JEE Physics Basic Mathematics Used in Physics MCQs Set 03

Answer: (b) \( \cos^{-1}\left(-\frac AB\right) \)

Explanation:
1. \( (\vec A+\vec B)\cdot\vec A=0 \) gives \( A^2+AB\cos\theta=0 \).
2. So \( \cos\theta=-\frac AB \) and \( \theta=\cos^{-1}(-A/B) \).

Teacher's Note:
1. Use the dot product of the resultant with A.
2. The angle is obtuse.

Answer: (d) \( \frac72(\hat j+\hat i) \)

Explanation:
1. The unit vector is \( \frac{\hat i+\hat j}{\sqrt2} \), and the scalar component is \( \frac{3+4}{\sqrt2}=\frac7{\sqrt2} \).
2. The vector component is \( \frac7{\sqrt2}\cdot\frac{\hat i+\hat j}{\sqrt2}=\frac72(\hat i+\hat j) \).

Teacher's Note:
1. Component along a direction = (A · n̂) n̂.
2. Do not forget to multiply by the unit vector again.

Answer: (d) 8.5

Explanation:
1. A · B = 15 + 2 − 2S = 0.
2. So S = 17/2 = 8.5.

Teacher's Note:
1. Perpendicular vectors have zero dot product.
2. Solve the linear equation in S.

Answer: (b) \( \vec A\cdot\hat B \)

Explanation:
1. The projection of \( \vec A \) on \( \vec B \) is \( A\cos\theta=\vec A\cdot\hat B \).

Teacher's Note:
1. The unit vector gives the direction of projection.
2. The other options are not scalar projections of A.

Answer: (d) 60°

Explanation:
1. cos θ = (1)/(√2·√2) = 1/2, so θ = 60°.

Teacher's Note:
1. Use cos θ = A·B / (AB).
2. The dot product is 1.

Answer: (c) 90°

Explanation:
1. A · B = 9 + 16 − 25 = 0.
2. So the angle is 90°.

Teacher's Note:
1. A zero dot product means perpendicular vectors.
2. Compute the products component-wise.

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

Explanation:
1. \( \vec A\cdot\vec B=AB(\cos\theta\sin\theta-\sin\theta\cos\theta)=0 \) for option 3.
2. The other options give non-zero dot products.

Teacher's Note:
1. Test the dot product for each option.
2. Swap the components and change one sign for a perpendicular vector.

Answer: (a) 3

Explanation:
1. P · Q = a² − 2a − 3 = 0.
2. (a − 3)(a + 1) = 0, so the positive value is a = 3.

Teacher's Note:
1. Set the dot product to zero.
2. Factorise the quadratic.

Answer: (d) 25 J

Explanation:
1. W = 3 × 3 + 4 × 4 = 25 J.

Teacher's Note:
1. Dot product: multiply matching components and add.
2. Force and displacement are parallel here.

Answer: (d) zero

Explanation:
1. The y-component of 3î + 4k̂ is zero.
2. So the projection on the y-axis is zero.

Teacher's Note:
1. The projection on an axis is the component along it.
2. The vector has no ĵ term.

Answer: (c) −1/2

Explanation:
1. Dot product: −8 + 12 + 8α = 0.
2. So α = −4/8 = −1/2.

Teacher's Note:
1. Set the dot product equal to zero.
2. Collect like terms carefully.

Answer: (c) \( \vec A\cdot\vec B=48 \)

Explanation:
1. B = 2A, so the vectors are parallel and the cross product is zero; A/B = 5/10 = 1/2 and A = 5.
2. A · B = 18 + 32 = 50, not 48, so statement 3 is false.

Teacher's Note:
1. Compute each statement.
2. The dot product of parallel vectors equals AB = 5 × 10.

Answer: (d) \( -4\hat i \)

Explanation:
1. A zero cross product means the vectors are parallel or anti-parallel.
2. Only −4î is along the x-axis.

Teacher's Note:
1. Parallel or antiparallel vectors have zero cross product.
2. Check which option has only an î component.

Answer: (d) \( \hat i\times\hat k=-\hat i \)

Explanation:
1. \( \hat i\times\hat k=-\hat j \) (not \( -\hat i \)).
2. The other three statements are correct.

Teacher's Note:
1. Use the cyclic order î → ĵ → k̂.
2. Reversing the order changes the sign.

Answer: (b) \( \frac{\hat P\times\hat Q}{\sin\theta} \)

Explanation:
1. \( \vec P\times\vec Q \) is perpendicular to both and has magnitude \( PQ\sin\theta \).
2. @
3. \hat P\times\hat Q
4. =\sin\theta\( , so dividing by sin θ gives a unit vector: \)\frac{\hat P\times\hat Q}{\sin\theta}@.

Teacher's Note:
1. A unit vector needs to be divided by its own magnitude.
2. The cross product of unit vectors has magnitude sin θ.

