Mathematics Class 12 — Part…
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Vector Dot Product, Cross Product and Areas

🎓 Class 12 Mathematics CBSE Theory Ch 10 — Vector Algebra ⏱ ~15 min
🌐 Language: [gtranslate]

This MCQ module is based on: Vector Dot Product, Cross Product and Areas

This mathematics assessment will be based on: Vector Dot Product, Cross Product and Areas
Targeting Class 12 level in General Mathematics, with Advanced difficulty.

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10.6 Product of Two Vectors

Two distinct multiplications combine vectors: the scalar (dot) product? returns a scalar, the vector (cross) product? returns a vector. Both have deep geometric meaning.

10.6.1 Scalar (Dot) Product

Scalar product
For vectors \(\vec a, \vec b\) with angle \(\theta\) between them (\(0\le\theta\le\pi\)): \[\boxed{\;\vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta\;}\] The result is a real number.

In components: \((a_1\hat\imath+a_2\hat\jmath+a_3\hat k)\cdot(b_1\hat\imath+b_2\hat\jmath+b_3\hat k)=a_1b_1+a_2b_2+a_3b_3\).
Properties of the dot product
  • Commutative: \(\vec a\cdot\vec b=\vec b\cdot\vec a\).
  • Distributive: \(\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c\).
  • \((\lambda\vec a)\cdot\vec b=\lambda(\vec a\cdot\vec b)=\vec a\cdot(\lambda\vec b)\).
  • \(\vec a\cdot\vec a=|\vec a|^2\).
  • Perpendicularity test: non-zero \(\vec a, \vec b\) are perpendicular iff \(\vec a\cdot\vec b=0\).
  • \(\hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1\), \(\hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0\) (orthonormal basis).

Angle between two vectors

Solving \(\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta\) for the angle:

\[\boxed{\;\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|}\;}\]

10.6.2 Projection of a vector

Projection
The scalar projection of \(\vec a\) onto \(\vec b\) is: \[\text{proj}_\vec b\vec a=|\vec a|\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec b|}.\] The vector projection is this scalar times the unit vector along \(\vec b\): \[\overrightarrow{\text{proj}}_{\vec b}\vec a=\left(\dfrac{\vec a\cdot\vec b}{|\vec b|^2}\right)\vec b.\]

10.6.3 Vector (Cross) Product

Vector product
For vectors \(\vec a, \vec b\) with angle \(\theta\) (\(0\le\theta\le\pi\)) between them: \[\boxed{\;\vec a\times\vec b=|\vec a|\,|\vec b|\sin\theta\,\hat n\;}\] where \(\hat n\) is the unit vector perpendicular to both \(\vec a\) and \(\vec b\), oriented by the right-hand rule: if the fingers of the right hand curl from \(\vec a\) to \(\vec b\), the thumb points along \(\hat n\).

Determinant formula: \[\vec a\times\vec b=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{vmatrix}=(a_2 b_3-a_3 b_2)\hat\imath-(a_1 b_3-a_3 b_1)\hat\jmath+(a_1 b_2-a_2 b_1)\hat k.\]
a⃗ b⃗ a⃗ × b⃗ (out of page) θ |a⃗ × b⃗| = area of parallelogram with sides a⃗, b⃗ = |a⃗| |b⃗| sin θ
Cross product geometry: magnitude = parallelogram area; direction by right-hand rule.
Properties of the cross product
  • Anti-commutative: \(\vec a\times\vec b=-(\vec b\times\vec a)\).
  • Distributive: \(\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c\).
  • \((\lambda\vec a)\times\vec b=\lambda(\vec a\times\vec b)=\vec a\times(\lambda\vec b)\).
  • \(\vec a\times\vec a=\vec 0\).
  • Parallel test: non-zero \(\vec a, \vec b\) are parallel iff \(\vec a\times\vec b=\vec 0\).
  • \(\hat\imath\times\hat\jmath=\hat k,\ \hat\jmath\times\hat k=\hat\imath,\ \hat k\times\hat\imath=\hat\jmath\) (cyclic; reverse → minus).

Areas via the cross product

Area formulas
For vectors \(\vec a, \vec b\) representing two sides of a parallelogram:
• Parallelogram area = \(|\vec a\times\vec b|\).
• Triangle area (with same two sides) = \(\dfrac{1}{2}|\vec a\times\vec b|\).

