Applications of Vectors

Q: How do I find the shortest distance if the two lines turn out to be parallel?

Switch formulas. The skew line expression has the magnitude of the cross product of the directions in the denominator, which is zero for parallel lines. Use the parallel line formula instead, which is just the distance of a point on one line from the other line.

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Applications of Vectors

Maths · Vector Algebra · Concept 5

Everything built so far now gets used. Dot products give angles and distances, cross products give normals and areas, and box products detect coplanarity. Put together they describe lines and planes in space completely, and they translate directly into the physics of work, torque and rotation.

Tests for collinearity and coplanarity

Collinearity of three points

$A (a), B (b), C (c)$ are collinear if any one of these holds:

$A B = λ A C$ for some scalar $λ$
$A B \times A C = 0$
$a \times b + b \times c + c \times a = 0$
There exist scalars $x, y, z$ not all zero with $x a + y b + z c = 0$ and $x + y + z = 0$

Coplanarity

Three vectors $a, b, c$ are coplanar if $[a b c] = 0$ .

Four points $A, B, C, D$ are coplanar if $[A B A C A D] = 0$ , or equivalently if there exist $x, y, z, w$ not all zero with

$x a + y b + z c + w d = 0$ and $x + y + z + w = 0$ .

Example 1

Given non-coplanar $a, b, c$ , prove that the points with position vectors $2 a + 3 b - c$ , $a - 2 b + 3 c$ , $3 a + 4 b - 2 c$ and $a - 6 b + 6 c$ are coplanar.

Solution. Call them $P, Q, R, S$ . Then

$P Q = - a - 5 b + 4 c$ , $P R = a + b - c$ , $P S = - a - 9 b + 7 c$ .

Seek $P Q = x P R + y P S$ . Comparing coefficients (valid since $a, b, c$ are non-coplanar):

$x - y = - 1$ , $x - 9 y = - 5$ , $- x + 7 y = 4$ .

The first two give $x = - \frac{1}{2}$ , $y = \frac{1}{2}$ , and these satisfy the third. A consistent solution exists, so $P, Q, R, S$ are coplanar.

Vector equation of a straight line

Two standard forms

Through $A (a)$ , parallel to $b$ :

r = a + t b, t \in R

Through two points $A (a)$ and $B (b)$ :

r = a + t (b - a), t \in R

Non-parametric form: $(r - a) \times b = 0$

Cartesian form for a line through $(x_{1}, y_{1}, z_{1})$ with direction ratios $a, b, c$ :

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}

Angle between two lines

If the lines have direction vectors $b_{1}$ and $b_{2}$ , the acute angle $θ$ between them satisfies

cos θ = \frac{b _{1} \cdot b _{2}}{b _{1} b _{2}}

The modulus in the numerator is what forces the answer to be the acute angle. The lines are perpendicular if $b_{1} \cdot b_{2} = 0$ and parallel if $b_{1} \times b_{2} = 0$ .

Distance of a point from a line

Let the line be $r = a + t b$ and let $P$ have position vector $p$ . Write $A P = p - a$ . Splitting $A P$ into a part along $b$ and a part perpendicular to it, the perpendicular part is the distance.

Perpendicular distance

d = \frac{( p - a ) \times b}{b} = (p - a) \times \hat{b}

Equivalently, using the dot product,

d = ∣ p - a ∣^{2} - \frac{( p - a ) \cdot b}{b}^{2}

Foot of the perpendicular and the image

The foot $N$ is obtained by projecting: $n = a + \frac{( p - a ) \cdot b}{b ^{2}} b$

The image (mirror point) of $P$ in the line is $p^{'} = 2 n - p$ .

Example 2

Find the distance of $P (1, 2, 3)$ from the line $r = (\hat{i} + \hat{j}) + t (2 \hat{i} + \hat{j} + 2 \hat{k})$ .

Solution. $p - a = \hat{j} + 3 \hat{k}$ and $b = 2 \hat{i} + \hat{j} + 2 \hat{k}$ with $∣ b ∣ = 3$ .

$(p - a) \times b = \hat{i} \hat{j} \hat{k} 013212 = (2 - 3) \hat{i} - (0 - 6) \hat{j} + (0 - 2) \hat{k} = - \hat{i} + 6 \hat{j} - 2 \hat{k}$ .

Its magnitude is $1 + 36 + 4 = 41$ , so $d = \frac{41}{3}$ .

