Vector Algebra

The chapter moves in a deliberate sequence, and each stage depends on the one before it. It opens with basic concepts — the definition of a vector as a directed line segment, the position vector of a point, and the direction cosines and direction ratios that pin a vector's orientation to the coordinate axes (Sections 10.2 and part of 10.5). Next come the types of vectors — zero, unit, coinitial, collinear, equal, negative — vocabulary you will use in every proof that follows (Section 10.3). The algebra proper begins with addition and scalar multiplication (Sections 10.4–10.5): the triangle law and the equivalent parallelogram law for adding two vectors, the properties of commutativity and associativity, multiplication by a scalar, and the resulting component form that turns every vector into an ordered triple of numbers. Inside this stretch sits the section formula (Section 10.5.3), which finds the position vector of a point dividing a segment in a given ratio — the single most reused tool from this chapter in the coordinate geometry that follows. The chapter closes with products of vectors (Section 10.6): the scalar (dot) product, which yields a number and underlies projection and the angle between two vectors, and the vector (cross) product, which yields a new vector and underlies area calculations, torque, and the normal to a plane. The worked examples and derivations below follow this same order — concepts and components first, then the two products, each proved from first principles and then drilled with board- and JEE/NEET-level problems.

Vectors, magnitude and position vectors

A directed line segment $\overrightarrow{AB}$, with initial point $A$ and terminal point $B$, is a vector: it carries a magnitude (the length $AB$) and a direction (the way the arrowhead points). Vectors are written with an arrow, $\vec{a}$, or in bold, and the magnitude of $\vec{a}$ is written $|\vec{a}|$. Since a length is never negative, the expression $|\vec{a}| < 0$ has no meaning — magnitude is always $\ge 0$, with equality only for the zero vector. Fixing a right-handed coordinate system with origin $O$, the position vector of a point $P(x, y, z)$ is the vector $\overrightarrow{OP}$ from the origin to $P$. Using the distance formula, its magnitude follows immediately:

where: $x, y, z$ are the coordinates of $P$, and the formula is simply the three-dimensional Pythagoras theorem applied twice — once in the horizontal plane, once lifting up to $P$. Position vectors of points $A, B, C, \ldots$ are conventionally written $\vec{a}, \vec{b}, \vec{c}, \ldots$ (or $\vec{r}_A$ and so on), a shorthand used throughout this chapter and the next. Throughout this chapter, vectors are treated as free vectors: a vector may be translated parallel to itself, without changing its magnitude or direction, and it still represents the same vector. This is why "vector $\vec{a}$" makes sense as an object in its own right, independent of where you choose to draw it.

Direction cosines and direction ratios

For the position vector $\vec{r}$ of a point $P(x, y, z)$, let $\alpha, \beta, \gamma$ be the angles $\vec{r}$ makes with the positive $x$, $y$ and $z$ axes respectively — the direction angles. Their cosines are the direction cosines of $\vec{r}$, usually written $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$. Because $\vec{r}$ makes a right-angled triangle with each axis, $l = x/r$, $m = y/r$, $n = z/r$ where $r = |\vec{r}|$, and consequently

where: $l, m, n$ are the direction cosines and this identity is simply $x^2+y^2+z^2 = r^2$ divided through by $r^2$ — a fact every vector's direction cosines obey, with no exceptions. Any three numbers $a, b, c$ proportional to $l, m, n$ (that is, $a = \lambda l$, $b=\lambda m$, $c=\lambda n$ for some nonzero $\lambda$) are called direction ratios of the vector; the scalar components $a_1, a_2, a_3$ of a vector in component form always serve as one valid set of direction ratios. Unlike direction cosines, direction ratios need not satisfy $a^2+b^2+c^2=1$ — mixing this up with equation (2) is one of the chapter's most common slips.

Types of vectors

Before building the algebra, the chapter fixes vocabulary for special vectors and for relationships between them. The table collects every term the NCERT text defines.

Two collinear vectors are not automatically equal (they may differ in magnitude, or point opposite ways along the same line), and two vectors of equal magnitude are not automatically collinear (they may point in entirely different directions) — a pair of true/false traps NCERT sets deliberately, and JEE still enjoys asking.

