68 free MCQs on Vector Algebra with worked answers and explanations. Vectors, dot product, cross product, and geometric applications
Take the timed Vector Algebra chapterwise test →Below are 68 practice questions on Vector Algebra, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Vector Algebra notes.
Triangle law of vector addition: placing vector b at the head of vector a, the diagonal from the start to the final head gives the resultant a + b.
Q1.
A quantity that has both magnitude and direction is called a:
Answer: B. Vector
Why: A vector has both magnitude (size) and direction, unlike a scalar which has only magnitude.
Q2.
The magnitude of vector a = 3i + 4j is:
Answer: C. 5
Why: |a| = sqrt(3<sup>2</sup> + 4<sup>2</sup>) = sqrt(9 + 16) = sqrt(25) = 5.
Q3.
A unit vector has magnitude equal to:
Answer: D. 1
Why: A unit vector has magnitude exactly 1. It only indicates direction.
Q4.
The unit vector along vector a is:
Answer: A. a / |a|
Why: To find a unit vector in the direction of a, divide the vector by its magnitude: a<sub>hat</sub> = a / |a|.
Q5.
Two vectors are equal if they have the same:
Answer: D. Magnitude and direction
Why: Two vectors are equal if and only if they have the same magnitude and the same direction.
Q6.
The zero vector has magnitude:
Answer: A. 0
Why: The zero vector (null vector) has magnitude 0 and no definite direction.
Q7.
If a = 2i - 3j + k and b = i + j - 2k, then a + b =
Answer: D. 3i - 2j - k
Why: Add components: (2+1)i + (-3+1)j + (1-2)k = 3i - 2j - k.
Q8.
The dot product (scalar product) of i and j is:
Answer: C. 0
Why: i . j = 0 because i and j are perpendicular unit vectors. For perpendicular vectors, a.b = |a||b|cos90 = 0.
Q9.
The dot product i . i equals:
Answer: A. 1
Why: i . i = |i||i|cos0 = 1 x 1 x 1 = 1. Any unit vector dotted with itself equals 1.
Q10.
If a . b = 0 and neither a nor b is zero, then a and b are:
Answer: B. Perpendicular
Why: a.b = |a||b|cos(theta) = 0 implies cos(theta) = 0, so theta = 90 degrees. The vectors are perpendicular.
Q11.
The cross product i x i equals:
Answer: D. 0 (zero vector)
Why: The cross product of any vector with itself is the zero vector because sin(0) = 0.
Q12.
i x j equals:
Answer: C. k
Why: i x j = k (follows the right-hand rule and cyclic order: i x j = k, j x k = i, k x i = j).
Q13.
The position vector of point P(3, -2, 5) from the origin is:
Answer: B. 3i - 2j + 5k
Why: Position vector of P(x,y,z) = xi + yj + zk = 3i - 2j + 5k.
Q14.
Scalar multiplication: if k = 3 and a = 2i - j, then ka =
Answer: A. 6i - 3j
Why: Multiply each component by the scalar: 3(2i - j) = 6i - 3j.
Q15.
Collinear vectors are vectors that:
Answer: D. Are parallel (lie along the same or parallel lines)
Why: Collinear vectors lie along the same line or parallel lines. One is a scalar multiple of the other.
Q16.
If a = i + 2j + 3k, then |a|<sup>2</sup> =
Answer: D. 14
Why: |a|<sup>2</sup> = 1<sup>2</sup> + 2<sup>2</sup> + 3<sup>2</sup> = 1 + 4 + 9 = 14.
Q17.
The dot product formula a.b = |a||b|cos(theta). If theta = 0, then a.b =
Answer: A. |a||b|
Why: cos(0) = 1, so a.b = |a||b| x 1 = |a||b|. Parallel vectors have the maximum positive dot product.
Q18.
The triangle law of vector addition states that the third side of a triangle represents:
Answer: B. a + b
Why: Triangle law: place vectors head to tail; the closing side (from tail of a to head of b) is a + b.
Q19.
A vector whose initial and terminal points are the same is called a:
Answer: C. Zero vector
Why: When start and end points coincide, the vector has zero magnitude and is called the zero vector.
Q20.
The parallelogram law of vector addition: |a + b|<sup>2</sup> + |a - b|<sup>2</sup> =
Answer: C. 2(|a|<sup>2</sup> + |b|<sup>2</sup>)
Why: Parallelogram law of cosines: |a+b|<sup>2</sup> + |a-b|<sup>2</sup> = 2(|a|<sup>2</sup> + |b|<sup>2</sup>). The sum of squares of diagonals equals twice the sum of squares of the sides.
Q21.
Find a.b if a = 2i + 3j - k and b = i - 2j + 4k.
Answer: B. -8
Why: a.b = (2)(1) + (3)(-2) + (-1)(4) = 2 - 6 - 4 = -8.
