Vector Analysis and Tensors Questions and Answers

Answer: C

The Laplacian of the field is 0.

The field is the position vector, \(\overline{A} = x\hat{e}_{x} + y\hat{e}_{y} + z\hat{e}_{z}\) (the paper prints the last unit vector as \(\hat{e}_{j}\)). In Cartesian coordinates the vector Laplacian acts on each component separately, and each component is linear in the coordinates:

\[ \nabla^{2}x = \nabla^{2}y = \nabla^{2}z = 0 \]

so \(\nabla^{2}\overline{A} = 0\). The value 3 is the divergence \(\nabla \cdot \overline{A}\), a common confusion; 1 and -3 do not arise from either operation.

Answer: A

The double curl is Zero.

Each component of \(\overline{F} = x\hat{i} + 2y\hat{j} + 3z\hat{k}\) depends only on its own coordinate, so every cross derivative vanishes; the x-component of the curl, for example, is

\[ \begin{aligned} (\nabla \times \overline{F})_{x} &= \frac{\partial (3z)}{\partial y} - \frac{\partial (2y)}{\partial z} \\ &= 0 - 0 = 0 \end{aligned} \]

The curl of the zero vector is again zero, so \(\nabla \times (\nabla \times \overline{F}) = 0\). The options \(\hat{j}\), \(2\hat{j}\) and \(3\hat{k}\) would need a non-zero curl to begin with; the field is conservative (it is the gradient of \(x^{2}/2 + y^{2} + 3z^{2}/2\)).

Answer: A

The unit vector is \(\frac{-3\hat{i} + 5\hat{j} + 11\hat{k}}{\sqrt{155}}\).

A vector perpendicular to both is their cross product:

\[ \begin{aligned} \bar{A} \times \bar{B} &= \hat{i}[(-1)(-1) - (1)(4)] \\ &\quad - \hat{j}[(2)(-1) - (1)(3)] \\ &\quad + \hat{k}[(2)(4) - (-1)(3)] \\ &= -3\hat{i} + 5\hat{j} + 11\hat{k} \end{aligned} \]

Its magnitude is \(\sqrt{9 + 25 + 121} = \sqrt{155}\), and dividing by it gives the unit vector. A quick check: its dot product with \(\bar{A}\) is \(-6 - 5 + 11 = 0\).

The other options fail that check; for example \((\hat{i} - \hat{j} + 2\hat{k})\cdot\bar{A} = 2 + 1 + 2 = 5\), so it is not perpendicular to \(\bar{A}\).

Answer: A

The true statement is Trace (AB) = Trace (BA).

\[ \begin{aligned} \mathrm{Tr}(AB) &= \sum_{i}\sum_{j} A_{ij}B_{ji} \\ &= \sum_{j}\sum_{i} B_{ji}A_{ij} = \mathrm{Tr}(BA) \end{aligned} \]

More generally the trace is unchanged by cyclic permutations: Tr(ABC) = Tr(BCA) = Tr(CAB). Swapping two neighbouring factors in a longer product, as in Tr(BAC) or Tr(BACD), is not a cyclic permutation, so those equalities fail for general matrices. Tr(AB) = Tr(A) Tr(B) is false too: for A = B = the 2 × 2 identity, Tr(AB) = 2 while Tr(A) Tr(B) = 4.

Answer: B

The vectors are linearly dependent.

Adding the first two vectors cancels the \(\hat{j}\) parts:

\[ \vec{a} + \vec{b} = 2\hat{i} + 2\hat{k} = 2\vec{c} \]

so \(\vec{a} + \vec{b} - 2\vec{c} = 0\), a non-trivial linear combination that vanishes. Equivalently, the determinant of their components is zero.

They are therefore not independent. They are not orthogonal either: \(\vec{a}\cdot\vec{b} = 1 - 25 + 1 = -23\). And none is normalised, since \(|\vec{a}| = \sqrt{27}\) and \(|\vec{c}| = \sqrt{2}\).

Answer: B

The line integral is 11.33.

On the path \(x = y^{2}\), so \(dx = 2y\,dy\), and \(y\) runs from 1 to 2. The integrand \(\vec{A}\cdot d\vec{r} = (x+y)\,dx + (y-x)\,dy\) becomes a function of \(y\) alone:

\[ \begin{aligned} I &= \int_{1}^{2} [(y^{2}+y)2y + (y - y^{2})]\,dy \\ &= \int_{1}^{2} (2y^{3} + y^{2} + y)\,dy \\ &= \left[\tfrac{y^{4}}{2} + \tfrac{y^{3}}{3} + \tfrac{y^{2}}{2}\right]_{1}^{2} \\ &= \tfrac{38}{3} - \tfrac{4}{3} = \tfrac{34}{3} \approx 11.33 \end{aligned} \]

The field is not conservative (\(\partial A_y/\partial x = -1\) while \(\partial A_x/\partial y = 1\)), so the value depends on the path and must be computed along the parabola. The other options differ from 34/3 by whole units and do not come out of this integral.