Shortest distance between two lines

Two lines in space either intersect, or are parallel, or are skew, meaning they neither meet nor run parallel. For skew lines the shortest joining segment is perpendicular to both, so its direction is along $b_{1} \times b_{2}$ , and its length is the projection of the join of any two points onto that direction.

LM is the common perpendicular of two skew lines. Its length is the projection of AB on the direction of p cross q.

Shortest distance between skew lines

For $r = a_{1} + t p$ and $r = a_{2} + s q$ ,

d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( p \times q ) ∣}{∣ p \times q ∣} = \frac{∣ [ a _{2} - a _{1} p q ] ∣}{∣ p \times q ∣}

Condition to intersect: the shortest distance is zero, that is

[a_{2} - a_{1} p q] = 0

which says $a_{2} - a_{1}$ , $p$ and $q$ are coplanar, so the two lines are coplanar.

Distance between parallel lines

For $r = a_{1} + t b$ and $r = a_{2} + s b$ ,

d = \frac{b \times ( a _{2} - a _{1} )}{b}

The skew formula fails here because $b \times b = 0$ makes the denominator zero, so this separate form is needed.

Example 3

Find the shortest distance between $r = \hat{i} + \hat{j} + t (2 \hat{i} - \hat{j} + \hat{k})$ and $r = 2 \hat{i} + \hat{j} - \hat{k} + s (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$ .

Solution. $a_{2} - a_{1} = \hat{i} - \hat{k}$ .

$p \times q = \hat{i} \hat{j} \hat{k} 2 - 11 3 - 52 = (- 2 + 5) \hat{i} - (4 - 3) \hat{j} + (- 10 + 3) \hat{k} = 3 \hat{i} - \hat{j} - 7 \hat{k}$ .

$(a_{2} - a_{1}) \cdot (p \times q) = 3 - 0 + 7 = 10$ and $∣ p \times q ∣ = 9 + 1 + 49 = 59$ .

Hence $d = \frac{10}{59}$ .

Equation of a plane

Standard forms

Through $r_{0}$ with normal $n$ : $(r - r_{0}) \cdot n = 0$ , that is $r \cdot n = d$ where $d = r_{0} \cdot n$

Normal form: $r \cdot \overset{n}{^} = p$ , where $p \geq 0$ is the perpendicular distance of the plane from the origin

Through three points $A (a), B (b), C (c)$ :

[r - a b - a c - a] = 0 or r \cdot (a \times b + b \times c + c \times a) = [a b c]

Parametric form through $a$ parallel to $b$ and $c$ : $r = a + s b + t c$

Cartesian form: $a x + b y + cz = d$ , where $a, b, c$ are direction ratios of the normal

Intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

A plane is fixed by one point on it and its normal direction. The distance of any external point

P

is measured along that normal:

N

is the foot of the perpendicular and

P^{'}

is the image of

P

, the same distance beyond the plane.

Distance, foot and image for a plane

For the plane $r \cdot n = d$ and a point $P (p)$ :

distance = \frac{∣ p \cdot n - d ∣}{∣ n ∣}

Foot of the perpendicular: $f = p - (\frac{p \cdot n - d}{∣ n ∣ ^{2}}) n$

Image of $P$ in the plane: $p^{'} = p - 2 (\frac{p \cdot n - d}{∣ n ∣ ^{2}}) n$

Angles involving planes

Between two planes with normals $n_{1}, n_{2}$ : $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣ ∣ n _{2} ∣}$

They are perpendicular if $n_{1} \cdot n_{2} = 0$ and parallel if $n_{1} \times n_{2} = 0$ .

Between a line of direction $b$ and a plane with normal $n$ : the angle with the plane is the complement of the angle with the normal, so

sin θ = \frac{b \cdot n}{b ∣ n ∣}

The line is parallel to the plane if $b \cdot n = 0$ , and additionally lies in the plane if a point of the line also satisfies the plane equation. The line is perpendicular to the plane if $b \times n = 0$ .

Sine or cosine?

Two lines: cosine. Two planes: cosine. A line and a plane: sine. The reason is that the plane is represented by its normal, so the angle you compute from the dot product is the angle with the normal, and the required angle is its complement.

Line of intersection of two planes

If $(r - a) \cdot n_{1} = 0$ and $(r - a) \cdot n_{2} = 0$ are two planes through the common point $a$ , their line of intersection is

r = a + λ (n_{1} \times n_{2})

The direction is $n_{1} \times n_{2}$ because the line lies in both planes and so is perpendicular to both normals.

Example 4

Find the equation of the plane through $A (1, 1, 0)$ , $B (1, 2, 1)$ and $C (- 2, 2, - 1)$ , and hence the distance of the origin from it.

