Vector Calculus MCQ Quiz - Objective Question with Answer for Vector Calculus - Download Free PDF

Explanation:

v . curl v = v . ∇ × v = [v ∇ v] = 0

Option (4) is true. 

Concept:

The divergence of a vector field \( \vec{B} = B_x \hat{a}_x + B_y \hat{a}_y + B_z \hat{a}_z \) is given by:

\( \nabla \cdot \vec{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} \)

Given:

\( \vec{B} = 2x^2 y \hat{a}_x + 3x^2 y^2 \hat{a}_y \), and the cube lies in \( x, y, z \in [1,3] \)

Calculation:

Since \( \vec{B} \) has no \( \hat{a}_z \) component, \( \frac{\partial B_z}{\partial z} = 0 \)

Find partial derivatives at the center of the cube: \( x = 2, y = 2 \)

\( B_x = 2x^2 y \Rightarrow \frac{\partial B_x}{\partial x} = 4xy \)

At \( x = 2, y = 2 \Rightarrow \frac{\partial B_x}{\partial x} = 4 \times 2 \times 2 = 16 \)

\( B_y = 3x^2 y^2 \Rightarrow \frac{\partial B_y}{\partial y} = 6x^2 y \)

At \( x = 2, y = 2 \Rightarrow \frac{\partial B_y}{\partial y} = 6 \times 4 \times 2 = 48 \)

Final Result:

\( \nabla \cdot \vec{B} = 16 + 48 = 64 \)

Correct Answer: 4) 64

Explanation:

(a) \(\vec{p} \times(\vec{q} \times \vec{r})=(\vec{p} \cdot \vec{r}) \vec{q}-(\vec{p} \cdot \vec{q}) \vec{r}\)

(This is always true for any three given vectors)

(b) We know that \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) is always true but \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\) because \(\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}\)

This can be true only when \(\vec{a} \times \vec{b}=0\)

So, \(\quad \vec{r} \cdot(\vec{p} \times \vec{q})=\vec{r} .(\vec{q} \times \vec{p})\)

\(\vec{r} \cdot(\vec{p} \times \vec{q})=-\vec{r} \cdot(\vec{p} \times \vec{q})\)

This can be true if

\(\vec{r} \cdot(\vec{p} \times q) =0\)

\(\vec{p} \times \vec{q} =\hat{i}-2 \hat{j}+j\)

\((\vec{r} \cdot \vec{p} \times \vec{q})=0\) this is true in this case

(c) \(\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})\) can't be equal to \((\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times \overrightarrow{\mathrm{r}}\) because \(\vec{p} \times(\vec{q} \times \vec{r}) \perp \vec{p}\) and \((\vec{p} \times \vec{q}) \times \vec{r}) \perp \vec{r}\)

So, \((\vec{p} \times(\vec{q} \times \vec{r}) \neq(\vec{p} \times \vec{q})) \times \vec{r}\)

(d) \(\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})+\overrightarrow{\mathrm{q}} \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}})+\overrightarrow{\mathrm{r}} \times(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})=\overrightarrow{0}\)

\((\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{q}}-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{r}}+(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{p}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{p}}+(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{q}} \)

0 = 0

\(\because \quad \vec{p} \cdot \vec{r} =\vec{r} \cdot \vec{p} \)

\(\vec{p} \cdot \vec{q} =\vec{q} \cdot \vec{p} \)

\(\vec{q} \cdot \vec{r} =\vec{r} \cdot \vec{q}\)

(Hence this is proved) 

Explanation:

To evaluate, ∯_s z dx dy

where S(sphere) = x² + y² + z² = R² 

= ∯z dxdy
\(\int_S \vec{F} \cdot \hat{n} \, ds = \iint_R \left(F_1 \, dy \, dz + F_2 \, dx \, dz + F_3 \, dx \, dy\right)\)

\( F_1 = 0, F_2 = 0, F_3 = z\)

\(\vec{F} = Z \hat{K}\)

\(\nabla \cdot \vec{F} = 1\)

Using Gauss Divergence Theorem:

\(\int_S \vec{F} \cdot \hat{n} \, ds = \int_V \nabla \cdot \vec{F} \, dV = \int_V dV\)

= Volume of sphere = \(\frac{4 \pi}{3} R^{3}\)