Q22.
The angle between a = i + j and b = j + k satisfies:
Answer: A. cos(theta) = 1/2
Why: a.b = 0+1+0 = 1. |a| = sqrt(2), |b| = sqrt(2). cos(theta) = 1/(sqrt(2)*sqrt(2)) = 1/2. theta = 60 degrees.
Q23.
If theta = 90 degrees and |a| = 3, |b| = 4, then |a x b| =
Answer: C. 12
Why: |a x b| = |a||b|sin(90) = 3 x 4 x 1 = 12. Cross product is maximum when vectors are perpendicular.
Q24.
The cross product satisfies:
Answer: B. a x b = -(b x a)
Why: Cross product is anti-commutative: a x b = -(b x a). Swapping the order reverses direction.
Q25.
The scalar projection of vector a onto vector b is:
Answer: B. (a.b)/|b|
Why: Scalar projection of a onto b = (a.b)/|b|. The vector projection = [(a.b)/|b|<sup>2</sup>] * b.
Q26.
The area of a parallelogram with adjacent sides a and b is:
Answer: A. |a x b|
Why: Area of parallelogram = |a x b|. The area of the triangle formed by these vectors = (1/2)|a x b|.
Q27.
If a = i + j + k, the unit vector in the direction of a is:
Answer: D. (1/sqrt(3))(i+j+k)
Why: |a| = sqrt(1+1+1) = sqrt(3). Unit vector = a/|a| = (1/sqrt(3))(i + j + k).
Q28.
Three vectors a, b, c are coplanar if and only if:
Answer: C. a.(b x c) = 0
Why: Vectors a, b, c are coplanar if and only if their scalar triple product a.(b x c) = 0.
Q29.
If a x b = 0 and both a and b are non-zero, then:
Answer: B. a is parallel to b
Why: a x b = 0 means sin(theta) = 0, so theta = 0 or 180 degrees. The vectors are parallel (or anti-parallel).
Q30.
The section formula: point dividing AB in ratio m:n internally (position vectors a, b) has position vector:
Answer: B. (na + mb)/(m+n)
Why: Internal division: position vector = (mb + na)/(m+n). For midpoint (m=n=1): (a+b)/2.
Q31.
The scalar triple product a.(b x c) gives:
Answer: D. The volume of the parallelepiped formed by a, b, c
Why: The scalar triple product [a b c] = a.(b x c) equals the volume of the parallelepiped formed by vectors a, b, c as edges.
Q32.
The vector component of a perpendicular to b is:
Answer: D. a - (a.b/|b|<sup>2</sup>)b
Why: The projection of a along b is (a.b/|b|<sup>2</sup>)b. The perpendicular component = a - (a.b/|b|<sup>2</sup>)b.
Q33.
If a = 3i + 2j - k, then |2a| =
Answer: B. 2sqrt(14)
Why: |a| = sqrt(9+4+1) = sqrt(14). |2a| = 2|a| = 2sqrt(14).
Q34.
Work done by force F over displacement d is:
Answer: C. F . d
Why: Work W = F . d = |F||d|cos(theta). Work is a scalar equal to the dot product of force and displacement vectors.
Q35.
If a + b + c = 0, then a.b + b.c + c.a =
Answer: C. -(|a|<sup>2</sup>+|b|<sup>2</sup>+|c|<sup>2</sup>)/2
Why: |a+b+c|<sup>2</sup> = 0. Expanding: |a|<sup>2</sup>+|b|<sup>2</sup>+|c|<sup>2</sup>+2(a.b+b.c+c.a) = 0. So a.b+b.c+c.a = -(|a|<sup>2</sup>+|b|<sup>2</sup>+|c|<sup>2</sup>)/2.
Q36.
The resultant of two equal vectors of magnitude F at angle theta has magnitude:
Answer: D. 2F cos(theta/2)
Why: |R|<sup>2</sup> = F<sup>2</sup>+F<sup>2</sup>+2F<sup>2</sup> cos(theta) = 2F<sup>2</sup>(1+cos(theta)) = 4F<sup>2</sup> cos<sup>2</sup>(theta/2). So |R| = 2F cos(theta/2).
Q37.
The vectors |a| = 3, |b| = 2, a.b = 4. The angle theta satisfies:
Answer: A. cos(theta) = 2/3
Why: cos(theta) = a.b / (|a||b|) = 4/(3 x 2) = 4/6 = 2/3. theta = cos<sup>-1</sup>(2/3).
Q38.
The midpoint of segment joining A(a) and B(b) has position vector:
Answer: C. (a + b)/2
Why: Midpoint divides in ratio 1:1. By section formula: (1*b + 1*a)/2 = (a+b)/2.
Q39.
The moment (torque) of force F about a point is:
Answer: D. r x F
Why: Moment (torque) = r x F, where r is the position vector from the point to the force application point. It is a vector.