Solution. $A B = \hat{j} + \hat{k}$ and $A C = - 3 \hat{i} + \hat{j} - \hat{k}$ .

$n = A B \times A C = \hat{i} \hat{j} \hat{k} 011 - 31 - 1 = (- 1 - 1) \hat{i} - (0 + 3) \hat{j} + (0 + 3) \hat{k} = - 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ .

With $a = \hat{i} + \hat{j}$ : $d = a \cdot n = - 2 - 3 = - 5$ , so the plane is $r \cdot (- 2 \hat{i} - 3 \hat{j} + 3 \hat{k}) = - 5$ , or $2 x + 3 y - 3 z = 5$ .

Distance from the origin $= \frac{∣0 - 5∣}{4 + 9 + 9} = \frac{5}{22}$ .

Reciprocal system of vectors

Two sets of non-coplanar vectors $a, b, c$ and $a^{'}, b^{'}, c^{'}$ form a reciprocal system if

a \cdot a^{'} = b \cdot b^{'} = c \cdot c^{'} = 1

together with every cross pairing being zero, for example $a \cdot b^{'} = 0$ .

Construction and properties

a^{'} = \frac{b \times c}{[ a b c ]}, b^{'} = \frac{c \times a}{[ a b c ]}, c^{'} = \frac{a \times b}{[ a b c ]}

$a \cdot a^{'} + b \cdot b^{'} + c \cdot c^{'} = 3$ and $a \times a^{'} + b \times b^{'} + c \times c^{'} = 0$

$[a^{'} b^{'} c^{'}] = \frac{1}{[ a b c ]}$ . The system $\hat{i}, \hat{j}, \hat{k}$ is its own reciprocal.

JEE Advanced

Resolving any vector in a non-orthogonal basis. If $a, b, c$ are non-coplanar, any $r$ can be written as

r = \frac{[ r b c ]}{[ a b c ]} a + \frac{[ r c a ]}{[ a b c ]} b + \frac{[ r a b ]}{[ a b c ]} c

and equivalently, using the reciprocal system,

r = (r \cdot a^{'}) a + (r \cdot b^{'}) b + (r \cdot c^{'}) c

This is the general version of the familiar $r = (r \cdot \hat{i}) \hat{i} + (r \cdot \hat{j}) \hat{j} + (r \cdot \hat{k}) \hat{k}$ , which is the special case where the basis is its own reciprocal.

Physical applications

The four standard formulas

Work done by a constant force $F$ through displacement $s$ : $W = F \cdot s$ (a scalar)

Moment (torque) of $F$ acting at $P$ about a point $O$ : $M = r \times F$ , where $r = O P$

Moment of a couple of equal and opposite forces $F$ and $- F$ applied at $r_{1}$ and $r_{2}$ :

$M = (r_{1} - r_{2}) \times F$ , which is independent of the choice of origin

Velocity in rotation: $v = ω \times r$ , with $ω$ the angular velocity vector

The moment of a force about O. Its magnitude equals the force times the perpendicular distance d from O to the line of action. Here M points into the page, marked by the crossed circle, since r and F both lie in the page.

JEE Advanced

Moment of a force about a line. The moment of $F$ about a line through $A$ with unit direction $\overset{u}{^}$ is the component of the moment about $A$ along that line:

M_{line} = (r \times F) \cdot \overset{u}{^} = [r F \overset{u}{^}]

where $r$ joins $A$ to any point on the line of action of $F$ . Being a box product, it vanishes exactly when $r$ , $F$ and $\overset{u}{^}$ are coplanar, which is the precise statement that a force whose line of action meets or is parallel to the axis produces no turning effect about it.

Example 5

Forces of magnitudes $5$ and $3$ act along $6 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ on a particle displaced from $(2, 2, - 1)$ to $(4, 3, 1)$ . Find the work done.

Solution. Both direction vectors have magnitude $7$ , so the resultant force is

$F = \frac{5}{7} (6 \hat{i} + 2 \hat{j} + 3 \hat{k}) + \frac{3}{7} (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) = \frac{1}{7} (39 \hat{i} + 4 \hat{j} + 33 \hat{k})$ .

The displacement is $d = 2 \hat{i} + \hat{j} + 2 \hat{k}$ , so

$W = F \cdot d = \frac{1}{7} (78 + 4 + 66) = \frac{148}{7}$ units.

