Mathematics Class 12 — Part…
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Vector Algebra Exercises and Summary

🎓 Class 12 Mathematics CBSE Theory Ch 10 — Vector Algebra ⏱ ~15 min
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This MCQ module is based on: Vector Algebra Exercises and Summary

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Targeting Class 12 level in General Mathematics, with Advanced difficulty.

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End-of-Chapter Exercises

Exercise 10.1 — Direction cosines, position vectors
1. Compute magnitude of \(\vec a=\hat\imath+\hat\jmath+\hat k\), \(\vec b=2\hat\imath-7\hat\jmath-3\hat k\), \(\vec c=\dfrac{1}{\sqrt 3}\hat\imath+\dfrac{1}{\sqrt 3}\hat\jmath-\dfrac{1}{\sqrt 3}\hat k\).
|a|=√3; |b|=√(4+49+9)=√62; |c|=√(1/3+1/3+1/3)=1 (already a unit vector).
3. Find a unit vector along \(\vec a=2\hat\imath+3\hat\jmath+\hat k\).
|a|=√14. \(\hat a=\dfrac{1}{\sqrt{14}}(2\hat\imath+3\hat\jmath+\hat k)\).
5. Find direction cosines of \(\vec r=\hat\imath+2\hat\jmath+3\hat k\).
|r|=√14. l=1/√14, m=2/√14, n=3/√14. Sum of squares=14/14=1 ✓.
Exercise 10.2 — Vector operations and section formula
1. Find \(|\vec{PQ}|\) for P(2,3,5), Q(4,6,7).
PQ=2î+3ĵ+2k̂. |PQ|=√(4+9+4)=√17.
5. Find sum, difference, scalar multiples for \(\vec a=\hat\imath-2\hat\jmath+\hat k,\ \vec b=-2\hat\imath+4\hat\jmath+5\hat k,\ \vec c=\hat\imath-6\hat\jmath-7\hat k\).
a+b+c=0î−4ĵ−1k̂=−4ĵ−k̂. (Various combinations follow similarly.)
10. Find the position vector of the midpoint of A(2,3,4) and B(4,1,2).
Midpoint = (3,2,3) = 3î+2ĵ+3k̂.
11. Find R dividing \(P=(2,1,3),\ Q=(4,5,-1)\) internally in ratio 2:1.
\(\vec{OR}=(2(4,5,-1)+1(2,1,3))/3=((8+2)/3,(10+1)/3,(-2+3)/3)=(10/3,11/3,1/3)\).
Exercise 10.3 — Scalar product
1. Find angle between \(\vec a=\hat\imath-2\hat\jmath+3\hat k,\ \vec b=3\hat\imath-2\hat\jmath+\hat k\).
a·b=3+4+3=10. |a|=√14, |b|=√14. cos θ=10/14=5/7. θ=cos⁻¹(5/7).
3. Find the projection of \(\vec a=\hat\imath-\hat\jmath\) on \(\vec b=\hat\imath+\hat\jmath\).
a·b=1−1=0. Projection = 0/√2 = 0. (a perpendicular to b.)
7. Find \(|\vec a-\vec b|\) given |a|=2, |b|=3, angle 60°.
\(|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b=4+9-2\cdot 6\cdot\cos 60°=13-6=7\). So \(|\vec a-\vec b|=\sqrt 7\).
10. Show that \(\vec a=2\hat\imath-\hat\jmath+\hat k\) and \(\vec b=\hat\imath-3\hat\jmath-5\hat k\) are perpendicular.
a·b=2+3−5=0. ✓ Perpendicular.
Exercise 10.4 — Vector product
1. Find \(\vec a\times\vec b\) for \(\vec a=\hat\imath-7\hat\jmath+7\hat k,\ \vec b=3\hat\imath-2\hat\jmath+2\hat k\).
\(\vec a\times\vec b=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 1 & -7 & 7\\ 3 & -2 & 2\end{vmatrix}=\hat\imath(-14+14)-\hat\jmath(2-21)+\hat k(-2+21)=0\hat\imath+19\hat\jmath+19\hat k\).
3. Find a unit vector perpendicular to \(\vec a=3\hat\imath+\hat\jmath+2\hat k\) and \(\vec b=2\hat\imath-2\hat\jmath+4\hat k\).
