CBSE Class 11 Physics Vectors Competitive Exam MCQs Set 01

Answer: (c) \( \hat{i} - 4\hat{j} + 5\hat{k} \)

Explanation:
1. \( \overrightarrow{AB} \) = (position vector of B) − (position vector of A)
2. \( = (2 - 1)\hat{i} + (-3 - 1)\hat{j} + (4 - (-1))\hat{k} \)
3. \( = \hat{i} - 4\hat{j} + 5\hat{k} \)

Teacher's Note:
1. A vector from A to B is always 'head minus tail', i.e. coordinates of B minus coordinates of A.
2. Option (a) is the vector from B to A; reversing the order only changes the sign of every component.

Answer: (a) magnitude

Explanation:
1. Magnitude of resultant: \( c = \sqrt{a^2 + b^2 + 2ab\cos\theta} \); interchanging the directions keeps a, b and θ the same.
2. So the magnitude of the resultant is unchanged.
3. Its direction changes: the resultant now lies closer to the direction in which the larger vector points, so the angle it makes changes (unless a = b).

Teacher's Note:
1. The magnitude depends only on a, b and the angle between them, not on which vector lies along which line.
2. The resultant always leans towards the larger of the two vectors, so swapping their directions swings the resultant.

Answer: (b) cannot be zero

Explanation:
1. \( \vec{A} + \vec{B} \) lies in the plane of \( \vec{A} \) and \( \vec{B} \).
2. For the total to be zero, \( \vec{C} \) would have to equal \( -(\vec{A} + \vec{B}) \), which lies in that plane.
3. Since \( \vec{C} \) is outside the plane, the resultant cannot be zero (and it does not lie in that plane either).

Teacher's Note:
1. Three vectors can add up to zero only if they are coplanar, i.e. they can form a closed triangle.
2. The out-of-plane part of C has nothing in A or B to cancel it.

Answer: (b) 8 N

Explanation:
1. The resultant of two forces lies between their difference and their sum.
2. R_min = 10 − 6 = 4 N and R_max = 10 + 6 = 16 N
3. Only 8 N lies in the range 4 N ≤ R ≤ 16 N.

Teacher's Note:
1. |A − B| ≤ |A + B| ≤ A + B; any value outside this range is impossible.
2. The extreme values occur when the forces are antiparallel (minimum) or parallel (maximum).

Answer: (d) null vector

Explanation:
1. By the polygon law, the resultant is drawn from the tail of the first vector to the head of the last.
2. In a closed polygon the head of the last vector meets the tail of the first.
3. So the resultant has zero length, i.e. it is a null vector.

Teacher's Note:
1. A closed polygon of vectors always means the vectors add up to zero.
2. This idea is used for equilibrium: forces in equilibrium form a closed polygon.

Answer: (d) π

Explanation:
1. \( R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta} \)
2. R is least when cos θ is least, i.e. cos θ = −1.
3. So θ = π and R_min = |P − Q|.

Teacher's Note:
1. The resultant decreases steadily as the angle increases from 0 to π.
2. When the forces are opposite (antiparallel) they partly cancel, giving the smallest resultant.

Answer: (a) 0°

Explanation:
1. \( R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \)
2. R is maximum when cos θ = 1, i.e. θ = 0°.
3. Then R_max = A + B.

Teacher's Note:
1. Parallel vectors (θ = 0°) add up directly, giving the largest possible resultant.
2. Remember: maximum at 0°, minimum at 180°, and √(A² + B²) at 90°.

Answer: (c) \( \frac{\pi}{2} \)

Explanation:
1. C² = A² + B² + 2AB cos θ ⇒ 169 = 144 + 25 + 2(12)(5) cos θ
2. ⇒ 169 = 169 + 120 cos θ ⇒ cos θ = 0
3. ⇒ θ = π/2

Teacher's Note:
1. When C² = A² + B², the two vectors must be perpendicular (Pythagoras).
2. 12, 5, 13 is a Pythagorean triple, so you can spot the answer immediately.

Answer: (d) Q = 0

Explanation:
1. \( \vec{P} + \vec{Q} = \vec{P} - \vec{Q} \) ⇒ \( 2\vec{Q} = \vec{0} \)
2. ⇒ \( \vec{Q} = \vec{0} \), i.e. Q = 0

Teacher's Note:
1. This is an equation between vectors, so the vectors themselves can be cancelled and compared.
2. If only the magnitudes were equal, |P + Q| = |P − Q|, the answer would instead be θ = 90°.

Answer: (d) also perpendicular to each other and are of equal lengths

Explanation:
1. Let |a| = |b| = a with a ⊥ b. Then \( |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| = \sqrt{a^2 + a^2} = \sqrt{2}a \), so they are equal in length.
2. \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = a^2 - b^2 = 0 \), so they are perpendicular.
3. Hence the sum and difference are perpendicular and of equal lengths.