Addition of vectors: the triangle and parallelogram laws

A vector $\overrightarrow{AB}$ is simply the displacement from $A$ to $B$. If a girl walks from $A$ to $B$ and then from $B$ to $C$, her net displacement is the vector $\overrightarrow{AC}$, giving the triangle law of vector addition:

where: the rule is to place vectors so the terminal point of one coincides with the initial point of the next; the sum is the vector from the very first initial point to the very last terminal point. Taking all three sides of a triangle in order, $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA} = \vec{0}$, since the path returns to its own start. When two vectors $\vec{a}, \vec{b}$ share an initial point instead — the natural setup for, say, two forces acting at a point — they are added by the equivalent parallelogram law: place $\vec{a}$ and $\vec{b}$ as adjacent sides of a parallelogram, and their sum $\vec{a}+\vec{b}$ is the diagonal through their common point. The two laws agree because shifting $\vec{b}$ to start where $\vec{a}$ ends (to apply the triangle law) is exactly what constructing the parallelogram does. Vector addition inherits two familiar properties, both provable from the parallelogram/triangle picture: it is commutative, $\vec{a}+\vec{b} = \vec{b}+\vec{a}$, and associative, $(\vec{a}+\vec{b})+\vec{c} = \vec{a}+(\vec{b}+\vec{c})$, so a sum of three or more vectors can be written without brackets. The zero vector is the additive identity, $\vec{a}+\vec{0}=\vec{a}$, and every vector has an additive inverse, $\vec{a}+(-\vec{a}) = \vec{0}$, which is exactly how subtraction is defined: $\vec{a}-\vec{b} := \vec{a}+(-\vec{b})$.

Multiplication by a scalar and the component form

For a vector $\vec{a}$ and a real number $\lambda$ (a scalar), the product $\lambda\vec{a}$ is a vector collinear with $\vec{a}$: same direction if $\lambda > 0$, opposite direction if $\lambda < 0$, and magnitude

where: $|\lambda|$ is the absolute value of the scalar. Taking $\lambda = -1$ gives the negative $-\vec{a}$ met above; taking $\lambda = 1/|\vec{a}|$ (valid whenever $\vec{a} \ne \vec{0}$) gives the unit vector in the direction of $\vec{a}$, written $\hat{a} = \vec{a}/|\vec{a}|$ — the standard first move whenever a problem asks for "a vector in the direction of" something. Taking unit vectors $\hat{i}, \hat{j}, \hat{k}$ along the $x$, $y$, $z$ axes, any position vector resolves as $\overrightarrow{OP} = x\hat{i}+y\hat{j}+z\hat{k}$, and more generally every vector in space can be written in component form $\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, with magnitude $|\vec{a}| = \sqrt{a_1^2+a_2^2+a_3^2}$ by two applications of Pythagoras. Once vectors are components, the whole toolkit of coordinate algebra opens up: for $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$,

where: $\vec{a}=\vec{b}$ exactly when $a_1=b_1$, $a_2=b_2$, $a_3=b_3$ — componentwise equality, the tool behind Example 4-style "find $x, y, z$" questions. Two vectors given in component form are collinear exactly when one is a scalar multiple of the other, which reduces to the ratio test $\dfrac{b_1}{a_1}=\dfrac{b_2}{a_2}=\dfrac{b_3}{a_3}$ whenever no $a_i$ is zero — a fast collinearity check used constantly in coordinate geometry and worth memorising alongside the cross-product test met later.

Vector joining two points and the section formula

If $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ are two points, the vector from $P_1$ to $P_2$ follows by subtracting position vectors (using the triangle law through the origin):

where: the magnitude formula is exactly the familiar distance formula between two points — vector algebra and coordinate geometry are, at this level, the same statement in two notations. A point $R$ that divides the segment $PQ$, with position vectors $\vec{p}$ and $\vec{q}$, in the ratio $m:n$ has position vector given by the section formula, proved in full in the first derivation below:

where: internal division means $R$ lies between $P$ and $Q$; external division means $R$ lies on the line $PQ$ extended, outside the segment. Setting $m=n$ in the internal formula gives the midpoint of $PQ$ as $\vec{r} = (\vec{p}+\vec{q})/2$, the single most reused special case of this formula.

