Vectors for JEE: Dot, Cross and Triple Products with Every 2026 PYQ

A vector carries two pieces of information at once, a size and a direction. That is why a single vector equation can say what three ordinary equations say, and why so many JEE questions that look like heavy coordinate work collapse in two lines once they are written with vectors.

This note builds the chapter in the order the methods depend on each other: what a vector is, components and direction cosines, the dot product, vectors in the plane of two given vectors, the cross product and vector equations, the scalar triple product, the vector triple product, and geometry done with vectors. Every JEE Main 2026 vector question is here, placed where its method lives, along with a JEE Advanced 2026 question, and every answer was re-derived from scratch.

What a vector is

A scalar has magnitude only (length, mass, work). A vector is a quantity with magnitude and direction. It is drawn as a directed segment $\overrightarrow{AB}$, from the initial point $A$ to the terminal point $B$, and its magnitude $\lvert\overrightarrow{AB}\rvert$ is the length $AB$.

Two vectors are added by the triangle law: place the tail of $\vec b$ at the head of $\vec a$, and $\vec a+\vec b$ runs from the tail of $\vec a$ to the head of $\vec b$. The parallelogram law says the same thing: $\vec a+\vec b$ is the diagonal of the parallelogram on $\vec a$ and $\vec b$.

Left: triangle law with an and b placed head to tail and a plus b closing the triangle. Right: parallelogram on a and b with the diagonal a plus b from the common tail

Addition is commutative and associative, and $\vec a-\vec b=\vec a+(-\vec b)$. Multiplying by a scalar $k$ scales the length by $\lvert k\rvert$ and reverses the direction when $k<0$.

Position vectors and the section formula

Fix an origin $O$. The position vector of a point $A$ is $\vec a=\overrightarrow{OA}$. Then every segment is a difference of position vectors:

$$\overrightarrow{AB}=\vec b-\vec a$$

If $P$ divides $AB$ internally in the ratio $m:n$, then

$$\vec p=\frac{m\vec b+n\vec a}{m+n}$$

For external division replace $n$ by $-n$. The midpoint is $\dfrac{\vec a+\vec b}{2}$ and the centroid of a triangle is $\dfrac{\vec a+\vec b+\vec c}{3}$.

Segment AB with the point P dividing it in the ratio m to n, and the three position vectors a, p, b drawn from the origin O

Proof. $\overrightarrow{AP}:\overrightarrow{PB}=m:n$ and both point the same way, so $n\,\overrightarrow{AP}=m\,\overrightarrow{PB}$:

$$n(\vec p-\vec a)=m(\vec b-\vec p)$$ $$(m+n)\vec p=m\vec b+n\vec a$$

Collinear points. $A$, $B$, $C$ are collinear exactly when $\overrightarrow{AB}=\lambda\overrightarrow{AC}$ for some $\lambda$. Equivalently, there are scalars $x$, $y$, $z$, not all zero, with

$$x\vec a+y\vec b+z\vec c=\vec 0,\qquad x+y+z=0$$

Example 1 (JEE Main 2026, 5 April Shift 2). Let $O$ be the origin, $\overrightarrow{OP}=\vec a$ and $\overrightarrow{OQ}=\vec b$. If $R$ is the point on $OP$ with $\overrightarrow{OP}=5\overrightarrow{OR}$, and $M$ is the point with $\overrightarrow{OQ}=5\overrightarrow{RM}$, then $\overrightarrow{PM}$ is

(1) $\dfrac15(\vec a-4\vec b)$ (2) $\dfrac15(\vec b-4\vec a)$ (3) $\dfrac15(-\vec a+4\vec b)$ (4) $\dfrac15(-\vec b+4\vec a)$

▸ SHOW SOLUTION

Step 1. Position vectors of R and M

$$\overrightarrow{OR}=\frac{\vec a}{5},\qquad \overrightarrow{RM}=\frac{\vec b}{5}$$$$\overrightarrow{OM}=\overrightarrow{OR}+\overrightarrow{RM}=\frac{\vec a+\vec b}{5}$$

Step 2. Subtract

$$\overrightarrow{PM}=\overrightarrow{OM}-\overrightarrow{OP}=\frac{\vec a+\vec b}{5}-\vec a$$$$\overrightarrow{PM}=\frac15(\vec b-4\vec a)$$

Answer: (2)

Components, magnitude and direction cosines

In the plane every vector is $x\hat i+y\hat j$. In space, with unit vectors $\hat i$, $\hat j$, $\hat k$ along the axes, every vector is

$$\vec a=a_1\hat i+a_2\hat j+a_3\hat k$$ $$\lvert\vec a\rvert=\sqrt{a_1^2+a_2^2+a_3^2}$$

Vectors are added, subtracted and scaled component by component, and two vectors are equal exactly when their components are equal.

If $\vec a$ makes angles $\alpha$, $\beta$, $\gamma$ with the positive $x$, $y$, $z$ axes, then

$$\cos\alpha=\frac{a_1}{\lvert\vec a\rvert},\quad \cos\beta=\frac{a_2}{\lvert\vec a\rvert},\quad \cos\gamma=\frac{a_3}{\lvert\vec a\rvert}$$

These are the direction cosines $l$, $m$, $n$, and the components $a_1$, $a_2$, $a_3$ (or any multiple of them) are direction ratios. Since $\hat a=l\hat i+m\hat j+n\hat k$ is a unit vector,

$$l^2+m^2+n^2=1$$ Three axes x, y, z drawn in perspective, the vector a from the origin with its three components along the axes, and the angles alpha, beta, gamma it makes with the axes

Example 2 (JEE Main 2026, 22 January Shift 2). Let $\vec a=\sqrt2\,\hat i-\hat j+\lambda\hat k$, $\lambda>0$, make an obtuse angle with $\vec b=-\lambda^2\hat i+4\sqrt2\,\hat j+4\sqrt2\,\hat k$ and an angle $\theta$, $\dfrac\pi6<\theta<\dfrac\pi2$, with the positive $z$-axis. If the set of all possible $\lambda$ is $(\alpha,\beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is ______.

▸ SHOW SOLUTION

Step 1. Angle with the z-axis

$$\cos\theta=\frac{\vec a\cdot\hat k}{\lvert\vec a\rvert}=\frac{\lambda}{\sqrt{3+\lambda^2}}$$

$\dfrac\pi6<\theta<\dfrac\pi2$ means $0<\cos\theta<\dfrac{\sqrt3}{2}$:

$$\frac{\lambda^2}{3+\lambda^2}<\frac34\implies\lambda^2<9$$

So $0<\lambda<3$.

