Vectors — Complete Guide with Solved Examples

Complete guide to vectors for Class 12. Solved examples, exam tips, PYQs.

CBSE JEE-MAIN • 14 min read
Tags Vectors

What Exactly Is a Vector?

A vector is a quantity that has both magnitude and direction. Compare this to a scalar, which has only magnitude. When you say “60 km/h” — that’s speed, a scalar. When you say “60 km/h towards Delhi” — that’s velocity, a vector.

In Class 12 and JEE, vectors are the language of physics and 3D geometry. You’ll use them to describe forces, displacement, electric fields, and lines in space. The better your vector intuition, the faster you solve problems.

We represent a vector as a⃗\vec{a} or in bold as a. Geometrically, it’s a directed line segment — the length is the magnitude, the arrowhead gives the direction.

Vectors carries 8-10 marks in CBSE Class 12 boards and appears in 1-2 questions in JEE Main every year. The JEE Main 2024 Shift 1 paper had a cross product + unit vector problem worth 4 marks — exactly the type we’ll cover here.

Key Terms and Definitions

Position Vector: The vector OP→\overrightarrow{OP} from the origin OO to a point P(x,y,z)P(x, y, z) is called the position vector of PP. Written as r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}.

Magnitude (Modulus): The length of the vector. For a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}:

∣a⃗∣=a12+a22+a32|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}

Unit Vector: A vector of magnitude 1 in the direction of a⃗\vec{a}:

a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}

Equal Vectors: Same magnitude AND same direction. Note that position doesn’t matter — a free vector can be placed anywhere.

Negative Vector: −a⃗-\vec{a} has the same magnitude as a⃗\vec{a} but opposite direction.

Zero Vector: 0⃗\vec{0} — magnitude is zero, direction is undefined.

Collinear Vectors: Two or more vectors are collinear (or parallel) if they lie along the same or parallel lines. a⃗∥b⃗\vec{a} \parallel \vec{b} if a⃗=λb⃗\vec{a} = \lambda\vec{b} for some scalar λ\lambda.

Coplanar Vectors: Vectors lying in the same plane.

Component Form

Any vector in 3D space can be written as:

a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

where i^\hat{i}, j^\hat{j}, k^\hat{k} are unit vectors along the xx, yy, zz axes respectively.

Core Operations

Addition and Subtraction

Triangle Law: Place vectors head-to-tail. The resultant closes the triangle.

Parallelogram Law: If two vectors act from the same point, the diagonal of the parallelogram formed gives the resultant.

In component form:

a⃗+b⃗=(a1+b1)i^+(a2+b2)j^+(a3+b3)k^\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}

Scalar Multiplication

λa⃗=λa1i^+λa2j^+λa3k^\lambda\vec{a} = \lambda a_1\hat{i} + \lambda a_2\hat{j} + \lambda a_3\hat{k}

This scales the magnitude by ∣λ∣|\lambda| and reverses direction if λ<0\lambda < 0.

Section Formula

If r⃗\vec{r} divides the line joining AA (position vector a⃗\vec{a}) and BB (position vector b⃗\vec{b}) in ratio m:nm:n:

Internal Division:

r⃗=mb⃗+na⃗m+n\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}

External Division:

r⃗=mb⃗−na⃗m−n\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}

Midpoint (m=n=1m = n = 1):

r⃗=a⃗+b⃗2\vec{r} = \frac{\vec{a} + \vec{b}}{2}

Dot Product (Scalar Product)

The dot product of a⃗\vec{a} and b⃗\vec{b} gives a scalar:

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta

Component form:

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

Why the dot product matters: When a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0 and neither vector is zero, they are perpendicular. This is how we check orthogonality — used constantly in both boards and JEE.

Key Dot Product Results

  • i^⋅i^=j^⋅j^=k^⋅k^=1\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1
  • i^⋅j^=j^⋅k^=k^⋅i^=0\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0
  • a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2

Finding Angle Between Vectors

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

If the question asks whether vectors are perpendicular, never compute the angle — just check if the dot product is zero. Saves 30 seconds per problem.