Answer: (a) greater than AB

Explanation:
1. @
2. \vec A\times\vec B
3. =AB\sin\theta\( with \)\sin\theta\le1@.
4. So it can never exceed AB.

Teacher's Note:
1. The magnitude is AB sin θ.
2. It is zero for parallel vectors and AB for perpendicular vectors.

Answer: (d) \( \vec R\perp(\vec P\times\vec Q) \)

Explanation:
1. R is perpendicular to both P and Q, hence to their sum.
2. But \( \vec P\times\vec Q=\vec R \), so R is not perpendicular to itself: statement 4 is false.

Teacher's Note:
1. The cross product is perpendicular to the plane of the two vectors.
2. A non-zero vector is not perpendicular to itself.

Answer: (b) \( \vec P+\vec Q=\vec Q\times\vec P \)

Explanation:
1. Vector addition and the dot product are commutative.
2. The cross product is not commutative: P × Q = −Q × P.
3. A sum cannot equal a cross product, so statement 2 is false.

Teacher's Note:
1. Statement 2 mixes an addition with a cross product.
2. The cross product anticommutes.

Answer: (a) 0

Explanation:
1. \( \vec A\times\vec B \) is perpendicular to both \( \vec A \) and \( \vec B \), hence to \( \vec A+\vec B \).
2. So the dot product is zero.

Teacher's Note:
1. This is the scalar triple product of a coplanar set.
2. The cross product is perpendicular to the plane.

Answer: (a) zero

Explanation:
1. Both A and C are parallel (or antiparallel) to B, so they are parallel to each other.
2. The angle between them is 0 (or π); of the options only zero fits.

Teacher's Note:
1. Parallel to the same vector means parallel to each other.
2. Zero is the only option among the choices.

Answer: (c) \( \frac3{\sqrt2} \) unit

Explanation:
1. Area = @\frac12
2. (\hat i+\hat j+\hat k)\times3\hat i
3. =\frac12
4. 3(\hat k\times\hat i+\hat j\times\hat i)
5. =\frac12
6. 3(\hat j-\hat k)
7. @.
8. The magnitude is \( 3\sqrt2 \), so the area is \( \frac{3\sqrt2}{2}=\frac3{\sqrt2} \).

Teacher's Note:
1. Area of a triangle is half the magnitude of the cross product.
2. Use the cyclic rules for unit vectors.

Answer: (a) \( -5\hat i+2\hat j+3\hat k \)

Explanation:
1. \( \vec\omega\times\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&3\\1&1&1\end{vmatrix} \).
2. \( =\hat i(-2-3)-\hat j(1-3)+\hat k(1+2)=-5\hat i+2\hat j+3\hat k \).

Teacher's Note:
1. Expand the determinant carefully.
2. The middle term carries a minus sign.

Answer: (b) \( 5\sqrt3 \) unit

Explanation:
1. Area = @\frac12
2. \vec d₁\times\vec d₂
3. @.
4. \( \vec d_1\times\vec d_2=-2\hat i-14\hat j-10\hat k \), with magnitude \( \sqrt{300}=10\sqrt3 \).
5. Area = \( 5\sqrt3 \).

Teacher's Note:
1. The area of a parallelogram from its diagonals is half the cross product.
2. Compute the cross product with a determinant.

Answer: (a) \( \pi \) rad

Explanation:
1. \( \vec B\times\vec A=-\vec A\times\vec B \), so the vectors are opposite.
2. The angle between them is π rad.

Teacher's Note:
1. The cross product changes sign when the order is reversed.
2. Opposite vectors make 180°.

Answer: (b) along west

Explanation:
1. Take east = x, north = y and up = z. Then A × B = ẑ × ŷ = −x̂.
2. −x̂ points west.

Teacher's Note:
1. Use the right-hand rule or unit vector rules.
2. k̂ × ĵ = −î.

Answer: (b) 45°

Explanation:
1. AB sin θ = AB cos θ gives tan θ = 1.
2. So θ = 45°.

Teacher's Note:
1. Equate the magnitudes of the cross and dot products.
2. tan 45° = 1.

Answer: (c) \( \vec A=\vec0 \)

Explanation:
1. \( \vec A+\vec B=\vec A-\vec B \) gives \( 2\vec B=\vec0 \), so \( \vec B=\vec0 \).
2. Then A · B = 0 and A × B = 0 are true for any A, but A need not be zero, so statement 3 is incorrect.

Teacher's Note:
1. Cancel A from both sides.
2. B is zero, but A is arbitrary.

Answer: (c) \( \vec B\times\vec C \)

Explanation:
1. A is perpendicular to both B and C, and \( \vec B\times\vec C \) is perpendicular to both.
2. So A can be parallel to \( \vec B\times\vec C \).

Teacher's Note:
1. The cross product is perpendicular to both vectors.
2. B · C is a scalar and cannot be parallel to a vector.