• Revision:Further Vectors

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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 anti-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 a, 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 L_1 and any point B on L_2 and find the vector \vec{AB} = b – a
  • Find the common normal, i.e. the vector which is perpendicular to both d_1 and d_2. 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|}
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