Example 6

A force $3 \hat{i} + 2 \hat{k}$ acts through the point $5 \hat{i} + 4 \hat{j} - 3 \hat{k}$ . Find its moment about the point $\hat{i} + 3 \hat{j} + \hat{k}$ .

Solution. $r = (5 \hat{i} + 4 \hat{j} - 3 \hat{k}) - (\hat{i} + 3 \hat{j} + \hat{k}) = 4 \hat{i} + \hat{j} - 4 \hat{k}$ .

$M = r \times F = \hat{i} \hat{j} \hat{k} 41 - 4 302 = (2 - 0) \hat{i} - (8 + 12) \hat{j} + (0 - 3) \hat{k} = 2 \hat{i} - 20 \hat{j} - 3 \hat{k}$ .

Example 7

Find the moment of the couple formed by the forces $5 \hat{i} + \hat{k}$ and $- 5 \hat{i} - \hat{k}$ acting at $(9, - 1, 2)$ and $(3, - 2, 1)$ respectively.

Solution. Take $F = 5 \hat{i} + \hat{k}$ applied at $r_{1} = 9 \hat{i} - \hat{j} + 2 \hat{k}$ , with $- F$ at $r_{2} = 3 \hat{i} - 2 \hat{j} + \hat{k}$ .

$r_{1} - r_{2} = 6 \hat{i} + \hat{j} + \hat{k}$ , so

$M = (r_{1} - r_{2}) \times F = \hat{i} \hat{j} \hat{k} 611501 = (1 - 0) \hat{i} - (6 - 5) \hat{j} + (0 - 5) \hat{k} = \hat{i} - \hat{j} - 5 \hat{k}$ .

Mistakes that cost marks

Using the skew line formula on parallel lines. The cross product of the directions is zero, so you must switch to the parallel line formula.
Taking the angle between a line and a plane from the cosine. It comes from the sine, since the normal represents the plane.
Forgetting the modulus in the shortest distance formula and reporting a negative length.
Writing the plane through three points using only two of them. You need both $b - a$ and $c - a$ to build the normal.
Confusing the foot of the perpendicular with the image. The image is twice as far, so $p^{'} = 2 f - p$ .
Computing a moment with $F \times r$ . The order is $r \times F$ , and reversing it reverses the sense of rotation.
Using the position vector of the point of application measured from the wrong origin when a moment about a specific point is asked for.

Quick recap

Line: $r = a + t b$ ; distance of a point: $\frac{( p - a ) \times b}{b}$
Skew lines: $d = \frac{∣ [ a _{2} - a _{1} p q ] ∣}{∣ p \times q ∣}$ ; the lines meet exactly when this box product is zero
Plane: $r \cdot n = d$ ; distance of a point: $\frac{∣ p \cdot n - d ∣}{∣ n ∣}$
Angles: line to line and plane to plane use cosine, line to plane uses sine
Line of intersection of two planes runs along $n_{1} \times n_{2}$
Reciprocal system: $a^{'} = \frac{b \times c}{[ a b c ]}$ and its cyclic partners
$W = F \cdot s$ , $M = r \times F$ , $v = ω \times r$

Frequently asked questions

What exactly makes two lines skew?

They are skew if they neither intersect nor are parallel, which is only possible in three dimensions. The algebraic test is that $p \times q \neq = 0$ , so they are not parallel, while $[a_{2} - a_{1} p q] \neq = 0$ , so they are not coplanar and therefore cannot meet.

Why does the line and plane angle use sine instead of cosine?

A plane is represented by its normal, so the dot product between the line direction and the normal gives the angle with the normal. The angle with the plane itself is the complement of that, and the cosine of the complement is the sine of the angle you want.

How do I find the shortest distance if the two lines turn out to be parallel?

Switch formulas. The skew line expression has $∣ p \times q ∣$ in the denominator, which is zero for parallel lines. Use $d = \frac{b \times ( a _{2} - a _{1} )}{b}$ instead, which is just the distance of a point on one line from the other line.

What is the point of the reciprocal system?

It lets you read off components in a basis that is not orthogonal. Dotting a vector with the reciprocal basis vectors gives its coefficients directly, exactly as dotting with $\hat{i}, \hat{j}, \hat{k}$ does in the usual case. The same idea underlies reciprocal lattices in crystallography.

Does the moment of a couple depend on the point about which it is taken?

No, and that is its defining feature. The moment works out to $(r_{1} - r_{2}) \times F$ , an expression containing only the separation of the two points of application, so the origin cancels out entirely.

Previous year questions on Applications of Vectors

12 questions from past papers, each with a step-by-step solution.

Show all 12 questions

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