Teacher's Note:
1. The sum and difference are the two diagonals of the parallelogram; here the parallelogram is a square.
2. The diagonals of a square are equal and perpendicular, which gives the result directly.

Answer: (a) \( 2\hat{i} - 3\hat{j} \)

Explanation:
1. \( \vec{F} = F_x\hat{i} + F_y\hat{j} \)
2. With F_x = 2 N and F_y = −3 N, \( \vec{F} = 2\hat{i} - 3\hat{j} \) N

Teacher's Note:
1. The x-component multiplies î and the y-component multiplies ĵ, signs included.
2. A negative y-component means the force has a part pointing along −y.

Answer: (d) \( \vec{B} \times \vec{C} \)

Explanation:
1. \( \vec{A} \cdot \vec{B} = 0 \) and \( \vec{A} \cdot \vec{C} = 0 \) mean \( \vec{A} \) is perpendicular to both \( \vec{B} \) and \( \vec{C} \).
2. \( \vec{B} \times \vec{C} \) is also perpendicular to both \( \vec{B} \) and \( \vec{C} \).
3. Hence \( \vec{A} \) can be parallel to \( \vec{B} \times \vec{C} \).

Teacher's Note:
1. A dot product of zero between non-zero vectors means they are perpendicular.
2. B · C is a scalar, so a vector can never be 'parallel' to it.

Answer: (b) 60°

Explanation:
1. AB cos θ = 8 and AB sin θ = \( 8\sqrt{3} \)
2. Dividing: tan θ = √3
3. Since cos θ > 0 (scalar product positive), θ = 60°

Teacher's Note:
1. Dividing |A × B| by A · B gives tan θ directly, without knowing A and B.
2. The sign of the scalar product tells whether the angle is acute or obtuse.

Answer: (d) 21 m

Explanation:
1. Displacement = body diagonal = \( \sqrt{10^2 + 12^2 + 14^2} \)
2. \( = \sqrt{100 + 144 + 196} = \sqrt{440} \approx 20.98 \) m
3. ≈ 21 m

Teacher's Note:
1. Displacement depends only on the start and end points, not on the path the fly takes.
2. The diagonal of a cuboid with sides a, b, c is √(a² + b² + c²).

Answer: (a) it is displaced parallel to itself

Explanation:
1. A (free) vector is fixed by its magnitude and direction only.
2. Displacing it parallel to itself keeps both unchanged.
3. Rotation changes its direction; a cross product with a unit vector or multiplication by a scalar changes direction and/or magnitude.

Teacher's Note:
1. This is why we can slide vectors to place them head to tail or tail to tail while adding them.
2. Multiplying by a scalar leaves the vector unchanged only in the special case of the scalar 1.

Answer: (a) (i), (ii); (b) (iii), (iv); (c) (vi)

Explanation:
1. In a right-handed system \( \hat{i} \times \hat{j} = \hat{k} \): turning x towards y with the right hand, the thumb points along z.
2. Taking the diagonal end at the lower left as 'out of the page': (i) x × y gives up = z ✓; (ii) gives out of page = z ✓; (iii) ✓; (iv) ✓; (vi) ✓.
3. In (v), x is up and y is to the right, so x × y points into the page, opposite to z: left-handed.
4. Hence (i), (ii), (iii), (iv) and (vi) are right-handed: options (a), (b) and (c).

Teacher's Note:
1. Check any one cyclic pair, x × y = z (or y × z = x, z × x = y); it is enough to decide the handedness.
2. Swapping any two axis labels of a right-handed set makes it left-handed.

Answer: (a) decreases

Explanation:
1. \( R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \)
2. As θ increases from 0° to 180°, cos θ decreases from 1 to −1.
3. So R decreases.

Teacher's Note:
1. The angle between two vectors is always taken between 0° and 180°, where cos θ keeps decreasing.
2. The resultant goes from A + B (at 0°) down to |A − B| (at 180°).

Answer: (b) \( 50\sqrt{2} \) km h⁻¹ S–W direction

Explanation:
1. Initial velocity \( \vec{v}_1 \) = 50 km/h north; after turning left, \( \vec{v}_2 \) = 50 km/h west.
2. \( \Delta\vec{v} = \vec{v}_2 - \vec{v}_1 \) = 50 west + 50 south
3. Magnitude = \( \sqrt{50^2 + 50^2} = 50\sqrt{2} \) km/h, directed towards south-west.

Teacher's Note:
1. Change in velocity is final minus initial; subtracting a northward vector adds a southward one.
2. Speed can stay the same while velocity changes, because direction is part of velocity.