Explanation:

𝑆 is the portion of the plane 𝑧 = 5𝑥 + 3𝑦 − 100 which lies inside the cylinder 𝑥2 + 𝑦2 = 3.

z_x = 5, z_y = 3

then surface area of S

= \(\int\int_S √{1+z_x^2+z_y^2}dxdy\) 

= \(\int\int_S √{1+5^2+3^2}dxdy\) 

= \(\int\int_S √{35} dxdy\) 

= \(√{35} \int\int_S dxdy\)

Now, the cylinder is 𝑥2 + 𝑦2 = 3

then surface area of S

=\(√{35} \int\int_S dxdy\)  = √35 × π × \({√{3}} ^2 \) = 3√35 π

Given area S = απ

Hence α = \(3\sqrt{35} \)  = 17.74

17.74 is the answer.

Explanation

Given

Integral

∫∫∫v 8 xyz dv 

Limits for x, y and z is given as

[2, 3] × [1, 2] × [0, 1]

Volume of the integral

V = ∫∫∫v 8 xyz dv 

i.e. V = ∫ ∫ ∫_V 8 xyz dxdydz

\(V = 8 \times \mathop \smallint \limits_2^3 xdx\mathop \smallint \limits_1^2 ydy\mathop \smallint \limits_0^1 zdz\)

\(V = 8 \times \left[ {\frac{{{x^2}}}{2}} \right]_2^3 \times \left[ {\frac{{{y^2}}}{2}} \right]_1^2 \times \left[ {\frac{{{z^2}}}{2}} \right]_0^1\)

V = 5 × 3 × 1

V = 15 

∴ Volume is 15 

Concept:

If a triangle is formed by three vectors, then the sum of the vectors should be zero.

AB + BC + CA = 0 

Cross product of the vectors:

For two vectors \(\bar a = ai+bj+ck\) and \(\bar b = di+ej+fk\) the cross, the product is given by: \(\bar a\times \bar b = \begin{vmatrix} i & j & k\\ a & b & c\\ d & e & f \end{vmatrix}=pi+qj+rk\)

The magnitude of the cross product is:

\(A = |a\times b| = \sqrt{p^2+q^2+r^2}\)

Area of a triangle:

If the vectors \(\bar a\mbox{ and } \bar b\) form adjacent sides of the triangle then the area of the triangle is given by: \(A = \dfrac{1}{2}|\bar a\times \bar b|\)

Calculation:

Given:

Let, AB = 3i + 4j and CA = 5i +7j + k

If a triangle is formed by three vectors, then the sum of the vectors should be zero.

AB + BC + CA = 0 ⇒ 3i + 4j  + BC + 5i +7j + k = 0

BC = - 8i - 11j - k

Let the adjacent vectors be AB (a), AC (b)  \(\bar a = 3i+4j\) , and \(\bar b = 5i+7j+k\).

First, we will calculate the cross product as follow:

\(\begin{align*} \bar a\times \bar b &= \begin{vmatrix} i & j & k\\ 3 & 4 & 0\\ 5 & 7 & 1 \end{vmatrix}\\ &= i(4-0) - j(3-0) + k(21-20)\\ &= 4i-3j+1k \end{align*}\)

Therefore, the magnitude of the cross product is:

\(\begin{align*} |\bar a\times \bar b| &= \sqrt{16+9+1}\\ &= \sqrt {26} \end{align*}\)

Using the formula for the area of the triangle, the area is given by:

\(A = {1\over 2}|a\times b| = {1\over 2}\times \sqrt {26} = {\sqrt {26}\over 2}\)

Explanation:

Position vector \(\vec r = x\hat i + y\hat j + zk\)

\(\left| {\vec r} \right| = \sqrt {{x^2} + {y^2} + {z^2}}\)

\(\phi = \ln \left| {\vec r} \right|\)

\(Gradient\;\nabla \phi = \left( {i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}} + k\frac{\partial }{{\partial z}}} \right).\left\{ {\ln \sqrt {{x^2} + {y^2} + {z^2}} } \right\}\)

\(= \frac{1}{2}\left\{ {i\frac{\partial }{{\partial x}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + j.\frac{\partial }{{\partial y}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + k\frac{\partial }{{\partial z}}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right\}\)