\(\vec a\times\vec b=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 3 & 1 & 2\\ 2 & -2 & 4\end{vmatrix}=\hat\imath(4+4)-\hat\jmath(12-4)+\hat k(-6-2)=8\hat\imath-8\hat\jmath-8\hat k\). |·|=8√3. Unit vector: \(\dfrac{1}{\sqrt 3}(\hat\imath-\hat\jmath-\hat k)\).
10. Find the area of the parallelogram with adjacent sides \(\vec a=\hat\imath-\hat\jmath+3\hat k,\ \vec b=2\hat\imath-7\hat\jmath+\hat k\).
a×b=\(\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 1 & -1 & 3\\ 2 & -7 & 1\end{vmatrix}=\hat\imath(-1+21)-\hat\jmath(1-6)+\hat k(-7+2)=20\hat\imath+5\hat\jmath-5\hat k\). Area = √(400+25+25)=√450=15√2 sq. units.
Miscellaneous Exercise — selected
2. Find scalar components of the vector \(\vec{AB}\) with A(1,0,1) and B(2,3,5).
AB=(2−1,3−0,5−1)=(1,3,4). Components: 1, 3, 4.
5. If \(\vec a=\hat\imath+\hat\jmath+\hat k,\ \vec b=2\hat\imath-\hat\jmath+3\hat k,\ \vec c=\hat\imath-2\hat\jmath+\hat k\), find a unit vector parallel to \(2\vec a-\vec b+3\vec c\).
2a=2î+2ĵ+2k̂; 2a−b=0î+3ĵ−k̂; 2a−b+3c=0+3î+3ĵ−6ĵ−k̂+3k̂=3î−3ĵ+2k̂. |·|=√(9+9+4)=√22. Unit vector: \(\dfrac{1}{\sqrt{22}}(3\hat\imath-3\hat\jmath+2\hat k)\).
7. If \(\vec a, \vec b, \vec c\) are mutually perpendicular unit vectors, find \(|\vec a+\vec b+\vec c|\).
|a+b+c|² = a·a + b·b + c·c + 2(a·b+b·c+c·a) = 1+1+1+0 = 3. So |a+b+c| = √3.
9. Find \(\vec a\cdot\vec b\) if \(\vec a+\vec b+\vec c=\vec 0\) and \(|\vec a|=3,\ |\vec b|=5,\ |\vec c|=7\).
From \(\vec c=-\vec a-\vec b\): \(|\vec c|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b\). 49=9+25+2·a·b ⇒ a·b=15/2. Note: a·b+b·c+c·a = (1/2)(|a+b+c|²−|a|²−|b|²−|c|²) = (0−83)/2 = −83/2 (alternate identity).
Activity: Detect Coplanar Vectors
L4 Analyse
Materials: Pen, paper.
Predict: Three vectors \(\vec a, \vec b, \vec c\) are coplanar iff their scalar triple product \(\vec a\cdot(\vec b\times\vec c)=0\). Test \(\vec a=(1,1,0), \vec b=(1,0,1), \vec c=(0,1,1)\).
  1. Compute \(\vec b\times\vec c=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 1 & 0 & 1\\ 0 & 1 & 1\end{vmatrix}=(0-1)\hat\imath-(1-0)\hat\jmath+(1-0)\hat k=-\hat\imath-\hat\jmath+\hat k\).
  2. \(\vec a\cdot(\vec b\times\vec c)=(1)(-1)+(1)(-1)+(0)(1)=-2\ne 0\). Not coplanar.
  3. Now try \(\vec a=(1,1,1),\ \vec b=(2,1,1),\ \vec c=(3,2,2)\). \(\vec b\times\vec c=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 2 & 1 & 1\\ 3 & 2 & 2\end{vmatrix}=(2-2)\hat\imath-(4-3)\hat\jmath+(4-3)\hat k=0\hat\imath-\hat\jmath+\hat k\). \(\vec a\cdot=0-1+1=0\). Coplanar ✓.
  4. Lesson: scalar triple product = 0 is the algebraic test for coplanarity (3 vectors all lie in one plane through origin).
The scalar triple product \([\vec a,\vec b,\vec c]=\vec a\cdot(\vec b\times\vec c)\) equals the volume of the parallelepiped spanned by the three vectors. Volume zero means the parallelepiped is flat — the three vectors lie in a single plane.

