Class 12 Maths · Original EZT practice
Vector Algebra (Class 12 Maths, NCERT, Chapter 10)
Vector Algebra is Chapter 10 of the NCERT Class 12 Mathematics Part 2 textbook. 'What is your height?' needs one number; 'How should a football player hit the ball to give a pass to another player of his team?' needs direction too - 'a quantity that involves only one value (magnitude) which is a real number. Such quantities are called scalars', and the directed ones are vectors. This chapter builds vector arithmetic: types of vectors, addition by triangle and parallelogram laws, scalar multiplication, components and section formula, then the dot and cross products that run all of Class 12 physics. (Note: this is original practice, not a textbook answer key.)
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 26 October 2026:
1. What the chapter covers
10.1 Introduction: scalars vs vectors from everyday queries. 10.2 Some Basic Concepts: position vector, direction cosines and ratios, magnitude. 10.3 Types of Vectors: zero, unit, coinitial, collinear, equal and negative vectors. 10.4 Addition of Vectors: triangle and parallelogram laws, properties. 10.5 Multiplication of a Vector by a Scalar: components of a vector, vector joining two points, section formula. 10.6 Product of Two Vectors: scalar (dot) product and projection; vector (cross) product, its direction and magnitude as the parallelogram area. EXERCISES 10.1-10.4 plus the Miscellaneous Exercise close the chapter.
2. Key ideas at a glance
A vector carries magnitude and direction; two vectors are equal when both match, wherever they sit. Direction cosines l, m, n satisfy l^2 + m^2 + n^2 = 1. The section formula divides a join internally in ratio m:n as (m b + n a)/(m + n) - position vectors make it one line. Dot product a.b = |a||b|cos theta: zero exactly when the vectors are perpendicular; projection falls straight out of it. Cross product magnitude |a x b| = |a||b|sin theta is the parallelogram area; direction is perpendicular to both, right-hand rule.
3. Traps this chapter sets
Mixing up direction ratios and direction cosines: ratios are any proportional triple; cosines are the one triple whose squares sum to 1. Declaring vectors equal because the magnitudes match: direction has to match too, and position is irrelevant. Treating a zero dot product as a zero vector: a.b = 0 says perpendicular, not zero. Forgetting that the cross product is ordered: b x a = -(a x b), so swapping the inputs flips the answer's direction.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Ratios to cosines
Choose the method: Divide by the norm.
EZT practice question
A vector has direction ratios 1, -2, 2. Write its direction cosines, and then write one actual vector of length 6 with these direction ratios.
Show the worked solution
- Norm of the ratio triple = sqrt(1 + 4 + 4) = 3, so direction cosines are 1/3, -2/3, 2/3.
- Any vector with these ratios is a scalar multiple of i - 2j + 2k.
- For length 6 take 2(i - 2j + 2k) = 2i - 4j + 4k.
Answer: Direction cosines 1/3, -2/3, 2/3; one length-6 vector is 2i - 4j + 4k - ratios normalised to cosines, then scaled to the required length.
Check: Squares of the cosines sum to 1, and 2i - 4j + 4k has magnitude sqrt(4 + 16 + 16) = 6.
Watch for this: Do not report 1, -2, 2 as direction cosines - their squares sum to 9, not 1; ratios become cosines only after division by the norm.
Now try it yourself
Is 2i - 4j + 4k the only vector of length 6 with these direction ratios?
Check your answer
No - its negative -2i + 4j - 4k has the same ratios up to an overall sign and the same length. Direction ratios determine a line of directions, and both unit vectors on that line share them.
Ratios fix the line, cosines fix the sign convention, and the length fixes the vector up to that sign.
Join, then divide
Choose the method: Subtract, then weight.
EZT practice question
Points A and B have position vectors a = i + 2j - k and b = 3i - j + k. Find the vector from A to B and the position vector of the point dividing AB in the ratio 2:1 internally.