Q40.
If a.b = a.c and a is non-zero, it necessarily means:
Answer: D. a is perpendicular to (b - c)
Why: a.b = a.c means a.(b-c) = 0. So a is perpendicular to (b-c). This does NOT mean b = c.
Q41.
The vector triple product a x (b x c) equals (BAC-CAB rule):
Answer: A. (a.c)b - (a.b)c
Why: BAC-CAB rule: a×(b×c) = b(a·c) − c(a·b). Mnemonic: the outer vector a dots with each inner vector. Result lies in the plane of b and c. Answer: (a·c)b − (a·b)c.
Q42.
If |a + b| = |a - b|, then a and b must be:
Answer: C. Perpendicular
Why: |a+b|<sup>2</sup> = |a-b|<sup>2</sup> gives |a|<sup>2</sup>+2a.b+|b|<sup>2</sup> = |a|<sup>2</sup>-2a.b+|b|<sup>2</sup>, so 4a.b = 0. Hence a.b = 0, meaning they are perpendicular.
Q43.
The scalar triple product satisfies:
Answer: B. [a b c] = -[b a c] but [a b c] = [b c a]
Why: [abc] is cyclic: [abc]=[bca]=[cab]. Swapping any two adjacent vectors negates it: [abc]=-[bac]. Statement B captures both properties correctly.
Q44.
The area of triangle with vertices A(1,0,0), B(0,1,0), C(0,0,1) is:
Answer: D. sqrt(3)/2
Why: AB = B−A = (−1,1,0), AC = C−A = (−1,0,1). Cross product AB×AC = |i j k; −1 1 0; −1 0 1| = (1,1,1). |AB×AC| = √3. Area = ½√3 = √3/2. Answer: √3/2.
Q45.
The identity (a x b).(c x d) equals:
Answer: C. (a.c)(b.d) - (a.d)(b.c)
Why: Using the scalar quadruple product identity (Lagrange): (a×b)·(c×d) = det[[a·c, a·d],[b·c, b·d]] = (a·c)(b·d) − (a·d)(b·c). Answer: (a·c)(b·d) − (a·d)(b·c).
Q46.
If a, b, c are mutually perpendicular unit vectors, |a + b + c|<sup>2</sup> =
Answer: A. 3
Why: Expand: |a+b+c|² = |a|²+|b|²+|c|²+2(a·b+b·c+c·a). Unit vectors: |a|²=|b|²=|c|²=1. Mutually perpendicular: all dot products = 0. So |a+b+c|² = 1+1+1+0 = 3. Answer: 3.
Q47.
The volume of the tetrahedron with one vertex at origin, others at a, b, c is:
Answer: D. (1/6)|[a b c]|
Why: Volume of tetrahedron = (1/6)|a.(b x c)|. The parallelepiped formed by the same edges has volume |[abc]|, and the tetrahedron is 1/6 of that.
Q48.
The Lagrange identity states |a x b|<sup>2</sup> =
Answer: C. |a|<sup>2</sup>|b|<sup>2</sup> - (a.b)<sup>2</sup>
Why: Lagrange identity: |a x b|<sup>2</sup> = |a|<sup>2</sup>|b|<sup>2</sup> - (a.b)<sup>2.</sup> This follows since |a x b| = |a||b|sin(theta) and (a.b) = |a||b|cos(theta).
Q49.
If [a b c] = 5, then [2a 3b 4c] =
Answer: C. 120
Why: Scalar triple product is multilinear: [2a, 3b, 4c] = 2·3·4·[a, b, c] = 24×5 = 120. Each scalar factor on a vector multiplies out of the determinant independently. Answer: 120.
Q50.
If a x b = a x c (a is non-zero), then necessarily:
Answer: C. b - c is parallel to a
Why: a×b = a×c → a×(b−c) = 0. Cross product is zero iff vectors are parallel (or one is zero). So (b−c) is parallel to a. Note: b−c need not be zero, so b≠c in general. Answer: b−c is parallel to a.
Q51.
For non-coplanar vectors a, b, c forming a basis, every vector r has a:
Answer: C. Unique linear combination r = xa + yb + zc
Why: Three non-coplanar (linearly independent) vectors span ℝ³ and form a basis. By the definition of a basis, every vector r has a unique representation r = xa + yb + zc where x, y, z are scalars. Answer: unique linear combination r = xa + yb + zc.
Q52.
In the reciprocal system, if a<sub>prime</sub> = (b x c)/[abc], then a . a<sub>prime</sub> =
Answer: D. 1
Why: By definition of reciprocal vectors: a.a<sub>prime</sub> = 1, a.b<sub>prime</sub> = 0, a.c<sub>prime</sub> = 0. The reciprocal system is the dual basis.
Q53.