Consolidation Competency-Based Questions

Scenario: A 3-D mechanical system has a force \(\vec F=3\hat\imath+\hat\jmath-2\hat k\) acting at the point with position vector \(\vec r=\hat\imath+2\hat\jmath\) from the origin.
Q1. Compute the torque \(\vec\tau=\vec r\times\vec F\) about the origin.
L3 Apply
Answer: \(\vec\tau=\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ 1 & 2 & 0\\ 3 & 1 & -2\end{vmatrix}=\hat\imath(-4-0)-\hat\jmath(-2-0)+\hat k(1-6)=-4\hat\imath+2\hat\jmath-5\hat k\) N·m.
Q2. (T/F) "If a·b = 0 and a×b = 0 with a, b non-zero, then a and b are parallel and perpendicular simultaneously." Justify.
L5 Evaluate
Vacuously consistent — but impossible for non-zero vectors. Perpendicular means cos θ = 0; parallel means sin θ = 0. Both can't hold simultaneously (sin² + cos² = 1, so they can't both be 0). One of a or b must be zero.
Q3. Find the area of the triangle with vertices \(A(1,0,0), B(0,1,0), C(0,0,1)\).
L3 Apply
Solution: AB=(−1,1,0); AC=(−1,0,1). AB×AC=\(\begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ -1 & 1 & 0\\ -1 & 0 & 1\end{vmatrix}=(1-0)\hat\imath-(-1-0)\hat\jmath+(0+1)\hat k=\hat\imath+\hat\jmath+\hat k\). Magnitude = √3. Area = √3/2 sq. units.
Q4. Apply: prove that \((\vec a-\vec b)\times(\vec a+\vec b)=2(\vec a\times\vec b)\).
L4 Analyse
Proof: Distribute: \((\vec a-\vec b)\times(\vec a+\vec b)=\vec a\times\vec a+\vec a\times\vec b-\vec b\times\vec a-\vec b\times\vec b=0+\vec a\times\vec b-(-\vec a\times\vec b)-0=2\vec a\times\vec b\). \(\square\)
Q5. Design: a force of \(\vec F=4\hat\imath+\hat\jmath-3\hat k\) N moves a particle from origin to (1,1,1). Find the work done.
L6 Create
Solution: Displacement \(\vec d=\hat\imath+\hat\jmath+\hat k\). Work = F·d = 4+1−3 = 2 J.

Consolidation Assertion–Reason

Assertion (A): The vector \(\hat\imath\times\hat\jmath\) equals \(\hat k\).
Reason (R): The standard right-hand orientation gives the cyclic rule i→j→k.
(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). Cyclic rule directly gives A.
Assertion (A): The area of a triangle with sides \(\vec a, \vec b\) is \((1/2)|\vec a\times\vec b|\).
Reason (R): The triangle is half of the parallelogram with the same two sides.
(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 diagonal of a parallelogram splits it into two congruent triangles.

Chapter Summary

Key concepts
  • Vector: magnitude + direction. Scalar: magnitude only.
  • Position vector: \(\vec{OP}=x\hat\imath+y\hat\jmath+z\hat k\) for \(P(x,y,z)\).
  • Direction cosines: \(l=a/|\vec r|, m=b/|\vec r|, n=c/|\vec r|\); satisfy \(l^2+m^2+n^2=1\).
  • Types: zero, unit, equal, parallel, collinear, coplanar.
  • Addition: triangle and parallelogram laws; commutative, associative.
  • Components: \(\vec r=a\hat\imath+b\hat\jmath+c\hat k\). Sum: add components.
  • \(\vec{PQ}=\vec b-\vec a\); section formula internal: \((m\vec b+n\vec a)/(m+n)\).
  • Dot product: \(\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=a_1b_1+a_2b_2+a_3b_3\). Perp ⇔ \(\vec a\cdot\vec b=0\).
  • Cross product: \(\vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n\); determinant formula. Parallel ⇔ \(\vec a\times\vec b=\vec 0\).
  • Areas: parallelogram = \(|\vec a\times\vec b|\); triangle = \((1/2)|\vec a\times\vec b|\).
  • Projection: scalar = \((\vec a\cdot\vec b)/|\vec b|\); vector = \(((\vec a\cdot\vec b)/|\vec b|^2)\vec b\).

Historical Note

The notion of vectors in mathematics traces back to William Rowan Hamilton's discovery of quaternions in 1843 — a 4-dimensional non-commutative number system whose "vector" part captured rotations in 3-D space. Hamilton coined the words "vector" (Latin carrier) and "scalar". The Grassmann algebra (1844, 1862) developed the algebraic theory of higher-dimensional vectors and exterior products.

The modern 3-D vector calculus we use today was distilled by J. Willard Gibbs (1881, Yale University) and, independently, Oliver Heaviside (1893) — both seeking a clean notation for Maxwell's equations in electromagnetism. By 1900 their vector calculus had largely replaced quaternion methods in physics. The 20th century then generalised vectors to abstract vector spaces (Banach, Hilbert) — the framework underlying modern functional analysis, quantum mechanics, and machine learning.

Frequently Asked Questions — Vector Algebra Exercises and Summary

What is the chapter summary of Class 12 Maths Vector Algebra?
Vectors have magnitude and direction. Direction cosines satisfy l²+m²+n²=1. Addition: triangle/parallelogram law. Dot product gives angles and projections; cross product gives perpendiculars and areas.
Who developed vector algebra?
Hamilton (quaternions, 1843), Grassmann (general algebra, 1844), Gibbs and Heaviside (modern 3-D vector calculus, 1880s–1890s).
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