For a parallelogram with diagonals \(\vec d_1, \vec d_2\): area = \(\dfrac{1}{2}|\vec d_1\times\vec d_2|\).

Worked Examples

Example 11. Find \(\vec a\cdot\vec b\) where \(\vec a=2\hat\imath+\hat\jmath+\hat k,\ \vec b=\hat\imath-\hat\jmath+2\hat k\).
\(2\cdot 1+1\cdot(-1)+1\cdot 2=2-1+2=3\).
Example 12. Find the angle between \(\vec a=\hat\imath+\hat\jmath\) and \(\vec b=\hat\jmath+\hat k\).
\(\vec a\cdot\vec b=0+1+0=1\). \(|\vec a|=\sqrt 2,\ |\vec b|=\sqrt 2\). \(\cos\theta=1/2\Rightarrow\theta=\pi/3\) (60°).
Example 13. Find the projection of \(\vec a=\hat\imath+3\hat\jmath+7\hat k\) on \(\vec b=7\hat\imath-\hat\jmath+8\hat k\).
\(\vec a\cdot\vec b=7-3+56=60\). \(|\vec b|=\sqrt{49+1+64}=\sqrt{114}\). Scalar projection = \(60/\sqrt{114}\).
Example 14. Find \(\vec a\times\vec b\) for \(\vec a=2\hat\imath+\hat\jmath+3\hat k,\ \vec b=3\hat\imath+5\hat\jmath-2\hat k\).
Determinant: \(\vec a\times\vec b=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 2 & 1 & 3\\ 3 & 5 & -2\end{vmatrix}=\hat\imath(1\cdot(-2)-3\cdot 5)-\hat\jmath(2\cdot(-2)-3\cdot 3)+\hat k(2\cdot 5-1\cdot 3)=-17\hat\imath+13\hat\jmath+7\hat k\).
Example 15. Find the area of the parallelogram with adjacent sides \(\vec a=3\hat\imath+\hat\jmath+4\hat k\) and \(\vec b=\hat\imath-\hat\jmath+\hat k\).
\(\vec a\times\vec b=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 3 & 1 & 4\\ 1 & -1 & 1\end{vmatrix}=\hat\imath(1+4)-\hat\jmath(3-4)+\hat k(-3-1)=5\hat\imath+\hat\jmath-4\hat k\). Magnitude \(\sqrt{25+1+16}=\sqrt{42}\). Area \(=\sqrt{42}\) sq. units.
Example 16. Find the area of the triangle with vertices \(A(1,1,1),\ B(1,2,3),\ C(2,3,1)\).
\(\vec{AB}=(0,1,2)\); \(\vec{AC}=(1,2,0)\). \(\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 0 & 1 & 2\\ 1 & 2 & 0\end{vmatrix}=(0-4)\hat\imath-(0-2)\hat\jmath+(0-1)\hat k=-4\hat\imath+2\hat\jmath-\hat k\). Magnitude = \(\sqrt{16+4+1}=\sqrt{21}\). Area \(=\sqrt{21}/2\) sq. units.
Activity: Recognise Perpendicular and Parallel Vectors
L3 Apply
Materials: Pen, paper.
Predict: Are \(\vec a=(1,2,3)\) and \(\vec b=(4,-2,0)\) perpendicular? Are \(\vec c=(2,4,6)\) and \(\vec a\) parallel?
  1. \(\vec a\cdot\vec b=4-4+0=0\). Perpendicular ✓.
  2. \(\vec c=2\vec a\), so they are parallel (positive scalar = same direction).
  3. Try \(\vec d=(2,1,1)\) and \(\vec a=(1,2,3)\). \(\vec a\cdot\vec d=2+2+3=7\ne 0\). Not perpendicular. \(\vec a\times\vec d=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 1 & 2 & 3\\ 2 & 1 & 1\end{vmatrix}=(2-3)\hat\imath-(1-6)\hat\jmath+(1-4)\hat k=-\hat\imath+5\hat\jmath-3\hat k\ne\vec 0\). Not parallel.
  4. Lesson: dot = 0 ⇒ perp; cross = 0 ⇒ parallel; everything else is "in between" (general angle).
The dot test and cross test are complementary. Together they fully characterise the angle between vectors: \(\sin^2\theta+\cos^2\theta=1\), so \(|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2\) (Lagrange identity).