Show the worked solution
- AB = b - a = (3 - 1)i + (-1 - 2)j + (1 + 1)k = 2i - 3j + 2k.
- Section formula for ratio 2:1 internally: (2b + 1a)/(2 + 1).
- = (7i + 0j + k)/3.
Answer: AB = 2i - 3j + 2k, and the dividing point has position vector (7i + k)/3 - the vector joining two points is the difference, the section formula is the weighted average.
Check: Component-wise subtraction for AB; weights 2 and 1 sit on the opposite endpoints (2 multiplies b) in the section formula.
Watch for this: Do not compute a - b for the vector from A to B - the vector from A to B is b - a; and in the section formula the weight m multiplies the far endpoint b, not the near one.
Now try it yourself
What does the midpoint formula say in this language?
Check your answer
Ratio 1:1 gives (a + b)/2 - each endpoint weighted equally. Every section point is a weighted average of the two position vectors.
The section formula is one idea with one parameter; the midpoint is its symmetric case.
Test perpendicularity
Choose the method: Dot to zero.
EZT practice question
Show that the vectors i + j + k and i - j are perpendicular.
Show the worked solution
- Compute the dot product: (1)(1) + (1)(-1) + (1)(0).
- = 1 - 1 + 0 = 0.
- The dot product is zero, so the angle is 90 degrees.
Answer: The dot product is 1 - 1 + 0 = 0, so the vectors are perpendicular - the dot-product-zero test from EXERCISE 10.3.
Check: Component-wise products summed to zero.
Watch for this: Do not eyeball perpendicularity from the components - i + j + k has no zero component yet is still perpendicular to i - j; only the dot product decides.
Now try it yourself
If a.b = 0 and neither vector is zero, what does the formula say?
Check your answer
a.b = |a||b|cos theta = 0 with nonzero magnitudes forces cos theta = 0, so theta = 90 degrees - perpendicularity is the only way two nonzero vectors dot to zero.
The zero dot product is a statement about angle, not about size.
Reason with projections
Choose the method: Set the formulas equal.
EZT practice question
The projection of a on b and the projection of b on a are equal for a particular pair of nonzero vectors with a.b not zero. What must be true about a and b? Reason from the formulas.
Show the worked solution
- Projection of a on b = a.b/|b|; projection of b on a = a.b/|a|.
- With a.b nonzero, equality forces 1/|b| = 1/|a|.
- So |a| = |b| - equal magnitudes, not necessarily equal vectors.
Answer: The magnitudes must be equal: from a.b/|b| = a.b/|a| and a.b nonzero, cancel a.b to get |a| = |b| - a conclusion about size only, not direction.
Check: The two projection formulas, the cancellation of the common nonzero a.b, and that the conclusion mentions magnitudes only.
Watch for this: Do not conclude a = b - equal projections say nothing about direction; two vectors of equal length at any angle to each other satisfy the condition.
Now try it yourself
Why was a.b not zero needed in the argument?
Check your answer
If a.b = 0 both projections are zero and equal for any magnitudes - the condition then says nothing. Cancelling a factor is only legal when the factor is nonzero.
Every cancellation hides a hypothesis; state it before you use it.
Build the perpendicular
Choose the method: Cross, then normalise.
EZT practice question
Find a unit vector perpendicular to both a = 3i + j + 2k and b = 2i - 2j + 4k.
Show the worked solution
- a x b = i(4 - (-4)) - j(12 - 4) + k(-6 - 2) = 8i - 8j - 8k.
- Magnitude = sqrt(64 + 64 + 64) = 8 sqrt3.
- Unit perpendicular = (i - j - k)/sqrt3, and its negative also works.
Answer: (i - j - k)/sqrt3 (or its negative) - the cross product 8i - 8j - 8k supplies a vector perpendicular to both inputs, and dividing by its magnitude 8 sqrt3 makes it a unit vector.