Step 2. Obtuse angle with b

$$\vec a\cdot\vec b=\sqrt2(-\lambda^2+4\lambda-4)<0$$$$(\lambda-2)^2>0\implies\lambda\ne2$$

Step 3. Combine

$\lambda\in(0,3)-\{2\}$, so $\alpha=0$, $\beta=3$, $\gamma=2$.

$$\alpha+\beta+\gamma=5$$

Answer: $5$

Example 3 (JEE Main 2026, 4 April Shift 1). Let $\vec a_k=(\tan\theta_k)\hat i+\hat j$ and $\vec b_k=\hat i-(\cot\theta_k)\hat j$, where $\theta_k=\dfrac{2^{k-1}\pi}{2^n+1}$ for some $n\in\mathbb N$, $n>5$. Find the value of

$$\frac{\sum_{k=1}^{n}\lvert\vec a_k\rvert^2}{\sum_{k=1}^{n}\lvert\vec b_k\rvert^2}$$

▸ SHOW SOLUTION

Step 1. The two magnitudes

$$\lvert\vec a_k\rvert^2=\tan^2\theta_k+1=\sec^2\theta_k$$$$\lvert\vec b_k\rvert^2=1+\cot^2\theta_k=\operatorname{cosec}^2\theta_k$$

Step 2. One identity

$\cot\theta-\tan\theta=2\cot2\theta$. Squaring and writing everything with $\sec$ and $\operatorname{cosec}$:

$$\sec^2\theta=4\operatorname{cosec}^22\theta-\operatorname{cosec}^2\theta$$

Step 3. Sum over k

$2\theta_k=\theta_{k+1}$, and $\theta_{n+1}=\pi-\theta_1$, so the doubled angles run over the same $\operatorname{cosec}^2$ values:

$$\sum_{k=1}^{n}\operatorname{cosec}^22\theta_k=\sum_{k=1}^{n}\operatorname{cosec}^2\theta_k$$

With $S=\sum\operatorname{cosec}^2\theta_k$:

$$\sum\sec^2\theta_k=4S-S=3S$$

Answer: $3$

The vector part is only Step 1. The paper as circulated printed $\cos\theta_k$ in $\vec b_k$; with $\cos$ the ratio is not a constant, and the official solution uses $\cot\theta_k$, as here.

Dot product

The dot product (scalar product) of $\vec a$ and $\vec b$ at angle $\theta$ is the number

$$\vec a\cdot\vec b=\lvert\vec a\rvert\lvert\vec b\rvert\cos\theta$$

In components,

$$\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3$$

It is commutative and distributes over addition. Facts used constantly:

• $\vec a\cdot\vec a=\lvert\vec a\rvert^2$.
• $\vec a\perp\vec b\iff\vec a\cdot\vec b=0$ for non-zero vectors.
• $\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$, and $\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0$.
• The angle is acute, right or obtuse as $\vec a\cdot\vec b$ is positive, zero or negative.
• $\lvert\vec a\cdot\vec b\rvert\le\lvert\vec a\rvert\lvert\vec b\rvert$, with equality only for parallel vectors.

Example 4 (JEE Main 2026, 2 April Shift 2). Two adjacent sides of a parallelogram $PQRS$ are $\overrightarrow{PQ}=\hat j+\hat k$ and $\overrightarrow{PS}=\hat i-\hat j$. If side $PS$ is rotated about $P$ through an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to $PQ$, then $\sin^2\dfrac{5\alpha}{2}-\sin^2\dfrac\alpha2$ is

(1) $\dfrac12$ (2) $\dfrac{\sqrt3}{2}$ (3) $\dfrac{\sqrt3}{4}$ (4) $\dfrac{2\sqrt3}{5}$

▸ SHOW SOLUTION

Step 1. Angle QPS

$$\cos\theta=\frac{\overrightarrow{PQ}\cdot\overrightarrow{PS}}{\lvert\overrightarrow{PQ}\rvert\lvert\overrightarrow{PS}\rvert}=\frac{-1}{\sqrt2\cdot\sqrt2}=-\frac12$$

$\theta=\dfrac{2\pi}{3}$, so turning $PS$ by $\alpha=\dfrac{2\pi}{3}-\dfrac\pi2=\dfrac\pi6$ makes it perpendicular.

Step 2. The trigonometry

$$\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$$$$\sin3\alpha\,\sin2\alpha=\sin\frac\pi2\,\sin\frac\pi3=\frac{\sqrt3}{2}$$

Answer: (2)

Projection

The scalar projection of $\vec a$ on $\vec b$ is the signed length of the shadow of $\vec a$ on the line of $\vec b$:

$$\text{proj}_{\vec b}\,\vec a=\vec a\cdot\hat b=\frac{\vec a\cdot\vec b}{\lvert\vec b\rvert}$$

The vector projection is that length times $\hat b$:

$$\frac{\vec a\cdot\vec b}{\lvert\vec b\rvert^2}\,\vec b$$

and $\vec a$ minus its vector projection is the part of $\vec a$ perpendicular to $\vec b$.

Vectors a and b from a common point at angle theta, with a dropped perpendicularly onto the line of b and the shadow of length a cos theta marked along b

Example 5 (JEE Main 2026, 22 January Shift 1). Let $\overrightarrow{AB}=2\hat i+4\hat j-5\hat k$ and $\overrightarrow{AD}=\hat i+2\hat j+\lambda\hat k$. The projection of $\vec v=\hat i+\hat j+\hat k$ on the diagonal $\overrightarrow{AC}$ of the parallelogram $ABCD$ has length $1$. If $\alpha>\beta$ are the roots of $\lambda^2x^2-6\lambda x+5=0$, then $2\alpha-\beta$ is

(1) $1$ (2) $4$ (3) $3$ (4) $6$

▸ SHOW SOLUTION

Step 1. The diagonal

$$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=3\hat i+6\hat j+(\lambda-5)\hat k$$

Step 2. Projection equal to 1

$$\vec v\cdot\overrightarrow{AC}=\lambda+4$$$$(\lambda+4)^2=\lvert\overrightarrow{AC}\rvert^2=45+(\lambda-5)^2$$$$18\lambda=54\implies\lambda=3$$

Step 3. The quadratic

$$9x^2-18x+5=0\implies x=\frac53,\ \frac13$$$$2\alpha-\beta=\frac{10}{3}-\frac13=3$$

Answer: (3)

Expanding a squared magnitude

Most dot-product questions are settled by expanding a square:

$$\lvert\vec a\pm\vec b\rvert^2=\lvert\vec a\rvert^2+\lvert\vec b\rvert^2\pm2\,\vec a\cdot\vec b$$ $$\lvert\vec a+\vec b+\vec c\rvert^2=\sum\lvert\vec a\rvert^2+2\sum\vec a\cdot\vec b$$

where the second sum is $\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$.