The scalar (dot) product

For two nonzero vectors $\vec{a}, \vec{b}$ with angle $\theta$ between them ($0 \le \theta \le \pi$), the scalar product is defined as

where: the result is always a real number (hence "scalar" product), never a vector; if either vector is $\vec{0}$, the angle $\theta$ is undefined and the product is defined to be $0$ by convention. Two immediate consequences drive most of the chapter's applications: $\vec{a}\cdot\vec{a} = |\vec{a}|^2\cos 0 = |\vec{a}|^2$, so the dot product of a vector with itself recovers the square of its length; and $\vec{a}\cdot\vec{b}=0$ for two nonzero vectors exactly when $\cos\theta = 0$, that is, exactly when $\vec{a} \perp \vec{b}$ — the algebraic test for perpendicularity used everywhere from this chapter onward. For the standard basis, $\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1$ and $\hat{i}\cdot\hat{j}=\hat{j}\cdot\hat{k}=\hat{k}\cdot\hat{i}=0$, since each pair is either parallel unit vectors ($\theta=0$) or perpendicular unit vectors ($\theta=\pi/2$). The dot product is commutative, $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}$, and distributes over addition, $\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}$; combined with the basis relations above, these two facts are exactly what is needed to derive the component formula (first derivation below):

where: the second equation, obtained by solving (8) for $\theta$, is the standard method for finding the angle between two vectors given in component form. A closely related idea is projection: the projection of $\vec{a}$ on a directed line (or on a vector $\vec{b}$) is the scalar $\vec{a}\cdot\hat{b} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$, the "shadow length" $\vec{a}$ casts along that direction; multiplying this scalar by $\hat{b}$ gives the full projection vector.

The vector (cross) product and area

For two nonzero, non-parallel vectors $\vec{a}, \vec{b}$ with angle $\theta$ between them, the vector product is defined as

where: $\hat{n}$ is the unit vector perpendicular to both $\vec{a}$ and $\vec{b}$, chosen so that $\vec{a}, \vec{b}, \hat{n}$ form a right-handed system — curl the fingers of the right hand from $\vec{a}$ toward $\vec{b}$ and the thumb points along $\hat{n}$. The result is a genuine vector (hence "vector product"), not a number. As with the dot product, $\vec{a}\times\vec{b} := \vec{0}$ by convention if either vector is $\vec{0}$, and for any two nonzero vectors, $\vec{a}\times\vec{b} = \vec{0}$ exactly when $\sin\theta = 0$, that is, exactly when $\vec{a}$ and $\vec{b}$ are collinear (parallel or anti-parallel) — the cross-product test for collinearity, and in particular $\vec{a}\times\vec{a}=\vec{0}$ always. The single most important — and most examined — fact about this product is that it is not commutative: reversing the order reverses the direction of $\hat{n}$ (curling the fingers the other way flips the thumb), so

For the standard basis, $\hat{i}\times\hat{i}=\hat{j}\times\hat{j}=\hat{k}\times\hat{k}=\vec{0}$ and, cycling through the right-handed order $\hat{i}\to\hat{j}\to\hat{k}\to\hat{i}$, $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, $\hat{k}\times\hat{i}=\hat{j}$ (with the reverse order picking up a minus sign, e.g. $\hat{j}\times\hat{i}=-\hat{k}$). Expanding $(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\times(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$ term by term with these relations — carried out in full in the third derivation below — collapses into the compact determinant form

where: expanding this determinant along the top row reproduces the formula $\vec{a}\times\vec{b} = (a_2b_3-a_3b_2)\hat{i} - (a_1b_3-a_3b_1)\hat{j} + (a_1b_2-a_2b_1)\hat{k}$ — the practical recipe used in every worked problem below. Geometrically, $|\vec{a}\times\vec{b}|$ is exactly the area of the parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$: the base has length $|\vec{a}|$ and the perpendicular height is $|\vec{b}|\sin\theta$, so base $\times$ height reproduces $|\vec{a}||\vec{b}|\sin\theta$. Halving the parallelogram gives the area of the triangle on the same two sides:

where: for a triangle with vertices $A$, $B$, $C$, take $\vec{a}=\overrightarrow{AB}$ and $\vec{b}=\overrightarrow{AC}$ as the two sides meeting at $A$ — this is the standard setup for every "area of a triangle with given vertices" question, applied in the third derivation and in the worked examples.

Board level — a unit vector and a vector of prescribed magnitude (extends NCERT Example 6)

Given $\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}$, find (a) the unit vector in the direction of $\vec{a}$, and (b) a vector of magnitude $7$ units in the same direction.

Step 1. Find $|\vec{a}|$. Here $a_1=2$, $a_2=3$, $a_3=1$, so $|\vec{a}| = \sqrt{2^2+3^2+1^2} = \sqrt{4+9+1} = \sqrt{14}$.