Answer: (b) 5, 9, 9 and 16 km

Explanation:
1. To return to the start, the displacements must form a closed polygon.
2. This is possible only if the largest one is not more than the sum of the others.
3. (a) 50 > 45, (c) 200 > 170, (d) 90 > 70 fail; (b) 16 ≤ 5 + 9 + 9 = 23 works.

Teacher's Note:
1. For a closed polygon, no side can be longer than the sum of all the other sides.
2. The same check is used for three vectors: they can add to zero only if they can form a triangle.

Answer: (b) less than or equal to (a + b)

Explanation:
1. \( |\vec{a} + \vec{b}| = \sqrt{a^2 + b^2 + 2ab\cos\theta} \le \sqrt{a^2 + b^2 + 2ab} = a + b \)
2. Equality holds when θ = 0° (vectors parallel).
3. So the resultant is always less than or equal to (a + b).

Teacher's Note:
1. This is the triangle inequality: one side of a triangle cannot exceed the sum of the other two.
2. The 'equal' case happens only for parallel vectors, so 'less than' alone is not always true.

Answer: (c) \( \sqrt{58} \) units

Explanation:
1. \( \vec{A} + \vec{B} = (2 + 5)\hat{i} + (3 - 6)\hat{j} = 7\hat{i} - 3\hat{j} \)
2. \( |\vec{A} + \vec{B}| = \sqrt{49 + 9} = \sqrt{58} \) units

Teacher's Note:
1. Add the components first, then find the magnitude; do not add the magnitudes.
2. Squaring removes the negative sign, so −3 contributes +9.

Answer: (a) \( \frac{3\hat{i} + \hat{k}}{\sqrt{10}} \)

Explanation:
1. \( \vec{A} - \vec{B} = (2 + 1)\hat{i} + (-1 + 1)\hat{j} + (2 - 1)\hat{k} = 3\hat{i} + \hat{k} \)
2. \( |\vec{A} - \vec{B}| = \sqrt{9 + 1} = \sqrt{10} \)
3. Unit vector \( = \frac{3\hat{i} + \hat{k}}{\sqrt{10}} \)

Teacher's Note:
1. A unit vector is the vector divided by its own magnitude.
2. Option (d) is the unit vector of B − A, which points the opposite way.

Answer: (d) 120°

Explanation:
1. Let A = B = R = x. Then x² = x² + x² + 2x² cos θ
2. ⇒ 2x² cos θ = −x² ⇒ cos θ = −½
3. ⇒ θ = 120°

Teacher's Note:
1. Equal vectors at 120° give a resultant equal to each of them, like three equal forces in equilibrium.
2. In general, for equal vectors R = 2A cos(θ/2); R = A gives cos(θ/2) = ½, so θ = 120°.

Answer: (a) 90°, 135°, 135°

Explanation:
1. The three vectors form a closed triangle with sides x, x and √2x, which is a right-angled isosceles triangle.
2. Its interior angles are 45°, 45° and 90°.
3. The angle between two vectors (tail to tail) is 180° minus the interior angle, giving 135°, 135° and 90°.

Teacher's Note:
1. When vectors are drawn head to tail in a triangle, the angle between two of them is the exterior angle, not the interior angle.
2. Check: the three angles between the vectors must add up to 360° (90° + 135° + 135° = 360°).

Answer: (a) A vector cannot be divided by another vector; (b) Angular displacement can either be a scalar or a vector; (d) The resultant of two equal forces of magnitude F acting at a point is F if the angle between the two forces is 120°

Explanation:
1. (a) True: division by a vector is not defined.
2. (b) True: a small (infinitesimal) angular displacement behaves as a vector, while a large (finite) one does not obey the laws of vector addition and is treated as a scalar.
3. (c) False: \( \vec{A} - \vec{B} = -(\vec{B} - \vec{A}) \), so subtraction is not commutative.
4. (d) True: R = 2F cos(120°/2) = 2F × ½ = F.

Teacher's Note:
1. Subtraction of vectors, like subtraction of numbers, changes sign when the order is reversed.
2. For two equal vectors, R = 2F cos(θ/2) is a handy shortcut.

Answer: (b) 2 along + x axis

Explanation:
1. \( \vec{A} = 2\cos 60^\circ\,\hat{i} + 2\sin 60^\circ\,\hat{j} = \hat{i} + \sqrt{3}\hat{j} \)
2. \( \vec{B} = 2\cos 60^\circ\,\hat{i} - 2\sin 60^\circ\,\hat{j} = \hat{i} - \sqrt{3}\hat{j} \)
3. \( \vec{A} + \vec{B} = 2\hat{i} \): magnitude 2 along +x axis

Teacher's Note:
1. Vectors placed symmetrically about an axis have their perpendicular components cancel.
2. Here the angle between A and B is 120°, so R = 2 × 2 cos 60° = 2, along the bisector (x-axis).