JEE Physics Basic Mathematics Used in Physics MCQs Set 05

Answer: (c) \( |\vec A|=|\vec B| \)

Explanation:
1. For the resultant, \( \tan\frac\theta2=\frac{B\sin\theta}{A+B\cos\theta} \).
2. With \( \tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta} \), equality holds only for \( A=B \).
3. So the magnitudes are equal: \( |\vec A|=|\vec B| \).

Teacher's Note:
1. The resultant of two equal vectors lies along the angle bisector.
2. Equal magnitude does not mean equal vectors; the directions still differ by theta.

Answer: (a) parallel

Explanation:
1. \( |\vec A+\vec B|^2=A^2+B^2+2AB\cos\theta \).
2. If this equals \( (A+B)^2 \), then \( \cos\theta=1 \), so \( \theta=0^\circ \).
3. The vectors are parallel (same direction).

Teacher's Note:
1. The magnitude of a sum equals the sum of magnitudes only when the vectors point the same way.
2. Anti-parallel vectors give the smallest resultant, the difference of the magnitudes.

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

Explanation:
1. Take the forces along +x, +y, -x, -y with magnitudes \( P,2P,3P,4P \).
2. Net x: \( P-3P=-2P \). Net y: \( 2P-4P=-2P \).
3. Resultant \( =\sqrt{(2P)^2+(2P)^2}=2\sqrt2\,P \).

Teacher's Note:
1. Opposite sides of the square give opposite directions, so resolve along the two axes.
2. The resultant is directed along a diagonal.

Answer: (d) \( \vec v_1 \) and \( \vec v_2 \) are mutually perpendicular

Explanation:
1. Square both sides: \( v_1^2+v_2^2+2\vec v_1\cdot\vec v_2=v_1^2+v_2^2-2\vec v_1\cdot\vec v_2 \).
2. So \( \vec v_1\cdot\vec v_2=0 \), and the vectors are perpendicular.

Teacher's Note:
1. Equal sum and difference magnitudes are the diagonals of a rectangle.
2. A zero dot product means the vectors are perpendicular.

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

Explanation:
1. Required vector \( =\hat i-(\hat i-5\hat j+2\hat k)-(3\hat i+6\hat j-7\hat k) \).
2. \( =(1-1-3)\hat i+(5-6)\hat j+(-2+7)\hat k \).
3. \( =-3\hat i-\hat j+5\hat k \).

Teacher's Note:
1. Subtract the given vectors from the target vector component by component.
2. Add the three vectors back to check that the sum is the unit vector i.

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

Explanation:
1. Parallel vectors have proportional components: \( \vec A=\lambda\vec B \).
2. In (a): \( \frac31=\frac62=\frac93=3 \). All ratios are equal.
3. In (b), (c) and (d) the ratios of the components are unequal, so those pairs are not parallel.

Teacher's Note:
1. Compare the ratio of corresponding components for all three axes.
2. A single sign or value mismatch is enough to make the pair non-parallel.

Answer: (a) \( 15\hat i+20\hat j \)

Explanation:
1. \( |\vec B|=\sqrt{49+576}=25 \) and \( |\vec A|=5 \).
2. Unit vector along \( \vec A \): \( \hat A=\frac{3\hat i+4\hat j}{5} \).
3. Required vector \( =25\hat A=15\hat i+20\hat j \).

Teacher's Note:
1. Parallel to A with a given magnitude means magnitude times the unit vector of A.
2. Check: the magnitude of 15i + 20j is 25.

Answer: (c) \( -\frac A2 \) and \( -\sqrt3\frac A2 \)

Explanation:
1. \( A_x=A\cos240^\circ=-A\cos60^\circ=-\frac A2 \).
2. \( A_y=A\sin240^\circ=-A\sin60^\circ=-\frac{\sqrt3A}{2} \).

Teacher's Note:
1. 240 degrees lies in the third quadrant, where both components are negative.
2. Reduce the angle to its acute reference angle of 60 degrees first.

Answer: (c) \( \frac52(\hat i+\hat j) \)

Explanation:
1. Unit vector along \( \hat i+\hat j \) is \( \frac{\hat i+\hat j}{\sqrt2} \).
2. Scalar component \( =\vec a\cdot\hat n=\frac{2+3}{\sqrt2}=\frac5{\sqrt2} \).
3. Vector component \( =\frac5{\sqrt2}\cdot\frac{\hat i+\hat j}{\sqrt2}=\frac52(\hat i+\hat j) \).

Teacher's Note:
1. The component along a direction is the dot product with the unit vector in that direction.
2. Multiply by the unit vector again to write it as a vector.

Answer: (d) all of these

Explanation:
1. If the three unit vectors are at \( 120^\circ \) to each other, the resultant is zero.
2. If two of them are opposite and the third is free, the resultant is 1.
3. If all three point the same way, the resultant is 3.
4. So the magnitude can be 0, 1 or 3.

Teacher's Note:
1. The resultant ranges from the minimum to the maximum possible value.
2. Here the maximum is 1 + 1 + 1 = 3 and the minimum is 0.

Answer: (d) All of these

Explanation:
1. Three forces can balance if each is not larger than the sum of the other two and not less than their difference.
2. 10, 10, 10: forms an equilateral triangle. 10, 10, 20: collinear (10 + 10 = 20). 10, 20, 20: forms a triangle.
3. All three sets satisfy the condition.

Teacher's Note:
1. Check the triangle rule: the largest force must not exceed the sum of the other two.
2. Equality (10 + 10 = 20) is the limiting case where all three act along one line.