Projection of a⃗\vec{a} on b⃗\vec{b}

Projection of a⃗ on b⃗=a⃗⋅b⃗∣b⃗∣\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} Vector projection=a⃗⋅b⃗∣b⃗∣2b⃗\text{Vector projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}

Cross Product (Vector Product)

The cross product of a⃗\vec{a} and b⃗\vec{b} gives a vector perpendicular to both:

a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}

Component form (determinant):

a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣\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}

Why this formula works the way it does: The cross product uses the determinant because we need a systematic way to get a vector perpendicular to two given vectors. The cofactor expansion along the first row does exactly that.

Key Cross Product Results

  • i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j} (cyclic)
  • j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k} (anti-cyclic)
  • a⃗×a⃗=0⃗\vec{a} \times \vec{a} = \vec{0}
  • ∣a⃗×b⃗∣|\vec{a} \times \vec{b}| = area of parallelogram formed by a⃗\vec{a} and b⃗\vec{b}
Area of parallelogram=∣a⃗×b⃗∣\text{Area of parallelogram} = |\vec{a} \times \vec{b}| Area of triangle=12∣a⃗×b⃗∣\text{Area of triangle} = \frac{1}{2}|\vec{a} \times \vec{b}|

Solved Examples

Example 1 — Finding Unit Vector (CBSE Level)

Q: Find a unit vector in the direction of a⃗=2i^−3j^+6k^\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}.

Step 1: Find the magnitude.

∣a⃗∣=4+9+36=49=7|\vec{a}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7

Step 2: Divide by magnitude.

a^=a⃗∣a⃗∣=2i^−3j^+6k^7=27i^−37j^+67k^\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}

Example 2 — Angle Between Vectors (CBSE Level)

Q: Find the angle between a⃗=i^+j^−k^\vec{a} = \hat{i} + \hat{j} - \hat{k} and b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}.

a⃗⋅b⃗=(1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1 ∣a⃗∣=3,∣b⃗∣=3|\vec{a}| = \sqrt{3}, \quad |\vec{b}| = \sqrt{3} cos⁡θ=−13⋅3=−13\cos\theta = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3} θ=cos⁡−1(−13)\theta = \cos^{-1}\left(\frac{-1}{3}\right)

Example 3 — Cross Product and Area (JEE Main Level)

Q: Find the area of the triangle with vertices A(1,1,2)A(1, 1, 2), B(2,3,5)B(2, 3, 5), C(1,5,5)C(1, 5, 5).

Step 1: Find AB→\overrightarrow{AB} and AC→\overrightarrow{AC}.

AB→=(2−1)i^+(3−1)j^+(5−2)k^=i^+2j^+3k^\overrightarrow{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k} AC→=0i^+4j^+3k^\overrightarrow{AC} = 0\hat{i} + 4\hat{j} + 3\hat{k}

Step 2: Compute AB→×AC→\overrightarrow{AB} \times \overrightarrow{AC}.

AB→×AC→=∣i^j^k^123043∣\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} =i^(6−12)−j^(3−0)+k^(4−0)=−6i^−3j^+4k^= \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0) = -6\hat{i} - 3\hat{j} + 4\hat{k}

Step 3: Area = 12∣AB→×AC→∣\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|

=1236+9+16=612= \frac{1}{2}\sqrt{36 + 9 + 16} = \frac{\sqrt{61}}{2}

Example 4 — Vector Perpendicularity Condition (JEE Main Level)

Q: If a⃗=2i^+λj^+k^\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k} and b⃗=i^−2j^+3k^\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k} are perpendicular, find λ\lambda.

For perpendicularity, a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0:

(2)(1)+(λ)(−2)+(1)(3)=0(2)(1) + (\lambda)(-2) + (1)(3) = 0 2−2λ+3=02 - 2\lambda + 3 = 0 λ=52\lambda = \frac{5}{2}

Example 5 — Scalar Triple Product (JEE Advanced Level)

The scalar triple product [a⃗ b⃗ c⃗]=a⃗⋅(b⃗×c⃗)[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) gives the volume of the parallelepiped formed by the three vectors.

[a⃗ b⃗ c⃗]=∣a1a2a3b1b2b3c1c2c3∣[\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}

If [a⃗ b⃗ c⃗]=0[\vec{a}\ \vec{b}\ \vec{c}] = 0, the vectors are coplanar.

