P4.8 Vectors in two and three dimensions

How do you describe a vector and find its magnitude?

A vector has a size (magnitude) and a direction. In three dimensions we use three perpendicular unit vectors: \(\mathbf i\) along the \(x\)-axis, \(\mathbf j\) along the \(y\)-axis and \(\mathbf k\) along the \(z\)-axis. The same vector can be written in either of two forms: \(2\mathbf i+3\mathbf j+6\mathbf k=\begin{pmatrix}2\\3\\6\end{pmatrix}\). In two dimensions there is no \(\mathbf k\): \(5\mathbf i-12\mathbf j=\begin{pmatrix}5\\-12\end{pmatrix}\).

The magnitude comes from Pythagoras, used twice in 3D: once across the floor, once up the vertical edge. A unit vector has magnitude 1; the unit vector in the direction of \(\mathbf a\) is \(\hat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}\), and a vector of magnitude \(k\) in the direction of \(\mathbf a\) is \(k\hat{\mathbf a}\). In 2D the direction can also be given as an angle: \(5\mathbf i-12\mathbf j\) makes \(\tan^{-1}\frac{12}5=67.4°\) with \(\mathbf i\), measured below it.

How do position vectors give directions, distances and ratios?

The position vector of a point \(A\) is \(\overrightarrow{OA}=\mathbf a\), the vector from the origin to \(A\); its components are the coordinates of \(A\). To go from \(A\) to \(B\), go back to \(O\) and out to \(B\): \(\overrightarrow{AB}=-\mathbf a+\mathbf b=\mathbf b-\mathbf a\) ("end minus start"). The distance between the points is \(AB=|\mathbf b-\mathbf a|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\); for example, \((1,2,-2)\) to \((4,6,10)\) is \(\sqrt{9+16+144}=13\).

Vectors are added and multiplied by scalars component by component. Parallel vectors are scalar multiples: \(6\mathbf i-3\mathbf j\) and \(-2\mathbf i+\mathbf j\) are parallel (\(-3\) times). Points \(A\), \(B\), \(C\) are collinear if \(\overrightarrow{AC}\) is a multiple of \(\overrightarrow{AB}\); the shared point \(A\) rules out two separate parallel lines. The point \(P\) on \(AB\) with \(AP:PB=\lambda:\mu\) has \(\overrightarrow{OP}=\mathbf a+\frac{\lambda}{\lambda+\mu}(\mathbf b-\mathbf a)\); the midpoint is \(\frac12(\mathbf a+\mathbf b)\).

What is the vector equation of a line?

Start at a known point on the line (position vector \(\mathbf a\)) and move any multiple of a direction vector \(\mathbf d\). Every point on the line has position vector

The equation is not unique: any point on the line can replace \(\mathbf a\), and any non-zero multiple of \(\mathbf d\) is still a direction vector. To test whether a point is on the line, set \(\mathbf a+t\mathbf d\) equal to its position vector, find \(t\) from one component, then check that the same \(t\) works in the other two. Each value of \(t\) gives one point; if \(|\mathbf d|=3\), consecutive integer values of \(t\) are 3 units apart. Where the line crosses a coordinate plane, one component is zero: \(z=0\) on the \(xy\)-plane.

Are two lines parallel, intersecting or skew?

Use different parameters for the two lines (\(\lambda\) and \(\mu\)): the lines meet when the same point is reached, not at the same parameter value. Skew lines exist only in three dimensions; in a plane, two non-parallel lines always meet. After finding a point of intersection, substitute \(\mu\) into the second line as a check: both lines must give the same coordinates.

What is the scalar product, and how does it give angles?

For non-zero vectors, perpendicular means \(\mathbf a\cdot\mathbf b=0\) (\(\cos90°=0\)), the 3D version of \(m_1m_2=-1\). Also \(\mathbf a\cdot\mathbf a=|\mathbf a|^2\); \(\mathbf i\cdot\mathbf i=\mathbf j\cdot\mathbf j=\mathbf k\cdot\mathbf k=1\) and \(\mathbf i\cdot\mathbf j=\mathbf j\cdot\mathbf k=\mathbf k\cdot\mathbf i=0\).

