\begin{bold}d\end{bold}$

Angles

The direction vector of a line is given by d in the equation r= $a+\lambda$ d

The direction of the normal to a plane is given by n in the equation r.n = a.n

The angle between two planes is the angle between their 2 normals, n_1 and n_2. This is usually found by using the scalar product, a.b = $abcos\theta$

The angle between two lines is the angle between their 2 direction vectors, d_1 and d_2.

The angle between a line and a plane is found by finding the angle between d and n then subtracting from 90 degrees. Alternatively, if $\theta$ is the angle between d and n then as $\phi =90-\theta$ you may use $sin\phi=cos\theta=\frac{d.n}{|d||n|}$

Distances

The distance of a plane r.n = p from the origin is $\frac{p}{|n|}$ or $\frac{|p|}{|n|}$ where the sign is unimportant.

To find the distance between two planes, first find their distances from the origin. If the two p-values are the same sign, the planes are on the same side of the origin so subtract the 2 distances. If the two p-values are different signs, the origin lies between them so add the positive distances.

Distance between a plane and a point, A

1)*Express the equation in the form n_1x+n_2y+n_3z+d=0

If the point is $(\alpha, \beta, \gamma)$ , quote the result from the formula booklet giving:

Distance = $\frac{|n_1 \alpha+n_2 \beta+n_3 \gamma+d|}{\sqrt{(n_1^2+n_2^2+n_3^2 )}}$

2)*Find the equation of the plane through the given point A, parallel to the original plane, using r.n = a.n

Find the distance between these 2 planes.

3)*Locate any point, P on the plane and find the vector $\vec{AP}$

If $\theta$ is the angle between $\vec{AP}$ and n, then the required distance h is $|\vec{AP}|cos\theta=\frac{|\vec{AP}||n|cos\theta}{|n|}$ so h = $\frac{\vec{AP}.n}{|n|}$

4)*Find the co-ordinates of F, the foot of the perpendicular from A to the plane and find AF. This method should only really be used if F is specifically asked for because it’s a bit long winded. Anyway, F can be found as follows:

As AF is perpendicular to the plane, then $\vec{AF}=\lambda$ n, so $\vec{OF}=\vec{OA}+\vec{AF}$ = a + $\lambda$ n

Proceed as for and intersection of a line and a plane

Distance between a point A, from a line L

1)*Locate any point, P on L and find the vector $\vec{AP}$

If $\theta$ is the angle between $\vec{AP}$ and d, then the required distance h is $|\vec{AP}|sin\theta=\frac{|\vec{AP}||d|sin\theta}{|d|}$ so h = $\frac{\vec{AP}\times d}{|d|}$

2)* Find the co-ordinates of F, the foot of the perpendicular from A to L and find AF. F can be found as follows:

As F lies on L, express the position vector of F as a single vector involving $\lambda$

Use the fact that $\vec{OF}=\vec{OA}+\vec{AF}$ to find $\vec{AF}=\vec{OF}-\vec{OA}$

As $\vec{AF}$ is perpendicular to L, then $\vec{AF}.d=0$ so use this to find $\lambda$ and hence F

Distance between 2 skew lines

Locate any point, A on and any point B on and find the vector $\vec{AB}$ = b – a

Find the common normal, i.e. the vector which is perpendicular to both and . This is given by n = $d_1 \times d_2$ and may be simplified if appropriate

If $\theta$ is the angle between $\vec{AB}$ and n, then the required distance h is $|\vec{AB}|cos\theta=\frac{|\vec{AB}||n|cos\theta}{|n|}$ so h = $\frac{\vec{AB}.n}{|n|}$