Q: Show that a⃗=i^+2j^−3k^\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}, b⃗=2i^−j^+k^\vec{b} = 2\hat{i} - \hat{j} + \hat{k}, c⃗=3i^+j^−2k^\vec{c} = 3\hat{i} + \hat{j} - 2\hat{k} are coplanar.

[a⃗ b⃗ c⃗]=∣12−32−1131−2∣[\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 1 & 2 & -3 \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{vmatrix}

Expanding: 1(2−1)−2(−4−3)+(−3)(2+3)=1+14−15=01(2 - 1) - 2(-4 - 3) + (-3)(2 + 3) = 1 + 14 - 15 = 0

Since the scalar triple product is zero, the vectors are coplanar. ✓

Exam-Specific Tips

CBSE Class 12 Board Exam

The board exam tests vectors mainly in two ways: direct formula application (1-2 mark questions in VSA section) and proof/derivation questions (4-5 marks). The section formula and dot product for angle questions are almost guaranteed every year.

CBSE 2023 and 2024 both had a “find a vector perpendicular to both a⃗\vec{a} and b⃗\vec{b}” question — that’s a cross product. If you see this phrasing, go straight to the determinant formula. 4 marks, 3 minutes if you know it cold.

In CBSE marking, show each step clearly:

  • State the formula used
  • Substitute values
  • Simplify step-by-step

Even if your final answer is wrong due to arithmetic, you’ll get 2/4 marks for correct method.

JEE Main

JEE Main typically has 1-2 vector questions, often combined with 3D Geometry. High-yield topics:

  • Dot product conditions (perpendicular, parallel)
  • Cross product for area
  • Vector equations of lines (bridge to 3D chapter)

Scoring tip: The scalar triple product question in JEE Main is usually a 2-step problem. Set up the determinant quickly and expand along the row with the most zeros.

JEE Advanced

Expect vectors inside proofs — proving inequalities using the Cauchy-Schwarz identity, or geometry problems where you must translate geometric conditions into vector equations. Practice the vector form of the equation of a plane.

Common Mistakes to Avoid

Mistake 1: Confusing dot product and cross product outputs. a⃗⋅b⃗\vec{a} \cdot \vec{b} is a scalar (just a number). a⃗×b⃗\vec{a} \times \vec{b} is a vector. If you write a⃗⋅b⃗=3i^\vec{a} \cdot \vec{b} = 3\hat{i}, the examiner will mark it wrong even if your calculation is right.

Mistake 2: Wrong sign in cross product expansion. Remember: when expanding the determinant for the j^\hat{j} component, there’s a negative sign in front. Students regularly forget this. The pattern is: +i^+\hat{i}, −j^-\hat{j}, +k^+\hat{k}.

Mistake 3: ∣a⃗+b⃗∣2≠∣a⃗∣2+∣b⃗∣2|\vec{a} + \vec{b}|^2 \neq |\vec{a}|^2 + |\vec{b}|^2. The correct expansion is ∣a⃗+b⃗∣2=∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2. The middle term vanishes only when a⃗⊥b⃗\vec{a} \perp \vec{b}.

Mistake 4: Unit vector direction errors. a^=a⃗/∣a⃗∣\hat{a} = \vec{a}/|\vec{a}| gives a unit vector in the direction of a⃗\vec{a}. The unit vector in the direction opposite to a⃗\vec{a} is −a^-\hat{a}. CBSE sometimes asks for this and students write the same answer both ways.

Mistake 5: Using degree mode in dot product calculations. cos⁡θ\cos\theta must be computed correctly. If the question gives the angle in degrees (say 60°), write cos⁡60°=1/2\cos 60° = 1/2. Don’t just write “0.5” without justification — show the exact value for full marks.

Practice Questions

Q1 (CBSE): Find the value of xx for which x(i^+j^+k^)x(\hat{i} + \hat{j} + \hat{k}) is a unit vector.

We need ∣x(i^+j^+k^)∣=1|x(\hat{i} + \hat{j} + \hat{k})| = 1.