Adding the first pair gives the parallelogram identity:

$$\lvert\vec a+\vec b\rvert^2+\lvert\vec a-\vec b\rvert^2=2\lvert\vec a\rvert^2+2\lvert\vec b\rvert^2$$

Example 6 (JEE Main 2026, 2 April Shift 1). If $\lvert\vec a\rvert=2$ and $\lvert\vec b\rvert=3$, then the maximum value of $3\lvert3\vec a+2\vec b\rvert+4\lvert3\vec a-2\vec b\rvert$ is

(1) $30$ (2) $36$ (3) $60$ (4) $72$

▸ SHOW SOLUTION

Step 1. Expand both squares

With $\vec a\cdot\vec b=6\cos\theta$:

$$\lvert3\vec a+2\vec b\rvert^2=72+72\cos\theta=144\cos^2\frac\theta2$$$$\lvert3\vec a-2\vec b\rvert^2=72-72\cos\theta=144\sin^2\frac\theta2$$

Step 2. One trigonometric expression

$$E=36\cos\frac\theta2+48\sin\frac\theta2$$

Step 3. Maximum

$$E_{\max}=\sqrt{36^2+48^2}=60$$

Answer: (3) $60$

Example 7 (JEE Main 2026, 28 January Shift 1). For unit vectors $\vec a$, $\vec b$, $\vec c$ with $\lvert\vec a-\vec b\rvert^2+\lvert\vec b-\vec c\rvert^2+\lvert\vec c-\vec a\rvert^2=9$ and $\lvert2\vec a+k\vec b+k\vec c\rvert=3$, the positive value of $k$ is

(1) $3$ (2) $6$ (3) $4$ (4) $5$

▸ SHOW SOLUTION

Step 1. Expand the given sum

Let $S=\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$:

$$6-2S=9\implies S=-\frac32$$

Step 2. The sum of the three vectors

$$\lvert\vec a+\vec b+\vec c\rvert^2=3+2S=0$$

So $\vec b+\vec c=-\vec a$.

Step 3. Substitute

$$\lvert2\vec a-k\vec a\rvert=\lvert2-k\rvert=3$$

$k=5$ or $k=-1$; the positive value is $5$.

Answer: (4) $5$

Example 8 (IITianForum). Unit vectors $\hat a$, $\hat b$, $\hat c$ satisfy $\hat a+\hat b=\sqrt3\,\hat c$. Find $\lvert\hat a-\hat b+\hat c\rvert$.

▸ SHOW SOLUTION

Step 1. Angle between a and b

$$\lvert\hat a+\hat b\rvert^2=3\implies2+2\,\hat a\cdot\hat b=3$$$$\hat a\cdot\hat b=\frac12,\qquad \lvert\hat a-\hat b\rvert^2=2-1=1$$

Step 2. c is perpendicular to a minus b

$$(\hat a-\hat b)\cdot(\hat a+\hat b)=1-1=0$$

so $(\hat a-\hat b)\cdot\hat c=0$.

Step 3. Expand

$$\lvert\hat a-\hat b+\hat c\rvert^2=\lvert\hat a-\hat b\rvert^2+\lvert\hat c\rvert^2=1+1$$

Answer: $\sqrt2$

$\hat a+\hat b$ and $\hat a-\hat b$ are the two diagonals of a rhombus, and the diagonals of a rhombus are always perpendicular.

Vectors in the plane of two given vectors

If $\vec a$ and $\vec b$ are not parallel, every vector in their plane can be written in exactly one way as

$$\vec v=x\vec a+y\vec b$$

So "a vector in the plane of $\vec a$ and $\vec b$" means two unknowns $x$, $y$, and the question must supply two conditions to fix it.

Vectors a and b from the origin with a grid of their multiples, and a vector v reaching the corner of the parallelogram on x a and y b

Example 9 (JEE Main 2026, 2 April Shift 2). Let $\vec a=-\hat i+\hat j+3\hat k$, $\vec b=\hat i+3\hat j+\hat k$ and $\vec c=\lambda\vec a+\mu\vec b$. If $\vec c\cdot(3\hat i-6\hat j+2\hat k)=10$ and $\vec c\cdot(\hat i+\hat j+\hat k)=-2$, then $\lvert\vec c\rvert^2$ is

(1) $8$ (2) $12$ (3) $14$ (4) $15$

▸ SHOW SOLUTION

Step 1. Components of c

$$\vec c=(\mu-\lambda)\hat i+(\lambda+3\mu)\hat j+(3\lambda+\mu)\hat k$$

Step 2. Two conditions

$$-3\lambda-13\mu=10$$$$3\lambda+5\mu=-2$$

Adding, $\mu=-1$, then $\lambda=1$.

Step 3. Magnitude

$$\vec c=-2\hat i-2\hat j+2\hat k,\qquad\lvert\vec c\rvert^2=12$$

Answer: (2) $12$

Example 10 (JEE Main 2026, 24 January Shift 2). Let $\vec a=2\hat i-\hat j-\hat k$, $\vec b=\hat i+3\hat j-\hat k$ and $\vec c=2\hat i+\hat j+3\hat k$. Let $\vec v$ be the vector in the plane of $\vec a$ and $\vec b$ whose projection on $\vec c$ has length $\dfrac{1}{\sqrt{14}}$. Then $\lvert\vec v\rvert$ is

(1) $\dfrac{\sqrt{21}}{2}$ (2) $13$ (3) $\dfrac{\sqrt{35}}{2}$ (4) $7$

▸ SHOW SOLUTION

Step 1. Write v

$$\vec v=x\vec a+y\vec b$$$$\vec v=(2x+y)\hat i+(3y-x)\hat j-(x+y)\hat k$$

Step 2. Projection condition

$$\vec v\cdot\vec c=2y,\qquad \frac{\lvert2y\rvert}{\sqrt{14}}=\frac{1}{\sqrt{14}}$$

so $y^2=\dfrac14$, and $x$ is not fixed.

Step 3. Magnitude

$$\lvert\vec v\rvert^2=6x^2+11y^2=6x^2+\frac{11}{4}$$

Answer: not unique. The official key gives (3), which is the value at $x=\pm1$.

This is the counting rule above in action: a vector in a plane has two unknowns and the question gives only one condition. When a question seems to leave a free parameter, check whether it really does before forcing an answer.

Cross product

The cross product (vector product) of $\vec a$ and $\vec b$ at angle $\theta$ is the vector

$$\vec a\times\vec b=\lvert\vec a\rvert\lvert\vec b\rvert\sin\theta\,\hat n$$

where $\hat n$ is the unit vector perpendicular to both, chosen so that $\vec a$, $\vec b$, $\hat n$ form a right-handed system. In components,

$$\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}$$ A parallelogram on a and b drawn in perspective, with the vector a cross b standing perpendicular to it and the angle theta marked between a and b

Rules to know cold:

• $\vec b\times\vec a=-\vec a\times\vec b$, so order matters.
• $\vec a\times\vec a=\vec 0$, and $\vec a\times\vec b=\vec 0\iff\vec a\parallel\vec b$ for non-zero vectors.
• $\hat i\times\hat j=\hat k$, $\hat j\times\hat k=\hat i$, $\hat k\times\hat i=\hat j$.
• $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$.
• Lagrange's identity:

$$\lvert\vec a\times\vec b\rvert^2+(\vec a\cdot\vec b)^2=\lvert\vec a\rvert^2\lvert\vec b\rvert^2$$

Area

$\lvert\vec a\times\vec b\rvert$ is the area of the parallelogram on $\vec a$ and $\vec b$, and half of it is the area of the triangle:

$$\text{Area of }\triangle ABC=\frac12\lvert\overrightarrow{AB}\times\overrightarrow{AC}\rvert$$

For any quadrilateral with diagonals $\vec d_1$ and $\vec d_2$ (a parallelogram included), the area is $\dfrac12\lvert\vec d_1\times\vec d_2\rvert$.

