Vector operations

A vector carries two pieces of information at once — how much and which way — and everything in Class 12 vector algebra, from the dot product to the equation of a line in space, is built on top of the small toolkit covered here: writing a vector in component form using $\hat i, \hat j, \hat k$ , finding its magnitude and direction, and combining two or more vectors by addition, subtraction, and scalar multiplication. This is the opening theme of the optional Vector Algebra unit (Section B), and it is deliberately computational — proofs of geometry theorems using vectors sit outside the syllabus. Get comfortable here first: direction cosines, position vectors, and the scalar-multiple test for collinear points reappear constantly once products of vectors are introduced.

The core insight

A vector in component form $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$ is added, subtracted, and scaled term-by-term, its magnitude is $|\vec a| = \sqrt{a_1^2+a_2^2+a_3^2}$ , and its direction is fixed by direction cosines $l,m,n$ satisfying $l^2+m^2+n^2=1$ .

The core idea

A vector $\vec a$ in three dimensions is written in component form using the unit vectors $\hat i, \hat j, \hat k$ along the $x$ , $y$ , $z$ axes:

\vec a = a_1\hat i + a_2\hat j + a_3\hat k.

Its magnitude — a nonnegative scalar, never a vector — is

|\vec a| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

where: $a_1, a_2, a_3$ are the scalar components of $\vec a$ along the three axes.

Direction cosines and ratios. The direction cosines $l, m, n$ of $\vec a$ are the cosines of the angles it makes with the $x$ , $y$ , $z$ axes, so $l = a_1/|\vec a|$ , $m = a_2/|\vec a|$ , $n = a_3/|\vec a|$ , and they always satisfy $l^2+m^2+n^2=1$ (proved below). Any numbers merely proportional to $l, m, n$ — that is, proportional to $a_1, a_2, a_3$ — are direction ratios; they fix the direction but need not satisfy the identity.

Special vectors. $\vec a$ is a unit vector if $|\vec a| = 1$ , and the unit vector along any nonzero $\vec a$ is $\hat a = \vec a/|\vec a|$ . The zero vector $\vec 0$ has zero magnitude and no defined direction. Two vectors are equal only if they share both magnitude and direction. Two nonzero vectors are parallel (or collinear) if one is a scalar multiple of the other: $\vec b = k\vec a$ .

Points to vectors. If $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ are two points, $\vec{PQ} = (x_2-x_1)\hat i + (y_2-y_1)\hat j + (z_2-z_1)\hat k$ ; note $\vec{PQ} = -\vec{QP}$ . The position vector of $P$ is $\vec{OP}$ , measured from the origin $O$ .

Addition, subtraction, scalar multiplication all act component-wise: for $\vec a = a_1\hat i+a_2\hat j+a_3\hat k$ and $\vec b = b_1\hat i+b_2\hat j+b_3\hat k$ ,

\vec a \pm \vec b = (a_1\pm b_1)\hat i + (a_2\pm b_2)\hat j + (a_3\pm b_3)\hat k, \qquad k\vec a = ka_1\hat i + ka_2\hat j + ka_3\hat k.

where: $k$ is a real scalar; $k\vec a$ scales the magnitude by $|k|$ and flips the direction when $k<0$ . The negative $-\vec a$ has the same magnitude as $\vec a$ but points the opposite way.

Figure

Two vectors a and b added tip to tail to give the resultant vector a plus b by the triangle law

Derivation AWhy the direction cosines satisfy l squared plus m squared plus n squared equals 1

The goal is to show, starting only from the definitions $l=a_1/|\vec a|$ , $m=a_2/|\vec a|$ , $n=a_3/|\vec a|$ and the magnitude formula $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$ , that these three ratios always satisfy the identity $l^2+m^2+n^2=1$ for any nonzero vector $\vec a$ , which is why the identity can be used as a quick check on any set of direction cosines a student computes.

Step 1. Write each direction cosine as a ratio of a component to the magnitude: $l = \dfrac{a_1}{|\vec a|}$ , $m = \dfrac{a_2}{|\vec a|}$ , $n = \dfrac{a_3}{|\vec a|}$ . These are exactly the definitions given above — nothing new is assumed yet.

Step 2. Square each ratio and add them together: $l^2+m^2+n^2 = \dfrac{a_1^2}{|\vec a|^2} + \dfrac{a_2^2}{|\vec a|^2} + \dfrac{a_3^2}{|\vec a|^2} = \dfrac{a_1^2+a_2^2+a_3^2}{|\vec a|^2}$ . All three fractions share the same denominator $|\vec a|^2$ , so they combine into a single fraction.