Check: Each cross-product component recomputed, the magnitude 8 sqrt3, and both signs noted.
Watch for this: Do not stop at the cross product - 8i - 8j - 8k is perpendicular to both but is not a unit vector; and do not report only one sign when the question asks for a perpendicular without orientation.
Now try it yourself
Why does the cross product automatically give a perpendicular?
Check your answer
Its direction is defined to be perpendicular to both inputs by the right-hand rule, and its magnitude |a||b|sin theta is the parallelogram area - the geometric job and the computation are the same object.
The cross product exists precisely to manufacture perpendiculars.
One angle, two products
Choose the method: Cosine from dot, sine from cross.
EZT practice question
If a and b are unit vectors with a.b = 1/2, what angle do they make, and what is |a x b|?
Show the worked solution
- a.b = |a||b|cos theta = cos theta = 1/2, so theta = 60 degrees.
- |a x b| = |a||b|sin theta = sin 60 degrees.
- = sqrt3/2.
Answer: Theta = 60 degrees and |a x b| = sqrt3/2 - the dot product hands over the cosine, the cross product's magnitude uses the sine of the same angle.
Check: Unit magnitudes used in both formulas; cos 60 = 1/2 and sin 60 = sqrt3/2.
Watch for this: Do not compute |a x b| as 1/2 by confusing sine with cosine - the dot product reads the cosine, the cross magnitude reads the sine, and the same angle feeds both.
Now try it yourself
What would |a x b| be if the vectors were parallel?
Check your answer
Zero - sin 0 = 0, so the parallelogram collapses. Parallel vectors have the largest possible dot product and the smallest possible cross magnitude.
The two products split the angle's information: dot tracks agreement, cross tracks spread.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Direction cosines from ratios?
Show answer and reason
Divide the ratio triple by its norm; the resulting l, m, n satisfy l^2 + m^2 + n^2 = 1.
Ratios are many, cosines are the normalised one.
2. Section formula, ratio m:n internally?
Show answer and reason
(m b + n a)/(m + n) - the weight sits on the opposite endpoint.
Position vectors turn division into one weighted average.
3. When is a.b zero?
Show answer and reason
Exactly when two nonzero vectors are perpendicular.
Zero dot product is an angle statement, not a size statement.
4. Meaning of |a x b|?
Show answer and reason
The area of the parallelogram with a and b as adjacent sides.
Cross magnitude is geometry you can compute.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Explain in your own words why two vectors with the same components but different starting points are the same vector, and give an everyday example.
- For a = i + 2j + 2k, find its direction cosines, and check directly that their squares sum to 1.
- State in one sentence what changes and what stays the same when a vector is multiplied by the scalar -3.
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Questions about this practice page
- Which book is Class 12 Maths Chapter 10 Vector Algebra from?
- It is Chapter 10 of the NCERT Class 12 Mathematics Part 2 textbook (official NCERT chapter PDF lemh204.pdf). No printing-year footer was observed in the fetched chapter text; the front matter lists Edition January 2023 with reprints through 'Reprint 2026-27' - check your own copy, as printings can differ.
- How does Chapter 10 distinguish a scalar from a vector?
- Per the chapter, a scalar involves only one value (magnitude), a real number - like height. A vector carries both magnitude and direction - like a pass in football, which must be hit a certain way, not just a certain strength.
- What is the dot product of two vectors?
- a.b = |a||b|cos theta, a scalar. It is zero exactly when the vectors are perpendicular, and the projection of a on b is a.b / |b|.
- What is the cross product of two vectors?
- a x b is a vector of magnitude |a||b|sin theta - equal to the area of the parallelogram with a and b as adjacent sides - pointing perpendicular to both by the right-hand rule.
- What is the section formula in vectors?
- The point dividing the join of position vectors a and b internally in the ratio m:n has position vector (m b + n a)/(m + n); for the midpoint this is simply (a + b)/2.