FP3 Vectors Revision

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

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

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The vector product of any 2 parallel vectors is zero

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The vector product is not commutative as a x b = -b x a

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

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


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

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

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

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

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Find the direction vector of the line by using the fact that the line is perpendicular to both normal vectors, i.e. d = n_1 x n_2

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

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Equation of line is r= $a+\lambda$ d

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

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

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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 = $abcos\theta$

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

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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 $sin\phi=cos\theta=\frac{d.n}{|d||n|}$

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

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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.

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Express the equation in the form $n_1x+n_2y+n_3z+d=0$

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If the point is $(\alpha, \beta, \gamma)$, quote the result from the formula booklet giving:
Distance = $\frac{|n_1 \alpha+n_2 \beta+n_3 \gamma+d|}{\sqrt{(n_1^2+n_2^2+n_3^2 )}}$

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

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Find the distance between these 2 planes.

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

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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 = $\frac{\vec{AP}.n}{|n|}$

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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:

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

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Proceed as for and intersection of a line and a plane

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

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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 = $\frac{\vec{AP}\times d}{|d|}$

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

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

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

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

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Locate any point, A on $L_1$ and any point B on $L_2$ and find the vector $\vec{AB}$ = b – a

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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

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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 = $\frac{\vec{AB}.n}{|n|}$