∣x∣⋅12+12+12=1  ⟹  ∣x∣⋅3=1  ⟹  x=±13|x| \cdot \sqrt{1^2 + 1^2 + 1^2} = 1 \implies |x| \cdot \sqrt{3} = 1 \implies x = \pm\dfrac{1}{\sqrt{3}}

Q2 (CBSE): If a⃗=5i^−j^−3k^\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k} and b⃗=i^+3j^−5k^\vec{b} = \hat{i} + 3\hat{j} - 5\hat{k}, show that a⃗+b⃗\vec{a} + \vec{b} and a⃗−b⃗\vec{a} - \vec{b} are perpendicular.

a⃗+b⃗=6i^+2j^−8k^\vec{a} + \vec{b} = 6\hat{i} + 2\hat{j} - 8\hat{k}

a⃗−b⃗=4i^−4j^+2k^\vec{a} - \vec{b} = 4\hat{i} - 4\hat{j} + 2\hat{k}

(a⃗+b⃗)⋅(a⃗−b⃗)=(6)(4)+(2)(−4)+(−8)(2)=24−8−16=0(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = (6)(4) + (2)(-4) + (-8)(2) = 24 - 8 - 16 = 0 ✓

Q3 (CBSE): Find the area of the parallelogram whose adjacent sides are a⃗=3i^+j^+4k^\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k} and b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}.

a⃗×b⃗=∣i^j^k^3141−11∣=i^(1+4)−j^(3−4)+k^(−3−1)=5i^+j^−4k^\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1+4) - \hat{j}(3-4) + \hat{k}(-3-1) = 5\hat{i} + \hat{j} - 4\hat{k}

Area =∣5i^+j^−4k^∣=25+1+16=42= |5\hat{i} + \hat{j} - 4\hat{k}| = \sqrt{25 + 1 + 16} = \sqrt{42} sq. units

Q4 (JEE Main): If ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=4|\vec{b}| = 4 and ∣a⃗+b⃗∣=5|\vec{a} + \vec{b}| = 5, find ∣a⃗−b⃗∣|\vec{a} - \vec{b}|.

Use ∣a⃗+b⃗∣2=∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2:

25=9+2a⃗⋅b⃗+16  ⟹  a⃗⋅b⃗=025 = 9 + 2\vec{a}\cdot\vec{b} + 16 \implies \vec{a}\cdot\vec{b} = 0

The vectors are perpendicular!

∣a⃗−b⃗∣2=∣a⃗∣2−2a⃗⋅b⃗+∣b⃗∣2=9−0+16=25|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 9 - 0 + 16 = 25

∣a⃗−b⃗∣=5|\vec{a}-\vec{b}| = 5

Q5 (JEE Main): Find a vector of magnitude 5 perpendicular to both a⃗=i^−2j^\vec{a} = \hat{i} - 2\hat{j} and b⃗=3j^+k^\vec{b} = 3\hat{j} + \hat{k}.

a⃗×b⃗=∣i^j^k^1−20031∣=i^(−2−0)−j^(1−0)+k^(3−0)=−2i^−j^+3k^\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 0 & 3 & 1 \end{vmatrix} = \hat{i}(-2-0) - \hat{j}(1-0) + \hat{k}(3-0) = -2\hat{i} - \hat{j} + 3\hat{k}

∣a⃗×b⃗∣=4+1+9=14|\vec{a}\times\vec{b}| = \sqrt{4+1+9} = \sqrt{14}

Required vector =5⋅−2i^−j^+3k^14=−10i^−5j^+15k^14= 5 \cdot \dfrac{-2\hat{i} - \hat{j} + 3\hat{k}}{\sqrt{14}} = \dfrac{-10\hat{i} - 5\hat{j} + 15\hat{k}}{\sqrt{14}}

Q6 (JEE Main): The position vectors of AA and BB are i^+2j^−3k^\hat{i} + 2\hat{j} - 3\hat{k} and 3i^−j^+2k^3\hat{i} - \hat{j} + 2\hat{k}. Find the position vector of the point that divides ABAB in ratio 2:32:3 internally.

Using section formula with m=2m=2, n=3n=3:

r⃗=2(3i^−j^+2k^)+3(i^+2j^−3k^)5=6i^−2j^+4k^+3i^+6j^−9k^5\vec{r} = \frac{2(3\hat{i}-\hat{j}+2\hat{k}) + 3(\hat{i}+2\hat{j}-3\hat{k})}{5} = \frac{6\hat{i}-2\hat{j}+4\hat{k}+3\hat{i}+6\hat{j}-9\hat{k}}{5} =9i^+4j^−5k^5=95i^+45j^−k^= \frac{9\hat{i}+4\hat{j}-5\hat{k}}{5} = \frac{9}{5}\hat{i}+\frac{4}{5}\hat{j}-\hat{k}

Q7 (JEE Advanced): If a⃗\vec{a}, b⃗\vec{b}, c⃗\vec{c} are unit vectors such that a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}, find a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}.