The unit vectors perpendicular to both $\vec a$ and $\vec b$ are

$$\pm\frac{\vec a\times\vec b}{\lvert\vec a\times\vec b\rvert}$$

This is the usual opening line of a question that wants a normal direction.

Example 11 (JEE Main 2026, 6 April Shift 2). Let $\vec a=2\hat i+3\hat j+3\hat k$ and $\vec b=6\hat i+3\hat j+3\hat k$. The square of the area of the triangle with adjacent sides $2\vec a+3\vec b$ and $\vec a-\vec b$ is

(1) $450$ (2) $900$ (3) $1800$ (4) $2400$

▸ SHOW SOLUTION

Step 1. The two sides

$$2\vec a+3\vec b=22\hat i+15\hat j+15\hat k$$$$\vec a-\vec b=-4\hat i$$

Step 2. Cross product

$$(22\hat i+15\hat j+15\hat k)\times(-4\hat i)$$$$=-60\hat j+60\hat k$$

with magnitude $60\sqrt2$.

Step 3. Area

$$\text{Area}=30\sqrt2,\qquad\text{Area}^2=1800$$

Answer: (3) $1800$

A shortcut worth seeing: $(2\vec a+3\vec b)\times(\vec a-\vec b)=-5\,\vec a\times\vec b$, so you only ever need $\vec a\times\vec b$.

Example 12 (JEE Main 2026, 21 January Shift 2). In a triangle $ABC$, let $\vec p=\overrightarrow{BC}$, $\vec q=\overrightarrow{CA}$, $\vec r=\overrightarrow{BA}$. If $\lvert\vec p\rvert=2\sqrt3$, $\lvert\vec q\rvert=2$ and the angle $\theta$ between $\vec p$ and $\vec q$ has $\cos\theta=\dfrac{1}{\sqrt3}$, then $\lvert\vec p\times(\vec q-3\vec r)\rvert^2+3\lvert\vec r\rvert^2$ is

(1) $340$ (2) $220$ (3) $410$ (4) $200$

▸ SHOW SOLUTION

Step 1. The relation between the sides

$$\vec p+\vec q=\vec r$$$$\lvert\vec r\rvert^2=12+4+2\cdot2\sqrt3\cdot2\cdot\frac{1}{\sqrt3}=24$$

Step 2. Simplify the cross product

$$\vec q-3\vec r=-3\vec p-2\vec q$$$$\vec p\times(\vec q-3\vec r)=-2\,\vec p\times\vec q$$

Step 3. Magnitude

$$4\lvert\vec p\rvert^2\lvert\vec q\rvert^2\sin^2\theta=4\cdot12\cdot4\cdot\frac23=128$$$$128+3\cdot24=200$$

Answer: (4) $200$

Example 13 (JEE Advanced 2026, Paper 2). Let $P$, $Q$, $R$ have position vectors $\vec a$, $\vec b$, $\vec a+\vec b$ with respect to $O$. If $\lvert\vec a+\vec b\rvert=\sqrt{21}$, $\lvert\vec a-\vec b\rvert=3$ and $\vec a\perp(\vec a-\vec b)$, then the area of triangle $OPR$ is

(1) $\sqrt3$ (2) $\dfrac{\sqrt3}{2}$ (3) $\dfrac{3\sqrt3}{2}$ (4) $\dfrac32$

▸ SHOW SOLUTION

Step 1. Three dot products

$\vec a\cdot(\vec a-\vec b)=0$ gives $\lvert\vec a\rvert^2=\vec a\cdot\vec b$. Expanding the two given squares:

$$\lvert\vec a\rvert^2+\lvert\vec b\rvert^2+2\,\vec a\cdot\vec b=21$$$$\lvert\vec a\rvert^2+\lvert\vec b\rvert^2-2\,\vec a\cdot\vec b=9$$

So $\vec a\cdot\vec b=3$, $\lvert\vec a\rvert^2=3$, $\lvert\vec b\rvert^2=12$.

Step 2. The area

$$\text{Area}=\frac12\lvert\vec a\times(\vec a+\vec b)\rvert=\frac12\lvert\vec a\times\vec b\rvert$$$$\lvert\vec a\times\vec b\rvert^2=3\cdot12-3^2=27$$$$\text{Area}=\frac{3\sqrt3}{2}$$

Answer: (3)

Lagrange's identity turned three dot products into the area without a single component.

Triangle OPR with OP equal to a, OQ equal to b and OR equal to a plus b, the parallelogram OPRQ completed with dashed sides

Example 14 (JEE Main 2026, 24 January Shift 1). Let $\vec a=2\hat i+\hat j-2\hat k$, $\vec b=\hat i+\hat j$ and $\vec c=\vec a\times\vec b$. Let $\vec d$ satisfy $\lvert\vec d-\vec a\rvert=\sqrt{11}$, $\lvert\vec c\times\vec d\rvert=3$, and let the angle between $\vec c$ and $\vec d$ be $\dfrac\pi4$. Then $\vec a\cdot\vec d$ is

(1) $11$ (2) $3$ (3) $0$ (4) $1$

▸ SHOW SOLUTION

Step 1. The vector c

$$\vec c=2\hat i-2\hat j+\hat k,\qquad\lvert\vec c\rvert=3$$

Step 2. Length of d

$$3=3\lvert\vec d\rvert\sin\frac\pi4\implies\lvert\vec d\rvert=\sqrt2$$

Step 3. Expand the given square

$$\lvert\vec d\rvert^2+\lvert\vec a\rvert^2-2\,\vec a\cdot\vec d=11$$$$2+9-2\,\vec a\cdot\vec d=11\implies\vec a\cdot\vec d=0$$

Answer: (3) $0$

Example 15 (JEE Main 2026, 22 January Shift 2). Let $\vec a=2\hat i-\hat j+\hat k$ and $\vec b=\lambda\hat j+2\hat k$, $\lambda\in\mathbb Z$. Let $\vec c=\vec a\times\vec b$, and let $\vec d$ be a vector of magnitude $2$ in the $yz$-plane. If $\lvert\vec c\rvert=\sqrt{53}$, then the maximum possible value of $(\vec c\cdot\vec d)^2$ is

(1) $26$ (2) $104$ (3) $208$ (4) $52$

▸ SHOW SOLUTION

Step 1. The vector c

$$\vec c=-(2+\lambda)\hat i-4\hat j+2\lambda\hat k$$$$\lvert\vec c\rvert^2=5\lambda^2+4\lambda+20=53$$

$\lambda=-3$ or $\dfrac{11}{5}$; integer gives $\lambda=-3$ and $\vec c=\hat i-4\hat j-6\hat k$.

