The gradient operator applied to a scalar yields a vector, and the divergence applied to a vector yields
Answer: C
The three operations differ in what they take and what they return: gradient turns a scalar into a vector, divergence turns a vector into a scalar, and curl turns a vector into a vector. Keeping that straight prevents most sign and dimension errors.
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The magnitude of \( \mathbf{A} = 3\mathbf{a}_x + 4\mathbf{a}_y \) is
Answer: D
The magnitude is the square root of the sum of the squares of the components, that is \( \sqrt{9 + 16} = 5 \). Dividing the vector by five gives the unit vector \( 0.6\mathbf{a}_x + 0.8\mathbf{a}_y \), whose components are the direction cosines.
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For \( \mathbf{A} = 2\mathbf{a}_x - \mathbf{a}_y + 2\mathbf{a}_z \) and \( \mathbf{B} = \mathbf{a}_x + 3\mathbf{a}_y - 2\mathbf{a}_z \), the dot product is
Answer: A
Correct answer: \( -5 \)
\[ (2)(1) + (-1)(3) + (2)(-2) = 2 - 3 - 4 = -5. \] The negative sign means the angle between the two exceeds ninety degrees, since the dot product carries the cosine of that angle.
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For \( \mathbf{A} = \mathbf{a}_x + 2\mathbf{a}_y \) and \( \mathbf{B} = 3\mathbf{a}_x - \mathbf{a}_y \), the cross product is
Answer: B
Both vectors lie in the xy plane so the product is along z, and its magnitude is \( (1)(-1) - (2)(3) = -7 \). The negative sign means the product points along the negative z direction, which the right hand rule confirms.
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The angle between \( \mathbf{a}_x \) and \( \mathbf{a}_x + \mathbf{a}_y \) is
Answer: C
The dot product is one and the magnitudes are one and root two, so the cosine of the angle is \( 1/\sqrt{2} \) and the angle is forty-five degrees. The second vector is the diagonal of a unit square, which makes the answer obvious geometrically.
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The unit vector along \( 2\mathbf{a}_x - \mathbf{a}_y + 2\mathbf{a}_z \) is
Answer: D
The magnitude is \( \sqrt{4 + 1 + 4} = 3 \), so dividing each component by three gives the unit vector. Its components, two thirds, minus a third and two thirds, are the direction cosines and the sum of their squares is one.
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The point \( P(3, 4, 5) \) in Cartesian coordinates has a cylindrical radial coordinate of
Answer: A
Correct answer: 5
\[ \rho = \sqrt{x^2+y^2} = \sqrt{9+16} = 5. \] The cylindrical radius is measured from the axis, so the z coordinate plays no part. Including it would give the spherical radius, \( \sqrt{50} = 7.07 \), which is offered as the first distractor.
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For the same point \( P(3, 4, 5) \), the azimuthal angle \( \phi \) is about
Answer: B
Correct answer: 53.1 degrees
\[ \phi = \arctan\frac{y}{x} = \arctan\frac{4}{3} = 53.1^\circ. \] The quadrant must be checked from the signs of x and y, since the arctangent alone cannot distinguish the first quadrant from the third - both x and y being positive here places the point in the first.
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The divergence of \( \mathbf{A} = x\mathbf{a}_x + y\mathbf{a}_y + z\mathbf{a}_z \) is
Answer: C
Each component differentiated with respect to its own coordinate gives one, and the three add to three. The field is the position vector, which spreads out from the origin, so a positive divergence everywhere is what one would expect.
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The divergence of \( \mathbf{A} = \rho \, \mathbf{a}_\rho \) in cylindrical coordinates is
Answer: D
Correct answer: 2
\[ \nabla\cdot\mathbf{A} = \frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho A_\rho) = \frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho^2) = \frac{2\rho}{\rho} = 2. \] The \( \rho \) must go inside the derivative; differentiating \( A_\rho \) alone would give 1, which is the offered trap.
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