50 questions on this topic of Kinematics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (A) -5
The expression (A⃗ + B⃗) · (A⃗ - B⃗) expands to |A⃗|² - |B⃗|². Here |A⃗|² = 2² + (-1)² + 2² = 9, and |B⃗|² = 3² + 2² + (-1)² = 14. Therefore, |A⃗|² - |B⃗|² = 9 - 14 = -5.
Answer: (D) (A² + B² + AB)¹/²
From |A⃗ × B⃗| = √3 (A⃗ · B⃗), we have AB sin θ = √3 AB cos θ, so tan θ = √3, yielding θ = 60°. The magnitude of the resultant is |A⃗ + B⃗| = √(A² + B² + 2AB cos 60°) = √(A² + B² + 2AB(1/2)) = (A² + B² + AB)¹/².
Answer: (A) (6 î - 2 ĵ - k̂)/√41
A vector perpendicular to both A⃗ and B⃗ is given by A⃗ × B⃗. Evaluating the determinant: A⃗ × B⃗ = î(12 - 6) - ĵ(6 - 4) + k̂(3 - 4) = 6 î - 2 ĵ - k̂. The magnitude of this vector is √(6² + (-2)² + (-1)²) = √(36 + 4 + 1) = √41. Thus, the unit vector is (6 î - 2 ĵ - k̂)/√41.
Answer: (B) 60°
The dot product is A⃗ · B⃗ = (1)(-1) + (2)(1) + (-1)(-2) = -1 + 2 + 2 = 3. The magnitudes are |A⃗| = √(1² + 2² + (-1)²) = √6 and |B⃗| = √((-1)² + 1² + (-2)²) = √6. Thus, cos θ = (A⃗ · B⃗) / (|A⃗||B⃗|) = 3 / (√6 × √6) = 3/6 = 1/2, giving θ = 60°.
Answer: (C) 12/13
The magnitude of displacement is |d⃗| = √(4² + (-3)² + 12²) = √(16 + 9 + 144) = √169 = 13 m. The direction cosine with the z-axis is cos γ = z / |d⃗| = 12/13.
Answer: (B) (-5 î + 5 ĵ + 5 k̂) N m
The torque is given by the determinant of [î, ĵ, k̂; 1, -1, 2; 2, 3, -1]. Evaluating the components: τₓ = (-1)(-1) - (2)(3) = 1 - 6 = -5; τᵧ = -[(1)(-1) - (2)(2)] = -[-1 - 4] = +5; τ_z = (1)(3) - (-1)(2) = 3 + 2 = +5. Therefore, τ⃗ = -5 î + 5 ĵ + 5 k̂ N m.
Answer: (B) Displacement
A vector has both magnitude and direction and adds by the triangle/parallelogram law. Displacement is a vector; distance, speed and mass are scalars.
NCERT Class 11
Answer: (D) 1/√3
Let α, β, and γ be the angles made by A⃗ with the x, y, and z axes respectively. Since α = β = γ, we have cos²α + cos²β + cos²γ = 1, which gives 3 cos²α = 1. Therefore, cos α = 1/√3.
Answer: (C) Statement I is incorrect but Statement II is correct.
Statement I is incorrect because quantities like electric current have both magnitude and direction, yet they do not obey the laws of vector addition and are treated as scalars. Statement II is correct because finite rotations do not commute (R₁ + R₂ ≠ R₂ + R₁), so they cannot be represented as vectors.
Answer: (D) 60°
The dot product is A⃗ · B⃗ = (1)(0) + (1)(1) + (0)(1) = 1. The magnitudes are |A⃗| = √(1² + 1²) = √2 and |B⃗| = √(1² + 1²) = √2. Thus, cos θ = (A⃗ · B⃗) / (|A⃗||B⃗|) = 1 / (√2 × √2) = 1/2, which gives θ = 60°.
Answer: (B) 4
The projection of any vector A⃗ = A_x î + A_y ĵ + A_z k̂ along a Cartesian axis is simply its rectangular component along that axis. For vector A⃗, the component along the y-axis is A_y = 4.
Answer: (A) cos⁻¹(-1/3)
The magnitude of A⃗ is |A⃗| = √(2² + 2² + (-1)²) = √(4 + 4 + 1) = 3. The direction cosine with the z-axis is cos γ = A_z / |A⃗| = -1/3. Hence, the angle is θ = cos⁻¹(-1/3).