Step 2. Divide $\vec{a}$ by its own magnitude to get the unit vector: $\hat{a} = \dfrac{\vec{a}}{|\vec{a}|} = \dfrac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}} = \dfrac{2}{\sqrt{14}}\hat{i}+\dfrac{3}{\sqrt{14}}\hat{j}+\dfrac{1}{\sqrt{14}}\hat{k}$.

Step 3. Scale $\hat{a}$ to the required magnitude. A vector of magnitude $7$ in the direction of $\vec{a}$ is $7\hat{a} = \dfrac{7}{\sqrt{14}}\big(2\hat{i}+3\hat{j}+\hat{k}\big) = \dfrac{14}{\sqrt{14}}\hat{i}+\dfrac{21}{\sqrt{14}}\hat{j}+\dfrac{7}{\sqrt{14}}\hat{k}$, which simplifies (since $7/\sqrt{14} = \sqrt{14}/2$) to $\sqrt{14}\hat{i}+\dfrac{3\sqrt{14}}{2}\hat{j}+\dfrac{\sqrt{14}}{2}\hat{k}$.

Answer: the unit vector is $\hat{a} = \dfrac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}}$, and the vector of magnitude $7$ in the same direction is $\dfrac{7}{\sqrt{14}}\big(2\hat{i}+3\hat{j}+\hat{k}\big)$.

Board level — the section formula, internal and external division (extends NCERT Example 11)

Points $P$ and $Q$ have position vectors $\vec{p} = \hat{i}+2\hat{j}-\hat{k}$ and $\vec{q} = -\hat{i}+\hat{j}+\hat{k}$. Find the position vector of the point $R$ dividing $PQ$ in the ratio $2:1$, (a) internally and (b) externally.

Step 1. Apply the internal section formula with $m=2$, $n=1$: $\vec{r} = \dfrac{m\vec{q}+n\vec{p}}{m+n} = \dfrac{2\vec{q}+\vec{p}}{3}$.

Step 2. Substitute components. $2\vec{q} = -2\hat{i}+2\hat{j}+2\hat{k}$, so $2\vec{q}+\vec{p} = (-2+1)\hat{i}+(2+2)\hat{j}+(2-1)\hat{k} = -\hat{i}+4\hat{j}+\hat{k}$. Dividing by $3$ gives $\vec{r} = -\dfrac{1}{3}\hat{i}+\dfrac{4}{3}\hat{j}+\dfrac{1}{3}\hat{k}$.

Step 3. Apply the external section formula with $m=2$, $n=1$: $\vec{r} = \dfrac{m\vec{q}-n\vec{p}}{m-n} = \dfrac{2\vec{q}-\vec{p}}{1} = 2\vec{q}-\vec{p}$.

Step 4. Substitute components. $2\vec{q}-\vec{p} = (-2-1)\hat{i}+(2-2)\hat{j}+(2+1)\hat{k} = -3\hat{i}+0\hat{j}+3\hat{k}$.

Answer: internally, $R$ has position vector $-\tfrac{1}{3}\hat{i}+\tfrac{4}{3}\hat{j}+\tfrac{1}{3}\hat{k}$; externally, $R$ has position vector $-3\hat{i}+3\hat{k}$.

Competitive level (JEE / NEET) — collinearity and area from the cross product (extends NCERT Example 21 and Example 24)

(a) Determine whether the points $A(1,2,7)$, $B(2,6,3)$ and $C(3,10,-1)$ are collinear. (b) A separate triangle has vertices $A(1,1,2)$, $B(2,3,5)$, $C(1,5,5)$; confirm its area using the cross product, cross-checked against the derivation above.

Step 1. For part (a), form two vectors from the shared vertex $A$: $\overrightarrow{AB} = (2-1)\hat{i}+(6-2)\hat{j}+(3-7)\hat{k} = \hat{i}+4\hat{j}-4\hat{k}$, and $\overrightarrow{AC} = (3-1)\hat{i}+(10-2)\hat{j}+(-1-7)\hat{k} = 2\hat{i}+8\hat{j}-8\hat{k}$.

Step 2. Spot the fast test before computing a full determinant: $\overrightarrow{AC} = 2\overrightarrow{AB}$ exactly (each component of $\overrightarrow{AC}$ is double the corresponding component of $\overrightarrow{AB}$), so $\overrightarrow{AB}\times\overrightarrow{AC} = \overrightarrow{AB}\times(2\overrightarrow{AB}) = 2(\overrightarrow{AB}\times\overrightarrow{AB}) = \vec{0}$. Since the cross product vanishes, $A$, $B$, $C$ are collinear — they do not form a genuine triangle, and "area" for this triple is $0$.

