Previous year questions: Vector Algebra

Previous year questions: vector algebra

The set mixes six questions from actual board papers (2023 to 2026) with four from official sample papers. Every heading gives the paper and the marks. Attempt the question before reading the solution, and check which formula the solution starts with.

Q1 (CBSE Board 2023, 1 mark)

Assertion (A): If a line makes angles alpha, beta and gamma with the positive directions of the coordinate axes, then sin squared alpha plus sin squared beta plus sin squared gamma equals 2. Reason (R): The sum of the squares of the direction cosines of a line is 1. Choose the option: (A) both true, R explains A; (B) both true, R does not explain A; (C) A true, R false; (D) A false, R true.

Answer: The correct option is (A). Each sine squared is one minus the matching cosine squared, so the sum of the three sine squares is 3 minus the sum of the three cosine squares. The direction cosines of a line satisfy l squared plus m squared plus n squared equals one, which makes the sum 3 minus 1, that is 2. The reason gives exactly the identity used in the assertion, so it explains the assertion.

Q2 (CBSE Board 2023, 2 marks)

If a, b and c are three non-zero unequal vectors such that a dot b equals a dot c, find the angle between a and b minus c.

Answer: Bring both dot products to one side: a dot b minus a dot c equals zero, so a dot (b minus c) equals zero. A zero dot product between non-zero vectors means the vectors are perpendicular, so the angle between a and b minus c is ninety degrees, that is pi by 2 radians.

Q3 (CBSE Board 2024, 1 mark)

Let a be any vector such that the magnitude of a is a. The value of the sum of the squares of the magnitudes of a cross i, a cross j and a cross k is: (A) a squared, (B) 2 a squared, (C) 3 a squared, (D) 0.

Answer: Option (B). Each magnitude is the length of a times the sine of the angle that a makes with the corresponding axis. Squaring and adding gives a squared times the sum of the three sine squares, and that sum is 2 because the sum of the three cosine squares is 1. So the total is 2 a squared.

Q4 (CBSE Board 2024, 4 marks)

An instructor at an astronomical centre shows three bright stars with position vectors OD = 2 i + 3 j + 4 k, OA = 7 i + 5 j + 8 k and OV = -3 i + 7 j + 11 k from the telescope at O. (i) How far is star V from star A? (ii) Find a unit vector in the direction of DA. (iii) Find the measure of angle VDA, or find the projection of DV on DA.

Answer: (i) VA equals OA minus OV, which is 10 i - 2 j - 3 k, and its length is the square root of 100 + 4 + 9, that is root 113. (ii) DA equals OA minus OD, which is 5 i + 2 j + 4 k with length 3 root 5, so the unit vector along DA is (5 i + 2 j + 4 k) divided by 3 root 5. (iii) DV equals OV minus OD, which is -5 i + 4 j + 7 k, and its length is 3 root 10. For angle VDA use the vectors DV and DA: DV dot DA is -25 + 8 + 28, that is 11, and the lengths are 3 root 10 and 3 root 5. The cosine of the angle is therefore 11 divided by 45 root 2, and the angle is the inverse cosine of that value. The alternative part asks for the projection of DV on DA, which is 11 divided by 3 root 5.

Q5 (CBSE Board 2025, 2 marks)

Vectors a and b are co-initial and form the adjacent sides of a parallelogram with |a| = 10, |b| = 2 and a dot b = 12. Find the area of the parallelogram.

Answer: The area is the magnitude of a cross b. Use the identity linking the two products: |a cross b| squared equals |a| squared |b| squared minus (a dot b) squared. Substituting, the right side is 400 minus 144, which is 256, so the area is the square root of 256, that is 16 square units.

Q6 (CBSE Board 2026, 3 marks)

If a, b and c are unit vectors, prove that |a - b| squared plus |b - c| squared plus |c - a| squared is less than or equal to 9.

Answer: Expand each squared length. Since each vector is a unit vector, the sum becomes 2 + 2 + 2 minus twice the sum of the three dot products, that is 6 minus twice (a dot b + b dot c + c dot a). Now consider the square of a + b + c: it equals 3 plus twice the same sum of dot products, and a squared length cannot be negative, so the sum of the three dot products is at least minus three halves. Substituting this lower bound gives at most 6 plus 3, which is 9. This proves the inequality.

Q7 (CBSE SQP 2022-23, 1 mark)

Find the scalar projection of 3 i - j - 2 k on i + 2 j - 3 k.

Answer: The projection is the dot product divided by the magnitude of the second vector. The dot product is 3 - 2 + 6, which is 7, and the length of i + 2 j - 3 k is root 14. So the projection is 7 divided by root 14, which simplifies to root 14 divided by 2.

Q8 (CBSE SQP 2022-23, 2 marks)

Find the magnitude of x if (x - a) dot (x + a) = 12, where a is a unit vector.

Answer: Expand the product: (x - a) dot (x + a) equals x dot x plus x dot a minus a dot x minus a dot a. The two mixed terms cancel because the dot product is commutative, so the expression is |x| squared minus |a| squared. Since a is a unit vector, |a| squared is 1, so |x| squared minus 1 equals 12 and |x| squared is 13. The magnitude is therefore root 13.

Q9 (CBSE SQP 2024-25, 2 marks)

Take the co-initial sides 2 i - 4 j - 5 k and 2 i + 2 j + 3 k. Work out both diagonals of the parallelogram they form, then take its area.

Answer: The diagonals are the sum and the difference of the adjacent sides. The sum is 4 i - 2 j - 2 k and the difference is -6 j - 8 k. The area of a parallelogram equals half the magnitude of the cross product of its diagonals, so compute the cross product of 4 i - 2 j - 2 k and -6 j - 8 k. That gives 4 i + 32 j - 24 k, whose magnitude is 4 root 101, and half of that is 2 root 101 square units. The same answer comes from the magnitude of the cross product of the two adjacent sides.

Q10 (CBSE SQP 2025-26, 1 mark)

If a = 3 i + 2 j + 4 k, b = i + j - 3 k and c = 6 i - j + 2 k, then (2 a dot i) i - (b dot j) j + (c dot k) k equals: (A) a, (B) b + c, (C) a - b, (D) c.

Answer: The correct option is (D). The dot products with the unit vectors pick out the matching components: a dot i is 3, b dot j is 1 and c dot k is 2. The expression becomes 6 i - j + 2 k, which is exactly the vector c.

Six of the ten are printed board questions, spread over four board exam years. Notice how often a five-line solution is enough: replace the geometry with the dot or cross product, substitute components, and simplify. Keep the sign conventions straight and state which formula you are using. In the one-mark items the answer is often a single formula, so write the formula and the value. In the case-based and multi-mark items, draw no figure unless it helps, but do name the vectors and their components before substituting. The identity linking the two products and the squared-length expansion are the two algebraic moves that turn most of these questions into arithmetic. Before starting a long computation, look for the identity that collapses the question into one line, and quote the formula you use at the start of each answer.