Step 2. d in the yz-plane

$\vec d=y\hat j+z\hat k$ with $y^2+z^2=4$:

$$\vec c\cdot\vec d=-4y-6z$$

Step 3. Maximum

$$(4y+6z)^2\le(4^2+6^2)(y^2+z^2)=208$$

Answer: (3) $208$

The inequality in Step 3 is just $\lvert\vec u\cdot\vec v\rvert\le\lvert\vec u\rvert\lvert\vec v\rvert$ for $\vec u=4\hat j+6\hat k$ and $\vec v=\vec d$.

Example 16 (JEE Main 2026, 23 January Shift 2). Let $\vec a=\hat i-2\hat j+3\hat k$, $\vec b=2\hat i+\hat j-\hat k$, $\vec c=\lambda\hat i+\hat j+\hat k$ and $\vec v=\vec a\times\vec b$. If $\vec v\cdot\vec c=11$ and the length of the projection of $\vec b$ on $\vec c$ is $p$, then $9p^2$ is

(1) $9$ (2) $6$ (3) $4$ (4) $12$

▸ SHOW SOLUTION

Step 1. The vector v

$$\vec v=-\hat i+7\hat j+5\hat k$$$$\vec v\cdot\vec c=12-\lambda=11\implies\lambda=1$$

Step 2. Projection

$$p=\frac{\vec b\cdot\vec c}{\lvert\vec c\rvert}=\frac{2}{\sqrt3}$$$$9p^2=12$$

Answer: (4) $12$

Example 17 (JEE Main 2026, 4 April Shift 2). Let $\hat u$, $\hat v$ be unit vectors at an acute angle with $\lvert\hat u\times\hat v\rvert=\dfrac{\sqrt3}{2}$. If $\vec A=\lambda\hat u+\hat v+(\hat u\times\hat v)$, then $\lambda$ equals

(1) $\dfrac43(\vec A\cdot\hat u)-\dfrac23(\vec A\cdot\hat v)$ (2) $\dfrac23(\vec A\cdot\hat u)-\dfrac13(\vec A\cdot\hat v)$ (3) $\dfrac43(\vec A\cdot\hat u)+\dfrac23(\vec A\cdot\hat v)$ (4) $(\vec A\cdot\hat u)-\dfrac12(\vec A\cdot\hat v)$

▸ SHOW SOLUTION

Step 1. The angle

$\sin\theta=\dfrac{\sqrt3}{2}$ with $\theta$ acute, so $\hat u\cdot\hat v=\dfrac12$.

Step 2. Dot A with u and with v

$\hat u\times\hat v$ is perpendicular to both, so it drops out:

$$\vec A\cdot\hat u=\lambda+\frac12$$$$\vec A\cdot\hat v=\frac\lambda2+1$$

Step 3. Eliminate the constants

$$2(\vec A\cdot\hat u)-(\vec A\cdot\hat v)=\frac{3\lambda}{2}$$$$\lambda=\frac43(\vec A\cdot\hat u)-\frac23(\vec A\cdot\hat v)$$

Answer: (1)

Vector equations with a cross product

Three moves solve almost every equation of this kind.

1. Collect into one cross product: $\vec x\times\vec a=\vec y\times\vec a$ means $(\vec x-\vec y)\times\vec a=\vec 0$, so $\vec x-\vec y=\lambda\vec a$.
2. Use the extra scalar condition (a dot product or a magnitude) to find $\lambda$.
3. If the equation is $\vec a\times\vec c=\vec b$ with $\vec a\cdot\vec c$ given, either solve in components or cross with $\vec a$ (see the vector triple product).

Example 18 (JEE Main 2026, 5 April Shift 1). Let $\vec a=\sqrt7\,\hat i+\hat j-\hat k$ and $\vec b=\hat j+2\hat k$. If $\vec r\times\vec a+\vec a\times\vec b=\vec 0$ and $\vec r\cdot\vec a=0$, then $\lvert3\vec r\rvert^2$ is

(1) $44$ (2) $54$ (3) $86$ (4) $132$

▸ SHOW SOLUTION

Step 1. One cross product

$\vec a\times\vec b=-\vec b\times\vec a$:

$$(\vec r-\vec b)\times\vec a=\vec 0\implies\vec r=\vec b+\lambda\vec a$$

Step 2. Find λ

$$\vec a\cdot\vec b+\lambda\lvert\vec a\rvert^2=0\implies-1+9\lambda=0$$$$\lambda=\frac19$$

Step 3. Magnitude

$$\lvert\vec r\rvert^2=5+\frac29(-1)+\frac{9}{81}=\frac{44}{9}$$$$\lvert3\vec r\rvert^2=44$$

Answer: (1) $44$

Example 19 (JEE Main 2026, 8 April Shift 2). Let $\vec a=4\hat i-\hat j+3\hat k$, $\vec b=10\hat i+2\hat j-\hat k$ and let $\vec c$ satisfy $2(\vec a\times\vec b)+3(\vec b\times\vec c)=\vec 0$. If $\vec a\cdot\vec c=15$, then $\vec c\cdot(\hat i+\hat j-3\hat k)$ is

(1) $-6$ (2) $-5$ (3) $-4$ (4) $-3$

▸ SHOW SOLUTION

Step 1. One cross product

$3(\vec b\times\vec c)=-3(\vec c\times\vec b)$:

$$(2\vec a-3\vec c)\times\vec b=\vec 0\implies\vec c=\frac{2\vec a-\lambda\vec b}{3}$$

Step 2. Find λ

$\lvert\vec a\rvert^2=26$, $\vec a\cdot\vec b=35$:

$$\frac{52-35\lambda}{3}=15\implies\lambda=\frac15$$

Step 3. The required dot product

$$\vec c\cdot(\hat i+\hat j-3\hat k)=\frac{2(-6)-\frac15(15)}{3}=-5$$

Answer: (2) $-5$

Example 20 (JEE Main 2026, 24 January Shift 2). Let $\vec a=2\hat i-5\hat j+5\hat k$ and $\vec b=\hat i-\hat j+3\hat k$. If $2(\vec a\times\vec c)+3(\vec b\times\vec c)=\vec 0$ and $(\vec a-\vec b)\cdot\vec c=-97$, then $\lvert\vec c\times\hat k\rvert^2$ is

(1) $193$ (2) $233$ (3) $218$ (4) $205$

▸ SHOW SOLUTION

Step 1. One cross product

$$(2\vec a+3\vec b)\times\vec c=\vec 0$$$$\vec c=\lambda(7\hat i-13\hat j+19\hat k)$$

Step 2. Find λ

$$(\vec a-\vec b)\cdot\vec c=97\lambda=-97$$

$\lambda=-1$, so $\vec c=-7\hat i+13\hat j-19\hat k$.