Step 3. For part (b), reuse the vectors from the third derivation: $\overrightarrow{AB} = \hat{i}+2\hat{j}+3\hat{k}$, $\overrightarrow{AC} = 4\hat{j}+3\hat{k}$, and $\overrightarrow{AB}\times\overrightarrow{AC} = -6\hat{i}-3\hat{j}+4\hat{k}$.

Step 4. Take the magnitude: $|\overrightarrow{AB}\times\overrightarrow{AC}| = \sqrt{36+9+16} = \sqrt{61}$, so the area is $\tfrac{1}{2}\sqrt{61}$.

Answer: $A$, $B$, $C$ in part (a) are collinear (no triangle, area $0$); the triangle in part (b) has area $\dfrac{\sqrt{61}}{2}$ square units. Timing note: under ninety seconds for part (a) if you spot the component ratio first — $2/1 = 8/4 = -8/-4 = 2$ — rather than grinding through the full $3\times3$ determinant only to discover it equals zero. JEE frequently plants a collinear triple as a disguised "find the area" question specifically to reward students who check the ratio test from equation (5) before reaching for the cross product. Always test collinearity by inspection first; only compute the determinant in full once you know the answer is nonzero.

• Direction cosines versus direction ratios. Direction cosines always satisfy $l^2+m^2+n^2=1$; direction ratios are merely proportional to them and essentially never satisfy that identity themselves. Never write "direction cosines" when you mean the raw components $a_1, a_2, a_3$ of a non-unit vector.
• Sign in the section formula. Internal division is $\dfrac{m\vec{q}+n\vec{p}}{m+n}$; external division swaps one sign to $\dfrac{m\vec{q}-n\vec{p}}{m-n}$. Writing a plus where a minus belongs (or vice versa) is the most common numerical slip in section-formula questions.
• $\vec{a}\cdot\vec{b}=0$ needs both vectors nonzero. The zero vector is "perpendicular" to everything by convention, but $\vec{a}\cdot\vec{b}=0$ only proves $\vec{a}\perp\vec{b}$ when it is already known that neither vector is $\vec{0}$ — check this before concluding perpendicularity.
• The cross product is not commutative. $\vec{a}\times\vec{b} = -(\vec{b}\times\vec{a})$; dropping the sign when reordering a cross product silently flips the answer's direction, and in area problems (where only the magnitude is used) this error is invisible, so it is worth verifying order carefully whenever direction matters.
• Confusing $|\vec{a}|<0$ as meaningful. Magnitude is a length and is always $\ge 0$; there is no such thing as a "negative magnitude," only a negative scalar multiplying a vector.
• Collinear vectors are not automatically equal vectors, and vectors of equal magnitude are not automatically collinear — equality demands both same magnitude and same direction.
• Forgetting the triangle-versus-parallelogram area factor of $\tfrac{1}{2}$. $|\vec{a}\times\vec{b}|$ is the parallelogram's area; halve it for the triangle on the same two sides — a frequent one-mark-lost slip under time pressure.

• State the definition of a vector, and clearly separate its magnitude $|\vec{a}|$ from its direction.
• Write any vector in component form $a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, compute its magnitude, and find its direction cosines and a valid set of direction ratios.
• Distinguish zero, unit, coinitial, collinear, equal and negative vectors, and correctly judge true/false statements mixing these terms.
• Add and subtract vectors using the triangle law or the parallelogram law, and multiply a vector by a scalar, including finding a unit vector or a vector of prescribed magnitude in a given direction.
• Apply the section formula, both internal and external, and its midpoint special case, to find the position vector of a dividing point.
• Test two vectors for collinearity using either the ratio-of-components test or the vanishing cross product.
• Compute $\vec{a}\cdot\vec{b}$ from components, find the angle between two vectors, and test for perpendicularity.
• Compute $\vec{a}\times\vec{b}$ as a $3\times 3$ determinant, and use its magnitude to find the area of a triangle or a parallelogram.
• Prove and correctly apply the Cauchy–Schwarz inequality $|\vec{a}\cdot\vec{b}|\le|\vec{a}||\vec{b}|$ and the triangle inequality $|\vec{a}+\vec{b}|\le|\vec{a}|+|\vec{b}|$.

Ask the tutor: if you want to see how these same section-formula and cross-product tools reappear to find lines, planes, and the shortest distance between them, ask for a walkthrough of three-dimensional geometry next.