Answer: (c) \( 7\hat k \)

Explanation:
1. Two vectors are perpendicular when their dot product is zero.
2. \( (4\hat i+3\hat j)\cdot5\hat k=0 \), \( 6\hat i\cdot5\hat k=0 \), \( (3\hat i+4\hat j)\cdot5\hat k=0 \).
3. But \( 7\hat k\cdot5\hat k=35\neq0 \), so \( 7\hat k \) is not perpendicular (it is parallel).

Teacher's Note:
1. Any vector lying in the x-y plane is perpendicular to the z-axis.
2. A vector along the same axis gives the full product of the magnitudes.

Answer: (b) \( 10\hat i+10\sqrt2\hat j+10\hat k \)

Explanation:
1. Direction cosines: \( l=\cos60^\circ=\frac12 \) (east) and \( n=\cos60^\circ=\frac12 \) (up).
2. \( l^2+m^2+n^2=1 \Rightarrow m^2=\frac12 \), so \( m=\frac1{\sqrt2} \) (northward).
3. \( \vec v=20\left(\frac12\hat i+\frac1{\sqrt2}\hat j+\frac12\hat k\right)=10\hat i+10\sqrt2\hat j+10\hat k \).

Teacher's Note:
1. The squares of the three direction cosines always add to 1.
2. Check: 100 + 200 + 100 = 400, so the speed is 20 m/s.

Answer: (d) \( t=\frac{\pi}{3\omega} \)

Explanation:
1. \( \vec A\cdot\vec B=\cos\omega t\cos\frac{\omega t}2-\sin\omega t\sin\frac{\omega t}2=\cos\frac{3\omega t}{2} \).
2. Orthogonal means \( \vec A\cdot\vec B=0 \): \( \frac{3\omega t}{2}=\frac\pi2 \) (first instant).
3. So \( t=\frac{\pi}{3\omega} \).

Teacher's Note:
1. Orthogonal means the dot product is zero.
2. At t = 0 the dot product is 1, so the vectors are parallel, not orthogonal.

Answer: (a) \( \sin^{-1}\left(\frac1{\sqrt3}\right) \)

Explanation:
1. \( \cos\theta=\frac{(\hat i+\hat j+\hat k)\cdot(\hat i+\hat j)}{\sqrt3\cdot\sqrt2}=\frac2{\sqrt6}=\sqrt{\frac23} \).
2. Then \( \sin\theta=\sqrt{1-\frac23}=\frac1{\sqrt3} \).
3. So \( \theta=\sin^{-1}\frac1{\sqrt3} \).

Teacher's Note:
1. Use the dot product to get the cosine and then convert to the sine.
2. This is the angle between a body diagonal of a cube and a face diagonal.

Answer: (b) along west

Explanation:
1. Take east = \( \hat i \), north = \( \hat j \), up = \( \hat k \) (a right-handed set).
2. \( \hat k\times\hat j=-\hat i \), which points west.

Teacher's Note:
1. Use the right-hand rule: fingers from A to B, thumb gives the product direction.
2. The cross product is perpendicular to both vectors, so it lies in the horizontal plane.

Answer: (a) zero

Explanation:
1. \( \vec A\times\vec B=\vec 0 \) means \( \vec A\parallel\vec B \), and \( \vec B\times\vec C=\vec 0 \) means \( \vec B\parallel\vec C \).
2. Hence \( \vec A\parallel\vec C \), and the angle between them is \( 0 \) or \( \pi \). Of the given options only zero is possible.

Teacher's Note:
1. The cross product of two vectors vanishes only when they are parallel or anti-parallel.
2. Parallelism is transitive when the middle vector is non-zero.

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

Explanation:
1. \( \vec A\times\vec B \) is perpendicular to both \( \vec A \) and \( \vec B \).
2. So it is perpendicular to any combination of them, including \( \vec A+\vec B \).
3. The angle is \( \frac\pi2 \).

Teacher's Note:
1. The cross product is perpendicular to the plane containing the two vectors.
2. Equivalently, (A + B) . (A x B) = 0.

Answer: (b) \( (A^2+B^2-AB)^{1/2} \)

Explanation:
1. \( AB\sin\theta=\sqrt3AB\cos\theta \Rightarrow\tan\theta=\sqrt3 \Rightarrow\theta=60^\circ \).
2. \( |\vec A-\vec B|^2=A^2+B^2-2AB\cos60^\circ=A^2+B^2-AB \).
3. So \( |\vec A-\vec B|=(A^2+B^2-AB)^{1/2} \).

Teacher's Note:
1. Compare the cross product and dot product magnitudes to get tan of the angle.
2. Use the cosine rule for the difference of two vectors.

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

Explanation:
1. Area \( =\frac12|\vec d_1\times\vec d_2| \).
2. \( \vec d_1\times\vec d_2=(4-6)\hat i-(12+2)\hat j+(-9-1)\hat k=-2\hat i-14\hat j-10\hat k \).
3. \( |\vec d_1\times\vec d_2|=\sqrt{4+196+100}=10\sqrt3 \), so area \( =5\sqrt3 \) unit.

Teacher's Note:
1. The area is half the magnitude of the cross product of the two diagonals.
2. If the sides were given, the area would be the full magnitude of their cross product.

Answer: (a) 64

Explanation:
1. \( |\vec A_1+\vec A_2|^2=A_1^2+A_2^2+2\vec A_1\cdot\vec A_2 \Rightarrow 9=4+9+2\vec A_1\cdot\vec A_2 \), so \( \vec A_1\cdot\vec A_2=-2 \).
2. Expand: \( (\vec A_1+2\vec A_2)\cdot(3\vec A_1-4\vec A_2)=3A_1^2+2\vec A_1\cdot\vec A_2-8A_2^2 \).
3. \( =12-4-72=-64 \), so the modulus is 64.

Teacher's Note:
1. Expand dot products like ordinary algebra, using A . A = A squared.
2. Get A1 . A2 first from the given magnitude of the sum.