Step 3. Substitute the magnitude formula. Since $|\vec a|^2 = a_1^2+a_2^2+a_3^2$ by definition, the numerator and denominator of the fraction from Step 2 are identical: $l^2+m^2+n^2 = \dfrac{a_1^2+a_2^2+a_3^2}{a_1^2+a_2^2+a_3^2} = 1$ .

Step 4. Read the result. The identity holds for every nonzero vector regardless of its individual components, which is why it is used as a built-in consistency check: if a computed set of $l, m, n$ does not satisfy $l^2+m^2+n^2=1$ , an arithmetic error has been made somewhere upstream.

Worked example

Let $\vec a = 2\hat i - \hat j + 3\hat k$ and $\vec b = -\hat i + 4\hat j - 2\hat k$ , the same vectors used to introduce this topic. Find $\vec a - 2\vec b$ , the magnitude and direction cosines of $\vec a$ , and then use those results to test whether $\vec a$ is parallel to $\vec c = -4\hat i + 2\hat j - 6\hat k$ .

Step 1 — subtract a scalar multiple. $2\vec b = -2\hat i + 8\hat j - 4\hat k$ , so $\vec a - 2\vec b = (2-(-2))\hat i + (-1-8)\hat j + (3-(-4))\hat k = 4\hat i - 9\hat j + 7\hat k$ .

Step 2 — magnitude and direction cosines of $\vec a$ . $|\vec a| = \sqrt{2^2+(-1)^2+3^2} = \sqrt{14}$ , so $l = \dfrac{2}{\sqrt{14}}$ , $m = \dfrac{-1}{\sqrt{14}}$ , $n = \dfrac{3}{\sqrt{14}}$ . Check: $l^2+m^2+n^2 = \dfrac{4+1+9}{14} = 1$ , confirming the identity from the derivation.

Step 3 — test $\vec a$ against $\vec c$ for a scalar multiple. $\vec c = -4\hat i+2\hat j-6\hat k = -2(2\hat i-\hat j+3\hat k) = -2\vec a$ . Since $\vec c$ is a scalar multiple of $\vec a$ with $k=-2$ , the two vectors are parallel (in fact anti-parallel, since $k<0$ ).

Answer: $\vec a - 2\vec b = 4\hat i - 9\hat j + 7\hat k$ ; $|\vec a| = \sqrt{14}$ with direction cosines $\left(\dfrac{2}{\sqrt{14}}, \dfrac{-1}{\sqrt{14}}, \dfrac{3}{\sqrt{14}}\right)$ ; and $\vec c = -2\vec a$ , so $\vec a \parallel \vec c$ .

Common traps

Treating $\vec a$ and $|\vec a|$ as interchangeable — one is a vector, the other a nonnegative scalar; never add a vector to a number.
Reporting direction ratios as if they were direction cosines without dividing by $|\vec a|$ — only direction cosines satisfy $l^2+m^2+n^2=1$ .
Getting the order wrong in $\vec{PQ}$ : it is final point minus initial point, and $\vec{PQ} = -\vec{QP}$ .
Assuming equal magnitude means equal vectors — direction must match too, and for collinearity the scalar multiple must be checked component-by-component, not just approximately.

Ask the tutor: if you want to see how this same direction-cosine and collinearity toolkit extends once the dot and cross products are introduced, ask the AI tutor to walk through a mixed problem.

What is actually the difference between direction cosines and direction ratios?

Direction cosines are the specific numbers $l = a_1/|\vec a|$ , $m = a_2/|\vec a|$ , $n = a_3/|\vec a|$ obtained by dividing each component by the vector's magnitude, and they always satisfy $l^2+m^2+n^2=1$ . Direction ratios are any three numbers proportional to $a_1, a_2, a_3$ — for instance the raw components themselves — that fix the same direction but are not normalised, so they generally do not satisfy that identity. If a question just says find the direction, giving direction ratios is fine unless it specifically asks for direction cosines.

Why does the scalar-multiple test prove three points are collinear?

Two vectors are parallel exactly when one is a scalar multiple of the other, because scaling a vector only changes its length (and possibly flips it), never its line of action. If $\vec{AB} = k\,\vec{AC}$ for some scalar $k$ , then $B$ and $C$ both lie on the line through $A$ in the direction of that common vector, so all three points share one line — that is the definition of collinear. The test fails to prove collinearity if the ratios of corresponding components are not all equal, since then no single $k$ works.

Do I need to memorise the direction cosine identity proof for the exam?

The proof itself is short enough that it is worth knowing it can be derived, but most questions simply ask you to compute $l, m, n$ from a given vector and use $l^2+m^2+n^2=1$ as a self-check, not to reproduce the derivation. Where it does matter is in mixed problems that give you two of $l, m, n$ and ask for the third — there you substitute into the identity and solve, which only works if you trust it always equals exactly 1.

Vector operations

The core idea

Worked example

Common traps

Before this

Where this leads