Angle Between Two Lines & Shortest Distance Between Skew Lines

MATLines in Space

The angle between two lines depends only on their direction vectors and is found via the dot product. Lines in space may intersect, be parallel, or be skew (non-parallel and non-intersecting); the shortest distance between skew lines is measured along their common perpendicular.

ISC heavily examines the scalar-triple-product distance formula and the coplanarity/intersection test (shortest distance $=0$ ).

Angle between two lines

\cos\theta = \dfrac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}|\,|\vec{b_2}|}

\vec{b_1}, \vec{b_2}

are the direction vectors of the two lines; the modulus in the numerator gives the acute angle

\theta

Shortest distance between skew lines (vector form)

d = \dfrac{\left|\,(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\,\right|}{|\vec{b_1} \times \vec{b_2}|}

Lines

\vec{r} = \vec{a_1} + \lambda\vec{b_1}

and

\vec{r} = \vec{a_2} + \mu\vec{b_2}

; the numerator is the magnitude of the scalar triple product.

Coplanarity / intersection condition

(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0

When this scalar triple product is zero (so

d = 0

) and

\vec{b_1} \nparallel \vec{b_2}

, the lines are coplanar and intersect.

Distance between parallel lines

d = \dfrac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|}

Both lines have the same direction

\vec{b}

;

\vec{a_1}, \vec{a_2}

are points on each line.

Shortest distance (Cartesian determinant form)

d = \dfrac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}}

(x_i,y_i,z_i)

are points and

(a_i,b_i,c_i)

direction ratios of the two lines; take the modulus of the whole expression (the determinant may come out negative).

Use the modulus $|\vec{b_1}\cdot\vec{b_2}|$ to always report the acute angle; without it you may get the obtuse supplement.
Two lines are perpendicular when $\vec{b_1}\cdot\vec{b_2} = 0$ , i.e. $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ ; parallel when $\vec{b_1} \times \vec{b_2} = \vec{0}$ (ratios proportional).
Skew lines are non-parallel AND non-intersecting; for them $\vec{b_1}\times\vec{b_2} \ne \vec{0}$ and the scalar triple product $\ne 0$ .
If $d = 0$ and the lines are not parallel, they intersect — substitute back to find the point by equating parametric coordinates.
For parallel lines the scalar-triple-product formula fails (denominator $=0$ ); use the cross-product distance formula with the common direction $\vec{b}$ .
$\vec{b_1}\times\vec{b_2}$ gives the direction of the common perpendicular between the two skew lines.
To find the intersection point: set $x_1+a_1\lambda = x_2+a_2\mu$ etc., solve any two equations for $\lambda, \mu$ , then verify in the third.
Always simplify the final surd and state the distance in the given units (e.g. km, m).

Where the marks go

Omitting the modulus, giving a negative distance or the obtuse angle instead of the required acute angle.
Using the scalar-triple-product formula on parallel lines (where $\vec{b_1}\times\vec{b_2}=\vec{0}$ ) — switch to the parallel-line formula.
Forgetting to verify intersection in the third equation after solving for $\lambda$ and $\mu$ ; consistency in only two equations does not prove intersection.
Computing $(\vec{a_2}-\vec{a_1})$ in the wrong order, or mis-evaluating the cross-product/determinant signs, leading to a wrong magnitude.

How the board asks it

Numericalscalar triple product distance formula
Find the shortest distance between the lines $\vec{r} = (\hat{i}+2\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r} = (2\hat{i}-\hat{j}-\hat{k}) + \mu(2\hat{i}+\hat{j}+2\hat{k})$ .
Numericalangle between two lines via dot product
Find the acute angle between the lines $\dfrac{x-2}{2}=\dfrac{y-1}{5}=\dfrac{z+3}{-3}$ and $\dfrac{x+2}{-1}=\dfrac{y-4}{8}=\dfrac{z-5}{4}$ .
Give reasonscoplanarity and intersection condition
Show that the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-4}{5}=\dfrac{y-1}{2}=\dfrac{z}{1}$ intersect, and hence find their point of intersection.
Numericaldistance between parallel lines (cross-product formula)
Show that the lines $\vec{r} = (\hat{i}+2\hat{j}-4\hat{k}) + \lambda(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r} = (3\hat{i}+3\hat{j}-5\hat{k}) + \mu(2\hat{i}+3\hat{j}+6\hat{k})$ are parallel, and find the distance between them.
Applicationshortest distance between skew lines
Two paths are modelled by $\vec{r} = (3\hat{i}+8\hat{j}+3\hat{k}) + \lambda(3\hat{i}-\hat{j}+\hat{k})$ and $\vec{r} = (-3\hat{i}-7\hat{j}+6\hat{k}) + \mu(-3\hat{i}+2\hat{j}+4\hat{k})$ (distances in $\text{km}$ ); find the shortest distance between the two paths.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.