Competency-Based Questions

Scenario: A force \(\vec F=2\hat\imath+3\hat\jmath-\hat k\) N acts on a particle that undergoes displacement \(\vec d=4\hat\imath+\hat\jmath+2\hat k\) m. Work done = \(\vec F\cdot\vec d\) (definition).
Q1. Compute the work done.
L3 Apply
Answer: \(\vec F\cdot\vec d=8+3-2=9\) J.
Q2. (T/F) "If a·b = 0 and a ≠ 0, then b = 0." Justify.
L5 Evaluate
False. a·b = 0 with non-zero a means b is either zero OR perpendicular to a. Counter-example: a = (1,0), b = (0,1) — both non-zero, dot product 0.
Q3. Compute \(\hat\imath\times\hat\jmath\), \(\hat\jmath\times\hat k\), \(\hat k\times\hat\imath\), \(\hat\jmath\times\hat\imath\).
L3 Apply
Answer: \(\hat\imath\times\hat\jmath=\hat k\) (cyclic). \(\hat\jmath\times\hat k=\hat\imath\). \(\hat k\times\hat\imath=\hat\jmath\). \(\hat\jmath\times\hat\imath=-\hat k\) (anti-cyclic = minus). Mnemonic: i→j→k→i in a cycle; forward = +, backward = −.
Q4. Apply: torque on a wrench is \(\vec\tau=\vec r\times\vec F\). If position vector \(\vec r=0.3\hat\imath\) m and force \(\vec F=20\hat\jmath\) N, find the torque magnitude and direction.
L4 Analyse
Solution: \(\vec\tau=0.3\hat\imath\times 20\hat\jmath=6(\hat\imath\times\hat\jmath)=6\hat k\) N·m. Magnitude 6 N·m, direction along +z (out of the page if x = right, y = up).
Q5. Design: in 3-D, find a unit vector perpendicular to both \(\vec a=2\hat\imath+\hat\jmath\) and \(\vec b=\hat\jmath+\hat k\).
L6 Create
Solution: \(\vec a\times\vec b=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 2 & 1 & 0\\ 0 & 1 & 1\end{vmatrix}=(1-0)\hat\imath-(2-0)\hat\jmath+(2-0)\hat k=\hat\imath-2\hat\jmath+2\hat k\). Magnitude = 3. Unit vector: \(\dfrac{1}{3}(\hat\imath-2\hat\jmath+2\hat k)\).

Assertion–Reason Questions

Assertion (A): The dot product is commutative: \(\vec a\cdot\vec b=\vec b\cdot\vec a\).
Reason (R): The angle between \(\vec a,\vec b\) is the same as between \(\vec b,\vec a\), and \(\cos\theta\) is symmetric.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Geometric symmetry yields algebraic commutativity.
Assertion (A): The cross product is anti-commutative: \(\vec a\times\vec b=-\vec b\times\vec a\).
Reason (R): Reversing the order flips the right-hand-rule direction of \(\hat n\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Hand orientation flips, so the unit normal flips.
Assertion (A): \(|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2\).
Reason (R): \(\sin^2\theta+\cos^2\theta=1\), so \(|\vec a|^2|\vec b|^2(\sin^2\theta+\cos^2\theta)=|\vec a|^2|\vec b|^2\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). The "Lagrange identity" is just \(\sin^2+\cos^2=1\) dressed up.

Frequently Asked Questions — Vector Dot Product, Cross Product and Areas

What is the scalar (dot) product?
a·b = |a||b|cos θ. In components: a₁b₁ + a₂b₂ + a₃b₃.
What is the vector (cross) product?
a × b = |a||b|sin θ · n̂, where n̂ is by the right-hand rule perpendicular to both. The result is a vector.
How do you compute a × b in components?
Determinant of the 3×3 matrix with rows i,j,k; a₁,a₂,a₃; b₁,b₂,b₃.
What is the projection of a vector?
Scalar projection of a on b: (a·b)/|b|. Vector projection: ((a·b)/|b|²)b.
How do you find the area of a parallelogram with vectors?
|a × b| where a, b are adjacent sides. For a triangle: (1/2)|a × b|.
When are two vectors perpendicular / parallel?
Perp iff a·b = 0. Parallel iff a × b = 0.
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