JEE Physics Vectors and Calculus MCQs Set 05

Answer: (c) \( -\frac A2\text{ and }-\frac{\sqrt3A}{2} \)

Explanation:
1. A_x = A cos 240° = -A/2.
2. A_y = A sin 240° = -(√3/2)A.

Teacher's Note:
1. Take the angle from the positive x-axis, so signs follow the quadrant.
2. 240° lies in the third quadrant, where both components are negative.

Answer: (d) 15°

Explanation:
1. First ball: tan θ1 = √3/1, so θ1 = 60°.
2. Second ball: tan θ2 = 2/2 = 1, so θ2 = 45°.
3. Angle between the paths = 60° - 45° = 15°.

Teacher's Note:
1. The direction of a velocity is given by tan θ = vy/vx.
2. The angle between two paths is the difference of their directions.

Answer: (d) 120°

Explanation:
1. R² = A² + A² + 2A² cos θ = A², so cos θ = -1/2.
2. Hence θ = 120°.

Teacher's Note:
1. Use R² = A² + B² + 2AB cos θ.
2. The resultant of two equal vectors at 120° has the same magnitude.

Answer: (d) \( \sqrt3 \)

Explanation:
1. Resultant = 1 means 1 = 1 + 1 + 2cos θ, so cos θ = -1/2 and θ = 120°.
2. Difference: D² = 1 + 1 - 2cos θ = 3, so D = √3.

Teacher's Note:
1. First find the angle from the resultant condition.
2. Sum and difference squares differ only in the sign of the cosine term.

Answer: (a) \( \tan^{-1}\frac23 \)

Explanation:
1. cos α = Ry/R = 3/√13.
2. Then sin α = 2/√13, so tan α = 2/3 and α = tan⁻¹(2/3).

Teacher's Note:
1. The angle with an axis uses the component along that axis for the cosine.
2. Convert to tan using the other component.

Answer: (a) 3

Explanation:
1. Two vectors of different magnitude can never cancel.
2. Three vectors with different magnitudes can form a closed triangle, so the minimum is 3.

Teacher's Note:
1. Zero resultant means the vectors form a closed polygon.
2. A triangle is the smallest polygon.

Answer: (b) \( \vec A,\vec B\text{ and }\vec C \) each must be a null vector

Explanation:
1. A = -(B + C), so |A| = |B + C| and |A| ≤ |B| + |C|.
2. Three vectors with zero resultant are coplanar.
3. They need not be null vectors (for example, the sides of a triangle).

Teacher's Note:
1. Zero resultant means a closed triangle.
2. So the vectors need not vanish individually.

Answer: (b) Yes, when the two vectors are the same in magnitude but opposite in sense

Explanation:
1. Zero resultant needs R² = A² + B² + 2AB cos θ = 0.
2. This requires A = B and cos θ = -1, i.e. θ = 180°.
3. At 120° the resultant of equal vectors has the same magnitude, not zero.

Teacher's Note:
1. Two vectors cancel only when they are equal and opposite.
2. Check the 120° case separately; it gives a resultant equal to each vector.

Answer: (b) \( 2\sqrt2\,P \)

Explanation:
1. Opposite sides: P and 3P cancel to give 2P in one direction; 2P and 4P give 2P in the perpendicular direction.
2. R = √((2P)² + (2P)²) = 2√2 P.

Teacher's Note:
1. Combine the pair along each axis first.
2. The two net components are perpendicular.

Answer: (a) \( -2(\vec A\times\vec B) \)

Explanation:
1. Expand: A×A - A×B + B×A - B×B.
2. A×A = B×B = 0 and B×A = -A×B, so the result is -2(A×B).

Teacher's Note:
1. The cross product is not commutative.
2. The cross product of a vector with itself is zero.

Answer: (d) \( \vec{V_1} \) and \( \vec{V_2} \) are mutually perpendicular

Explanation:
1. Squaring both sides: V1² + V2² + 2V1·V2 = V1² + V2² - 2V1·V2.
2. So V1·V2 = 0, meaning the vectors are perpendicular.

Teacher's Note:
1. Equal magnitudes of sum and difference give a zero dot product.
2. Think of the diagonals of a rectangle.

Answer: (c) \( \vec{A} \) = \( 3\hat{i}+6\hat{j}+9\hat{k} \), \( \vec{B} \) = \( \hat{i}+2\hat{j}+3\hat{k} \)

Explanation:
1. Parallel vectors have proportional components.
2. In (a), A = 3B (3/1 = 6/2 = 9/3). The other pairs have unequal ratios.

Teacher's Note:
1. Compare the ratios of corresponding components.
2. Equal ratios mean the vectors are parallel.

Answer: (c) \( \vec{B} \) = 0

Explanation:
1. Cancelling A from both sides gives 2B = 0.
2. So B is the null vector.

Teacher's Note:
1. Add -A to both sides.
2. Only the null vector equals its own negative.

Answer: (b) \( \frac\pi2 \)

Explanation:
1. A×B is perpendicular to the plane of A and B.
2. A + B lies in that plane, so the angle is π/2.

Teacher's Note:
1. The cross product is normal to both factors.
2. Any combination of A and B lies in their plane.

Answer: (c) \( -\frac12(\hat i-\hat j) \)

Explanation:
1. Unit vector n = (i - j)/√2.
2. A·n = (2 - 3)/√2 = -1/√2.
3. Component = (A·n)n = -(1/2)(i - j).

Teacher's Note:
1. Multiply the scalar projection by the unit vector.
2. The negative sign shows the projection opposes (i - j).

Answer: (c) \( \hat iB\sin\theta-\hat jB\cos\theta \)

Explanation:
1. For perpendicularity A·B = 0.
2. AB(cos θ sin θ - sin θ cos θ) = 0 for option (b).

Teacher's Note:
1. Rotating (cos θ, sin θ) by 90° gives (sin θ, -cos θ).
2. Check by taking the dot product.

Answer: (c) 90°

Explanation:
1. P·Q = 0 means P ⟂ Q.
2. P×R = 0 means R is parallel (or antiparallel) to P.
3. So Q is perpendicular to R.

Teacher's Note:
1. Dot product zero means perpendicular; cross product zero means parallel.
2. Combine the two conditions.

Answer: (c) \( \vec a\parallel\vec b \)

Explanation:
1. (a·b)² = a²b²cos²θ.
2. Equating to a²b² gives cos²θ = 1, so θ = 0° or 180°.
3. The vectors are parallel.

Teacher's Note:
1. Write the dot product with the cosine.
2. cos² θ = 1 means collinear vectors.

Answer: (c) \( \frac\pi2 \)

Explanation:
1. 12² + 5² = 144 + 25 = 169 = 13².
2. So C² = A² + B², and A ⟂ B.

Teacher's Note:
1. Recognise the 5, 12, 13 triple.
2. C² = A² + B² means 2AB cos θ = 0.

Answer: (a) 15i + 20j

Explanation:
1. |B| = √(49 + 576) = 25 and |A| = 5.
2. Required vector = 25 × (A/5) = 5A = 15i + 20j.

Teacher's Note:
1. Use a unit vector along A.
2. Scale it by the magnitude of B.

Answer: (d) all of these

Explanation:
1. A + B + E: M to N, N to P, P to M closes the loop, so the sum is zero.
2. C - D = (O - N) - (O - M) = M - N = -A.
3. B + E - C = (P - N) + (M - P) - (O - N) = M - O = -D.

Teacher's Note:
1. Write each vector as final point minus initial point.
2. All three relations then follow directly.