Since a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}, take the dot product of both sides with itself:

∣a⃗+b⃗+c⃗∣2=0|\vec{a}+\vec{b}+\vec{c}|^2 = 0

∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0

Since each is a unit vector: 1+1+1+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=01 + 1 + 1 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0

a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−32\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = -\frac{3}{2}

Q8 (CBSE): Prove that (a⃗+b⃗)×(a⃗−b⃗)=−2(a⃗×b⃗)(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = -2(\vec{a}\times\vec{b}).

(a⃗+b⃗)×(a⃗−b⃗)=a⃗×a⃗−a⃗×b⃗+b⃗×a⃗−b⃗×b⃗(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) = \vec{a}\times\vec{a} - \vec{a}\times\vec{b} + \vec{b}\times\vec{a} - \vec{b}\times\vec{b}

=0⃗−a⃗×b⃗+b⃗×a⃗−0⃗= \vec{0} - \vec{a}\times\vec{b} + \vec{b}\times\vec{a} - \vec{0}

Since b⃗×a⃗=−a⃗×b⃗\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}:

=−a⃗×b⃗−a⃗×b⃗=−2(a⃗×b⃗)= -\vec{a}\times\vec{b} - \vec{a}\times\vec{b} = -2(\vec{a}\times\vec{b}) ✓

Frequently Asked Questions

What is the difference between a scalar and a vector? A scalar has only magnitude (temperature, mass, speed). A vector has both magnitude and direction (force, velocity, displacement). The mathematical test: if reversing direction changes the quantity, it’s a vector.

When is a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}? Either one of the vectors is the zero vector, OR the two vectors are parallel (collinear). The cross product vanishes when sin⁡θ=0\sin\theta = 0, i.e., θ=0°\theta = 0° or 180°180°.

Can the dot product be negative? Yes. a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta. When 90°<θ≤180°90° < \theta \leq 180°, cos⁡θ\cos\theta is negative, so the dot product is negative. This means the angle between the vectors is obtuse.

What does the scalar triple product represent geometrically? It gives the signed volume of the parallelepiped (a 3D parallelogram-box) formed by the three vectors. If it’s zero, the three vectors are coplanar — they all lie in the same plane.

How do I find a vector perpendicular to a given plane? If the plane contains vectors a⃗\vec{a} and b⃗\vec{b}, then a⃗×b⃗\vec{a}\times\vec{b} is perpendicular (normal) to the plane. This is the core idea behind the vector equation of a plane.

Is a⃗×b⃗=b⃗×a⃗\vec{a}\times\vec{b} = \vec{b}\times\vec{a}? No — the cross product is anti-commutative: a⃗×b⃗=−b⃗×a⃗\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}. Order matters. The direction reverses when you swap the vectors, following the right-hand rule.

What is the formula for the angle between two vectors in JEE Main? cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}. This is valid for 0°≤θ≤180°0° \leq \theta \leq 180°. JEE sometimes asks for the acute angle — in that case, take ∣cos⁡θ∣|\cos\theta|.

How is the projection formula used in JEE? If a⃗\vec{a} is resolved along b⃗\vec{b} and perpendicular to b⃗\vec{b}, the component along b⃗\vec{b} is a⃗⋅b⃗∣b⃗∣\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|} and the component perpendicular to b⃗\vec{b} is ∣a⃗∣2−(a⃗⋅b⃗∣b⃗∣)2\sqrt{|\vec{a}|^2 - \left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\right)^2}. This appears in work-energy problems in JEE Physics too.

Practice Questions

Cross Product — Find a⃗ × b⃗ Using Determinant Method

medium JEE Main 2024

Dot Product of Two Vectors — Find Angle Between Them

easy CBSE 2024 Board Exam

Find the area of parallelogram with adjacent sides a=2i+3j and b=i-4j

hard

If |a|=2, |b|=3 and a·b=4, find |a×b| and angle between vectors

easy NCERT Class 12

Find Magnitude of Vector 3î + 4ĵ

easy NCERT Class 12