If ABCD is a parallelogram with diagonal AC = p and BD = q, then AB =
Answer: A. (p + q)/2
Why: Setting up: p = AB + BC = AB + AD and q = -AB + AD. Adding: p + q = 2AD so AD = (p+q)/2. Subtracting: p - q = 2AB so AB = (p-q)/2. Wait: q = BD = BA + AD = -AB + AD. So p+q = 2AD and p-q = 2AB. Thus AB = (p-q)/2. Answer is index 3.
Q54.
The equation of a line through point a and perpendicular to both b and c is:
Answer: C. r = a + lambda*(b x c)
Why: A vector perpendicular to both b and c is their cross product n = b×c. The line through position vector a with direction n is r = a + λ(b×c), where λ∈ℝ. Answer: r = a + λ(b×c).
Q55.
Using a.b = |a||b|cos(theta): angle between a = 3i+4j and b = 4i+3j satisfies:
Answer: D. cos(theta) = 24/25
Why: Dot product: a·b = 3(4)+4(3) = 12+12 = 24. Magnitudes: |a| = √(9+16) = 5, |b| = √(16+9) = 5. cosθ = 24/(5×5) = 24/25. Answer: cosθ = 24/25.
Q56.
If a = i+j, b = j+k, c = k+i, the scalar triple product [a b c] =
Answer: D. 2
Why: b x c = (j+k) x (k+i) = j x k + j x i + k x k + k x i = i + (-j) + 0 + j = i. Wait: j x k = i, j x i = -k, k x k = 0, k x i = j. So b x c = i - k + j = i+j-k. a.(b x c) = (i+j).(i+j-k) = 1+1+0 = 2.
Q57.
The work done by F = 2i - j + k along displacement from A(1,0,0) to B(2,1,3) is:
Answer: A. 4
Why: Displacement d = B−A = (2−1, 1−0, 3−0) = (1,1,3). Work W = F·d = 2(1)+(−1)(1)+1(3) = 2−1+3 = 4 units. Answer: W = 4.
Q58.
If vectors (2,-1,k), (1,2,-1), (0,1,2) are coplanar, then k =
Answer: A. -12
Why: Coplanar → scalar triple product = 0. det[[2,−1,k],[1,2,−1],[0,1,2]] = 2(4+1)−(−1)(2−0)+k(1−0) = 10+2+k = 0. So k = −12. Answer: k = −12.
Q59.
The vector equation of the plane through three points with position vectors a, b, c is:
Answer: D. r = a + lambda*(b-a) + mu*(c-a)
Why: The plane through a, b, c contains two direction vectors (b−a) and (c−a). Parametric form: r = a + λ(b−a) + μ(c−a), where λ,μ∈ℝ. When λ=μ=0: r=a; λ=1,μ=0: r=b; λ=0,μ=1: r=c. Answer: r = a + λ(b−a) + μ(c−a).
Q60.
The magnitude of the vector 3î + 4ĵ is:
Answer: A. 5
Why: Magnitude = √(3² + 4²) = √25 = 5.
Q61.
If a = i + j + k and b = i − j + k, the magnitude of a × b is:
Answer: B. 2√2
Why: a × b = 2i − 2k, whose magnitude is √(4 + 4) = 2√2.
Q62.
The projection of a = 2i + 3j + 2k on b = i + 2j + k is:
Answer: A. 10/√6
Why: Projection = (a·b)/|b| = (2 + 6 + 2)/√6 = 10/√6.
Q63.
If |a| = 3, |b| = 4 and |a + b| = 5, then a·b equals:
Answer: A. 0
Why: |a + b|² = |a|² + |b|² + 2a·b gives 25 = 9 + 16 + 2a·b, so a·b = 0.
Q64.
The volume of the parallelepiped with edges i + j, j + k and k + i is:
Answer: B. 2
Why: The scalar triple product equals the determinant, which is 2.
Q65.
A unit vector perpendicular to both i + j and j + k is:
Answer: B. (i − j + k)/√3
Why: (i + j) × (j + k) = i − j + k, whose unit vector is (i − j + k)/√3.
Q66.
If a, b, c are mutually perpendicular unit vectors, then |a + b + c| equals:
Answer: C. √3
Why: |a + b + c|² = 1 + 1 + 1 = 3 (cross terms vanish), so the magnitude is √3.
Q67.
The value of λ for which 2i + λj + k is perpendicular to i − 2j + 3k is:
Answer: B. 5/2
Why: Dot product = 2 − 2λ + 3 = 0 gives λ = 5/2.
Q68.
The area of the triangle with vertices A(1,1,1), B(2,3,5) and C(1,5,5) is:
Answer: B. 2√6
Why: AB = (1,2,4), AC = (0,4,4); AB × AC = (−8,−4,4) with magnitude 4√6. Area = half of this = 2√6.