Step 3. Cross with k

$$\vec c\times\hat k=13\hat i+7\hat j$$$$\lvert\vec c\times\hat k\rvert^2=169+49=218$$

Answer: (3) $218$

Example 21 (JEE Main 2026, 21 January Shift 1). Let $\vec c$, $\vec d$ satisfy $\lvert\vec c+\vec d\rvert=\sqrt{29}$ and $\vec c\times(2\hat i+3\hat j+4\hat k)=(2\hat i+3\hat j+4\hat k)\times\vec d$. If $\lambda_1>\lambda_2$ are the possible values of $(\vec c+\vec d)\cdot(-7\hat i+2\hat j+3\hat k)$, then

$$K^2x^2+(K^2-5K+\lambda_1)xy$$ $$+\left(3K+\frac{\lambda_2}{2}\right)y^2-8x+12y+\lambda_2=0$$

is a circle for $K$ equal to

(1) $4$ (2) $1$ (3) $-1$ (4) $2$

▸ SHOW SOLUTION

Step 1. One cross product

$$(\vec c+\vec d)\times(2\hat i+3\hat j+4\hat k)=\vec 0$$$$\vec c+\vec d=\mu(2\hat i+3\hat j+4\hat k)$$

Step 2. Magnitude

$\lvert\mu\rvert\sqrt{29}=\sqrt{29}$, so $\mu=\pm1$ and the dot product is $4\mu=\pm4$. So $\lambda_1=4$, $\lambda_2=-4$.

Step 3. Circle conditions

No $xy$ term: $K^2-5K+4=0$, so $K=1$ or $4$.

Equal $x^2$ and $y^2$ coefficients: $K^2=3K-2$, so $K=1$ or $2$.

Answer: (2) $K=1$

Example 22 (JEE Main 2026, 23 January Shift 2). Let $\vec a\times\vec b=2(\vec a\times\vec c)$ with $\lvert\vec a\rvert=1$, $\lvert\vec b\rvert=4$, $\lvert\vec c\rvert=2$, and let the angle between $\vec b$ and $\vec c$ be $60^\circ$. Then $\lvert\vec a\cdot\vec c\rvert$ is

(1) $2$ (2) $4$ (3) $0$ (4) $1$

▸ SHOW SOLUTION

Step 1. One cross product

$$\vec a\times(\vec b-2\vec c)=\vec 0\implies\vec b-2\vec c=\lambda\vec a$$

Step 2. Find λ

$\vec b\cdot\vec c=4\cdot2\cdot\dfrac12=4$:

$$\lvert\vec b-2\vec c\rvert^2=16+16-16=16$$

So $\lambda^2=16$, $\lambda=\pm4$.

Step 3. Dot with c

$$\vec b\cdot\vec c-2\lvert\vec c\rvert^2=\lambda(\vec a\cdot\vec c)$$$$4-8=\pm4(\vec a\cdot\vec c)$$

Answer: (4) $1$

Example 23 (JEE Main 2026, 21 January Shift 1). Let $\vec a=-\hat i+2\hat j+2\hat k$, $\vec b=8\hat i+7\hat j-3\hat k$, and let $\vec c$ satisfy $\vec a\times\vec c=\vec b$. If $\vec c\cdot(\hat i+\hat j+\hat k)=4$, then $\lvert\vec a+\vec c\rvert^2$ is

(1) $33$ (2) $30$ (3) $35$ (4) $27$

▸ SHOW SOLUTION

Step 1. Components

With $\vec c=c_1\hat i+c_2\hat j+c_3\hat k$:

$$\vec a\times\vec c=(2c_3-2c_2)\hat i+(c_3+2c_1)\hat j$$$$\qquad-(c_2+2c_1)\hat k$$

Step 2. Compare with b

$$2c_3-2c_2=8$$$$c_3+2c_1=7,\quad c_2+2c_1=3$$

with $c_1+c_2+c_3=4$. Adding the last two, $c_2+c_3+4c_1=10$, so $3c_1=6$.

$$\vec c=2\hat i-\hat j+3\hat k$$

Step 3. Magnitude

$$\vec a+\vec c=\hat i+\hat j+5\hat k,\qquad\lvert\vec a+\vec c\rvert^2=27$$

Answer: (4) $27$

Only two of the three component equations are independent, because $\vec a\times\vec c$ is always perpendicular to $\vec a$. That is why the extra condition on $\vec c$ is needed.

Scalar triple product

The scalar triple product of $\vec a$, $\vec b$, $\vec c$ is the number

$$[\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c)$$

In components it is a determinant:

$$[\vec a\ \vec b\ \vec c]=\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}$$

Its absolute value is the volume of the parallelepiped with coterminous edges $\vec a$, $\vec b$, $\vec c$: the base has area $\lvert\vec b\times\vec c\rvert$ and the height is the projection of $\vec a$ on the normal $\vec b\times\vec c$.

A parallelepiped with edges a, b, c from one corner, the normal b cross c to the base, and the height of the box marked as the projection of a on that normal

Properties, all from the determinant:

• A cyclic shift changes nothing: $[\vec a\ \vec b\ \vec c]=[\vec b\ \vec c\ \vec a]=[\vec c\ \vec a\ \vec b]$.
• Swapping two vectors changes the sign: $[\vec a\ \vec c\ \vec b]=-[\vec a\ \vec b\ \vec c]$.
• Dot and cross can be interchanged: $\vec a\cdot(\vec b\times\vec c)=(\vec a\times\vec b)\cdot\vec c$.
• Two equal vectors give $0$.
• $\vec a$, $\vec b$, $\vec c$ are coplanar exactly when $[\vec a\ \vec b\ \vec c]=0$.
• Four points $A$, $B$, $C$, $D$ are coplanar exactly when $[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]=0$.
• The volume of a tetrahedron is $\dfrac16\lvert[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\rvert$.

Two results that turn long questions into one line:

If $\vec u$, $\vec v$, $\vec w$ are combinations of $\vec a$, $\vec b$, $\vec c$ with coefficient rows $(p_1,q_1,r_1)$, $(p_2,q_2,r_2)$, $(p_3,q_3,r_3)$, then

$$[\vec u\ \vec v\ \vec w]=D\,[\vec a\ \vec b\ \vec c]$$ $$D=\begin{vmatrix}p_1&q_1&r_1\\p_2&q_2&r_2\\p_3&q_3&r_3\end{vmatrix}$$

and

$$[\vec a\times\vec b\ \ \vec b\times\vec c\ \ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2$$

Linearly dependent and independent vectors

Vectors $\vec a_1,\ldots,\vec a_n$ are linearly dependent if some combination $x_1\vec a_1+\cdots+x_n\vec a_n=\vec 0$ holds with the scalars not all zero, and linearly independent if the only such combination has every $x_i=0$.