\(= \frac{1}{2}\left\{ {i.\frac{{2x}}{{{x^2} + {y^2} + {z^2}}} + j\frac{{2y}}{{{x^2} + {y^2} + {z^2}}} + k\frac{{2z}}{{{x^2} + {y^2} + {z^2}}}} \right\}\)

\(\nabla \phi = \left\{ {\frac{{xi + yj + zk}}{{{x^2} + {y^2} + {z^2}}}} \right\}\)

\(Gradient\;\nabla \phi = \frac{{\vec r}}{{\vec r.\vec r}}\)

Concept:

The curl of a vector is given by the expansion of the following matrix

If \(\vec{f}={{f}_{1}}\hat{i}+{{f}_{2}}\hat{j}+{{f}_{3}}\hat{k}\)

Then

\(\nabla \times ~\vec{f}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & \hat{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ {{f}_{1}} & {{f}_{2}} & {{f}_{3}} \\ \end{matrix} \right|\)

Calculation:

Given vector is

\(~\vec{u}=yz~\hat{i}+3zx~\hat{j}+z~\hat{k}\)

Than \(\nabla \times ~\vec{u}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ yz & 3zx & z \\ \end{matrix} \right|\)

\(\nabla \times ~\vec{u}=\hat{i}\left( \frac{\partial (z)}{\partial y}-\frac{\partial (3zx)}{\partial z} \right)-\hat{j}\left( \frac{\partial (z)}{\partial x}-\frac{\partial (yz)}{\partial z} \right)+~\hat{k}\left( \frac{\partial (3zx)}{\partial x}-\frac{\partial( yz)}{\partial y} \right)\)

Concept:

When two points (x₁, y₁. z₁) and (x₁, y₁. z₂) are mentioned find the relation in terms of the third variable in terms of x,y, and z:

\(\dfrac{{{{x}} - x_1}}{{x_2 - x_1}} = \dfrac{{{{y}} - y_1}}{{y_2 - y_1}} = \dfrac{{{{z}} - z_1}}{{z_2 - z_1}} = {{t}}\)

Put the value of z,y, and z and use the end-points of one variable.

Calculation:

Given:

\({\rm{I}} = \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right)\), A (0, 0, 0) and B(1, 1, 1).

Equation of line i.e. path

\( \frac{{{\rm{x}} - 0}}{{1 - 0}} = \frac{{{\rm{y}} - 0}}{{1 - 0}} = \frac{{{\rm{z}} - 0}}{{1 - 0}} = {\rm{t}} \)

\( \therefore {\rm{\;x\;}} = {\rm{\;y\;}} = {\rm{\;z\;}} = {\rm{\;t\;and\;t\;}}:{\rm{\;}}0{\rm{\;}} \to {\rm{\;}}1\)

\( \therefore {\rm{I}} = \mathop \smallint \limits_0^1 \left( {2{{\rm{t}}^3}{\rm{dt}} + 2{{\rm{t}}^3}{\rm{dt}} + {\rm{dt}}} \right) \)

\(= 4 \cdot \left[ {\frac{{{{\rm{t}}^4}}}{4}} \right]_0^1 + \left[ {\rm{t}} \right]_0^1 = 1 + 1 = 2\)

\( \therefore \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right) = 2 \)

Concept:

Revolution about x-axis: The volume of the solid generated by the revolution about the x-axis, of the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b is

\(V = \mathop \smallint \nolimits_a^b π {y^2}dx\)

similarly for revolution about y-axis:

\(V=\int \pi {{x}^{2}}dy\)

Calculation:

Given:

\(V = \mathop \smallint \limits_1^2 \pi{y^2}dx\)

\(V = \frac{{3\pi }}{2}\)

Hence the required volume will be \(\frac{3\pi}{2}\).