For the angle between two lines, use their direction vectors only. If the scalar product is negative, \(\theta\) is obtuse and the acute angle between the lines is \(180°-\theta\). For the angle at a vertex of a triangle, \(\angle BAC\), use \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\), both starting at \(A\); using \(\overrightarrow{BA}\) with \(\overrightarrow{AC}\) gives \(180°-\angle BAC\). Finally, \(|\mathbf b|\cos\theta\) is the length of the projection of \(\mathbf b\) onto the direction of \(\mathbf a\), the idea behind the foot of a perpendicular in the next section.

How do you find the foot of the perpendicular from a point to a line?

The foot \(F\) is the point on the line nearest to \(P\); there, \(\overrightarrow{PF}\) is perpendicular to the line. Method: (1) write the general point of the line, \(F=\mathbf a+t\mathbf d\), in components; (2) form \(\overrightarrow{PF}=\overrightarrow{OF}-\overrightarrow{OP}\) in terms of \(t\); (3) solve \(\overrightarrow{PF}\cdot\mathbf d=0\) for \(t\); (4) substitute back for \(F\). The shortest distance from \(P\) to the line is \(|\overrightarrow{PF}|\). The image (reflection) of \(P\) in the line is \(P'\), with \(F\) as the midpoint of \(PP'\): \(\overrightarrow{OP'}=2\overrightarrow{OF}-\overrightarrow{OP}\).

How do you find the area of a triangle with vectors?

Find two sides from the same vertex, \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\), and the angle \(\theta=\angle BAC\) from the scalar product. Then area \(=\frac12|\overrightarrow{AB}||\overrightarrow{AC}|\sin\theta\); for an exact answer, get \(\sin\theta\) from \(\cos\theta\) with \(\sin^2\theta=1-\cos^2\theta\). Alternatively, area \(=\frac12\times\) base \(\times\) perpendicular height, with the height from the foot of the perpendicular from \(C\) to the line \(AB\). A parallelogram with adjacent sides \(AB\) and \(AC\) has twice the area of the triangle; its fourth vertex is \(D\) with \(\overrightarrow{OD}=\mathbf b+\mathbf c-\mathbf a\).

Exam tips for P4.8

Common misconceptions

How this topic connects

P1 straight lines and trigonometry

The distance formula and midpoint extend to three dimensions with a \(z\) term; \(m_1m_2=-1\) becomes \(\mathbf a\cdot\mathbf b=0\); and the cosine rule gives the same angles as the scalar product when all three sides are known.

Mechanics

Displacement, velocity and force are vectors in \(\mathbf i\) and \(\mathbf j\); position = initial position + time × velocity, \(\mathbf r=\mathbf r_0+t\mathbf v\), is exactly the vector equation of a line with time as the parameter. Work done by a constant force is a scalar product, and closest-approach problems use the same perpendicular condition as the foot of a perpendicular.

IAL Physics

Resolving forces, work \(=\mathbf F\cdot\mathbf s\) and the components of velocity in projectile motion all use the vector ideas on this page.

Key formulae

Check your understanding

Specification checklist

• (P4 7.1) I can write vectors in two and three dimensions using \(\mathbf i\), \(\mathbf j\) and \(\mathbf k\) or as column vectors, and switch between the two forms.
• (P4 7.2) I can find the magnitude of a vector, the unit vector in its direction, and a vector of a given magnitude in a given direction.
• (P4 7.3) I can add and subtract vectors and multiply them by scalars, and I know that parallel vectors are scalar multiples of each other.
• (P4 7.4) I can use position vectors and \(\overrightarrow{AB}=\mathbf b-\mathbf a\), find the point dividing a segment in a given ratio, and show that three points are collinear.
• (P4 7.5) I can find the distance between two points in two or three dimensions.
• (P4 7.6) I can write a vector equation of a line through a point in a given direction, or through two points, and decide whether a given point lies on it.
• (P4 7.6) I can decide whether two lines are the same, parallel, intersecting or skew, and find the point where two lines intersect.
• (P4 7.7) I can calculate a scalar product and use it to find the angle between two vectors or the acute angle between two lines, and to show that vectors are perpendicular.
• (P4 7.7) I can find the foot of the perpendicular from a point to a line, the shortest distance from the point to the line, and the area of a triangle using vectors.