So three non-coplanar vectors $\vec a$, $\vec b$, $\vec c$ are independent, and every vector in space can be written in exactly one way as $x\vec a+y\vec b+z\vec c$. To find $x$, $y$, $z$, take triple products: $[\vec r\ \vec b\ \vec c]=x[\vec a\ \vec b\ \vec c]$, and similarly for $y$ and $z$.

Example 24 (IITianForum). Let $\vec p$, $\vec q$, $\vec r$ be unit vectors with $\vec p\times\vec q=\vec r$ and $\vec p\perp\vec q$. If $[\vec a\ \vec q\ \vec r]=4$, $[\vec a\ \vec r\ \vec p]=-1$ and $[\vec a\ \vec p\ \vec q]=2$, find $\vec a$ in terms of $\vec p$, $\vec q$, $\vec r$.

▸ SHOW SOLUTION

Step 1. The triad is right-handed and orthonormal

$$\vec q\times\vec r=\vec p,\quad\vec r\times\vec p=\vec q,\quad\vec p\times\vec q=\vec r$$

Step 2. Read off the components

$$\vec a\cdot\vec p=[\vec a\ \vec q\ \vec r]=4$$$$\vec a\cdot\vec q=[\vec a\ \vec r\ \vec p]=-1$$$$\vec a\cdot\vec r=[\vec a\ \vec p\ \vec q]=2$$

Answer: $\vec a=4\vec p-\vec q+2\vec r$

Example 25 (IITianForum). Let $\vec a$, $\vec b$, $\vec c$ be non-coplanar, and let $\vec u=\vec a+2\vec b$, $\vec v=\vec b+3\vec c$, $\vec w=\vec c+\vec a$. If

$$[\vec u\times\vec v\ \ \vec v\times\vec w\ \ \vec w\times\vec u]=k[\vec a\ \vec b\ \vec c]^2$$

find $k$.

▸ SHOW SOLUTION

Step 1. Use the square rule

The left side is $[\vec u\ \vec v\ \vec w]^2$.

Step 2. Triple product of the combinations

$$[\vec u\ \vec v\ \vec w]=\begin{vmatrix}1&2&0\\0&1&3\\1&0&1\end{vmatrix}[\vec a\ \vec b\ \vec c]$$$$\begin{vmatrix}1&2&0\\0&1&3\\1&0&1\end{vmatrix}=1(1)-2(0-3)+0=7$$

Step 3. Square

$$[\vec u\ \vec v\ \vec w]^2=49[\vec a\ \vec b\ \vec c]^2$$

Answer: $k=49$

Example 26 (IITianForum). $OABC$ is a tetrahedron with $O$ at the origin, and $M_1$, $M_2$, $M_3$ are the midpoints of $BC$, $CA$, $AB$. If $v_1$ is the volume of the tetrahedron and $v_2$ the volume of the parallelepiped with coterminous edges $OM_1$, $OM_2$, $OM_3$, find $\dfrac{12v_1}{v_2}$.

▸ SHOW SOLUTION

Step 1. The two volumes

$$v_1=\frac16\lvert[\vec a\ \vec b\ \vec c]\rvert$$$$v_2=\lvert[\vec m_1\ \vec m_2\ \vec m_3]\rvert,\qquad\vec m_1=\frac{\vec b+\vec c}{2},\ \ldots$$

Step 2. Triple product of the midpoints

$$[\vec m_1\ \vec m_2\ \vec m_3]=\frac18[\vec b+\vec c\ \ \vec c+\vec a\ \ \vec a+\vec b]$$$$=\frac18\begin{vmatrix}0&1&1\\1&0&1\\1&1&0\end{vmatrix}[\vec a\ \vec b\ \vec c]=\frac14[\vec a\ \vec b\ \vec c]$$

Step 3. Ratio

$$\frac{12v_1}{v_2}=\frac{12\cdot\frac16}{\frac14}=8$$

Answer: $8$

Vector triple product

For any three vectors,

$$\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c$$ $$(\vec a\times\vec b)\times\vec c=(\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a$$

The first lies in the plane of $\vec b$ and $\vec c$; the second in the plane of $\vec a$ and $\vec b$. So the cross product is not associative.

The plane of b and c with both vectors drawn in it, b cross c standing perpendicular to the plane, and a cross (b cross c) lying back in the plane of b and c

The special case to remember, with $\vec a=\vec b$:

$$\vec a\times(\vec a\times\vec c)=(\vec a\cdot\vec c)\vec a-\lvert\vec a\rvert^2\vec c$$

It solves $\vec a\times\vec c=\vec b$ in one stroke: cross both sides with $\vec a$ and the unknown $\vec c$ appears alone.

Example 27 (JEE Main 2026, 23 January Shift 1). Let $\vec a=-\hat i+\hat j+2\hat k$, $\vec b=\hat i-\hat j-3\hat k$, $\vec c=\vec a\times\vec b$ and $\vec d=\vec c\times\vec a$. Then $(\vec a-\vec b)\cdot\vec d$ is

(1) $4$ (2) $-4$ (3) $-2$ (4) $2$

▸ SHOW SOLUTION

Step 1. Expand d

$$\vec d=(\vec a\times\vec b)\times\vec a=\lvert\vec a\rvert^2\vec b-(\vec a\cdot\vec b)\vec a$$

$\lvert\vec a\rvert^2=6$, $\vec a\cdot\vec b=-8$, so $\vec d=6\vec b+8\vec a$.

Step 2. The dot product

$$(\vec a-\vec b)\cdot(8\vec a+6\vec b)$$$$=8\lvert\vec a\rvert^2-2\,\vec a\cdot\vec b-6\lvert\vec b\rvert^2$$$$=48+16-66=-2$$

Answer: (3) $-2$

Example 28 (JEE Main 2026, 6 April Shift 1). If $\vec a=\hat i+\hat j+\hat k$, $\vec b=\hat j-\hat k$ and $\vec c$ satisfy $\vec a\times\vec c=\vec b$ and $\vec a\cdot\vec c=3$, then $\vec c\cdot(\vec a-2\vec b)$ is ______.

▸ SHOW SOLUTION

Step 1. Cross both sides with a

$$\vec a\times(\vec a\times\vec c)=\vec a\times\vec b$$$$(\vec a\cdot\vec c)\vec a-\lvert\vec a\rvert^2\vec c=\vec a\times\vec b$$

Step 2. Numbers

$\vec a\times\vec b=-2\hat i+\hat j+\hat k$, $\vec a\cdot\vec c=3$, $\lvert\vec a\rvert^2=3$:

$$3\vec c=3\vec a-(\vec a\times\vec b)=5\hat i+2\hat j+2\hat k$$

Step 3. The dot product

$$\vec a-2\vec b=\hat i-\hat j+3\hat k$$$$\vec c\cdot(\vec a-2\vec b)=\frac{5-2+6}{3}=3$$

Answer: $3$

Example 29 (IITianForum). Vectors $\vec a$, $\vec b$, $\vec c$ have magnitudes $1$, $\sqrt3$, $2$ and satisfy $\vec a\times(\vec a\times\vec c)+\vec b=\vec 0$. Find the acute angle between $\vec a$ and $\vec c$, and the value of $\vec b\cdot\vec c$.

▸ SHOW SOLUTION

Step 1. Expand

$$\vec a\times(\vec a\times\vec c)=(\vec a\cdot\vec c)\vec a-\vec c$$

so $\vec b=\vec c-(\vec a\cdot\vec c)\vec a$, the part of $\vec c$ perpendicular to $\vec a$.

Step 2. Magnitude of b

$$\lvert\vec b\rvert^2=\lvert\vec c\rvert^2-(\vec a\cdot\vec c)^2$$$$3=4-(\vec a\cdot\vec c)^2\implies\vec a\cdot\vec c=\pm1$$

$\cos\theta=\pm\dfrac12$, so the acute angle is $\dfrac\pi3$.

Step 3. b dot c

$$\vec b\cdot\vec c=\lvert\vec c\rvert^2-(\vec a\cdot\vec c)^2=3$$

Answer: $\dfrac\pi3$ and $3$

Geometry with vectors

The angle bisector

The unit vectors $\hat a$ and $\hat b$ have equal length, so they span a rhombus, and the diagonal of a rhombus bisects its angle. Hence the internal bisector of the angle between $\vec a$ and $\vec b$ has direction

$$\hat a+\hat b=\frac{\vec a}{\lvert\vec a\rvert}+\frac{\vec b}{\lvert\vec b\rvert}$$

and the external bisector has direction $\hat a-\hat b$. A point equidistant from two lines through $A$ lies on a bisector of the angle at $A$.

Vectors a and b from a common point, their unit vectors forming a rhombus, and the diagonal a hat plus b hat bisecting the angle

Example 30 (JEE Main 2026, 28 January Shift 2). Let $P$ be a point in the plane of $\overrightarrow{AB}=3\hat i+\hat j-\hat k$ and $\overrightarrow{AC}=\hat i-\hat j+3\hat k$, equidistant from the lines $AB$ and $AC$. If $\lvert\overrightarrow{AP}\rvert=\dfrac{\sqrt5}{2}$, then the area of triangle $ABP$ is

(1) $2$ (2) $\dfrac32$ (3) $\dfrac{\sqrt{30}}{4}$ (4) $\dfrac{\sqrt{26}}{4}$

▸ SHOW SOLUTION

Step 1. Angle at A

$$\cos2\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{\sqrt{11}\cdot\sqrt{11}}=-\frac{1}{11}$$

Step 2. AP bisects the angle

$\angle BAP=\theta$, and $\cos2\theta=1-2\sin^2\theta$:

$$\sin^2\theta=\frac{6}{11}$$

Step 3. Area

$$\frac12\cdot\sqrt{11}\cdot\frac{\sqrt5}{2}\cdot\sqrt{\frac{6}{11}}=\frac{\sqrt{30}}{4}$$

Answer: (3)

Example 31 (JEE Main 2026, 28 January Shift 1). In triangle $PQR$, $\overrightarrow{PQ}=-2\hat i-\hat j+2\hat k$ and $\overrightarrow{PR}=a\hat i+b\hat j-4\hat k$ with $a,b\in\mathbb Z$. Let $S$ on $QR$ be equidistant from the lines $PQ$ and $PR$. If $\lvert\overrightarrow{PR}\rvert=9$ and $\overrightarrow{PS}=\hat i-7\hat j+2\hat k$, then $3a-4b$ is ______.

▸ SHOW SOLUTION

Step 1. PS bisects angle QPR

$$\frac{\overrightarrow{PQ}\cdot\overrightarrow{PS}}{\lvert\overrightarrow{PQ}\rvert}=\frac{\overrightarrow{PR}\cdot\overrightarrow{PS}}{\lvert\overrightarrow{PR}\rvert}$$$$\frac{9}{3}=\frac{a-7b-8}{9}\implies a-7b=35$$

Step 2. Length of PR

$$a^2+b^2+16=81\implies a^2+b^2=65$$

Step 3. Solve

$a=35+7b$ gives $5b^2+49b+116=0$, so $b=-4$ (integer) and $a=7$.

$$3a-4b=21+16=37$$

Answer: $37$

Only the direction of $\overrightarrow{PS}$ was used. With these numbers the bisector actually meets $QR$ at $\dfrac14(\hat i-7\hat j+2\hat k)$ from $P$, a quarter of the printed vector, but the direction and so the answer are unchanged.

Area ratios from a vector equation

If $O$ is inside triangle $PQR$ and

$$\overrightarrow{OP}+k_1\overrightarrow{OQ}+k_2\overrightarrow{OR}=\vec 0$$

with $k_1,k_2>0$, then the areas are in the ratio of the coefficients: $[OQR]:[ORP]:[OPQ]=1:k_1:k_2$, and $[PQR]=(1+k_1+k_2)[OQR]$.

Triangle PQR with a point O inside joined to the three vertices, splitting it into the three triangles OQR, ORP and OPQ

Example 32 (IITianForum). $O$ is a point inside triangle $PQR$ with $\overrightarrow{OP}+k_1\overrightarrow{OQ}+k_2\overrightarrow{OR}=\vec 0$. If the area of $\triangle PQR$ is $6$ times the area of $\triangle OQR$, find $k_1+k_2$.

▸ SHOW SOLUTION

Step 1. Take O as origin

$\vec p=-k_1\vec q-k_2\vec r$.

Step 2. Area of PQR

$$2[PQR]=\lvert(\vec q-\vec p)\times(\vec r-\vec p)\rvert$$$$=\lvert\vec q\times\vec r+\vec r\times\vec p+\vec p\times\vec q\rvert$$$$\vec r\times\vec p=k_1\,\vec q\times\vec r,\qquad\vec p\times\vec q=k_2\,\vec q\times\vec r$$$$[PQR]=(1+k_1+k_2)\cdot\frac12\lvert\vec q\times\vec r\rvert$$$$[PQR]=(1+k_1+k_2)[OQR]$$

Step 3. Solve

$$1+k_1+k_2=6\implies k_1+k_2=5$$

Answer: $5$

Which result to use

A decision chart: from what the question gives, arrows lead to the section formula, expanding a square, writing v as x a plus y b, the cross product for area, collecting into one cross product, the triple product determinant, the vector triple product, or the angle bisector direction

Quick recall

Practise these methods on real exam questions in the doMath Vectors chapter, and continue with the 3D Geometry note, where every line and plane is written with the vectors built here.