Concept:

If \(\vec a, \vec b, \vec c \) are any non-zero vector then

\([ \vec a \; \vec b\; \vec c]= \vec a. ( \vec b \times \vec c) ....(1)\)

Calculation: 

Given, \([ \vec a, \; \vec b + \vec c, \; \vec a + \vec b + \vec c]\)

\(= \vec a. [(\vec b + \vec c) \times ( \vec a + \vec b + \vec c)]\)

\(= \vec a. {[ \vec b \times \vec a + \vec b \times \vec b + \vec b \times \vec c + \vec c \times \vec a + \vec c \times \vec b + \vec c \times \vec c]}\)

\(= \vec a. {[ \vec b \times \vec a + \vec b \times \vec c + \vec c \times \vec a + \vec c \times \vec b ]}\)

\(=[ \vec a \; \vec b\; \vec a] + [ \vec a \; \vec b\; \vec c] + [ \vec a \; \vec c\; \vec a]+ [ \vec a \; \vec c\; \vec b]\)

\((since [ \vec a \; \vec b\; \vec a]= 0)\)

\([ \vec a \; \vec b\; \vec c] - [ \vec a \; \vec b\; \vec c]\)

= 0

Concept:

\(|\vec {a}+\vec {b}+\vec {c}|^2=(\vec {a}+\vec {b}+\vec {c}).(\vec {a}+\vec {b}+\vec {c})\)

Calculation:

Given:

\(|\vec a|=2, |\vec b|=3, |\vec c|=5\)

As  \(\vec {a}+\vec {b}+\vec {c}=0\)

∴\(|\vec {a}+\vec {b}+\vec {c}|=0\)

\(|\vec {a}+\vec {b}+\vec {c}|^2=(\vec {a}+\vec {b}+\vec {c}).(\vec {a}+\vec {b}+\vec {c})=0\)

\(\vec a.\vec a+\vec b.\vec b+\vec c.\vec c+\vec a.\vec b+\vec b.\vec a+\vec b.\vec c+\vec c.\vec b+\vec c.\vec a+\vec a.\vec c=0\)

\(\vec a.\vec a+\vec b.\vec b+\vec c.\vec c+2\vec a.\vec b+2\vec b.\vec c+2\vec c.\vec a=0\)

\(\vec a.\vec a+\vec b.\vec b+\vec c.\vec c+2(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a)=0\)

\(|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a.\vec b+ \vec b.\vec c+\vec c.\vec a)=0\)

\(2^2+3^2 +5^2+2(\vec a.\vec b+ \vec b.\vec c+\vec c.\vec a)=0\)

\(2(\vec a.\vec b+ \vec b.\vec c+\vec c.\vec a)=-38\)

Hence \(\vec a.\vec b+ \vec b.\vec c+\vec c.\vec a=-19\)

Given that,

F = a_x y - a_y x

y = x²

By differentiating on both sides,

⇒ dy = 2x dx

\(\smallint F.dl = \mathop \smallint \nolimits_C \left( {{a_x}y - {a_y}x} \right).\left( {dx\;{a_x} + dy\;{a_y}} \right)\)

\( = \mathop \smallint \nolimits_C \left( {y\;dx - x\;dy\;} \right)\)

In the given graph, the limits of x are: -1 to 2.

\(\smallint F.dl = \;\mathop \smallint \nolimits_{ - 1}^2 {x^2}\;dx - 2{x^2}dx\)

\( = - \mathop \smallint \nolimits_{ - 1}^2 {x^2}\;dx = - \frac{1}{3}\left[ {{x^3}} \right]_{ - 1}^2 = - 3\)

Concept:

∫_C F(r) dr can be solved by putting:

x = r cos θ

y = r sin θ

dx = -r sin θ dθ

dy = r cos θ dθ

Application:

Given r = 1

θ = 0 to 45°

The required integral can be written as:

\(\mathop \smallint \nolimits_C \vec F \cdot dr = \mathop \smallint \nolimits_0^{45^\circ } \left[ {\left( { - r\cos \theta } \right)\left( { - r\sin \theta } \right)d\theta + \left( {r\sin \theta } \right)\left( {r\cos \theta } \right)d\theta } \right]\)

\(\mathop \smallint \nolimits_0^{45^\circ } \left( {{r^2}\cos \theta \sin \theta + {r^2}\sin \theta \cos \theta } \right)d\theta \)

With r = 1, the above integral becomes:

\( = \frac{1}{2}\mathop \smallint \nolimits_0^{45^\circ } \left( {\sin 2\theta + \sin 2\theta } \right)d\theta \)

r = 1

\( = \mathop \smallint \nolimits_0^{45^\circ } \sin 2\theta \;d\theta \)

\( = \frac{{\left( { - \cos 2\theta } \right)_0^{45^\circ }}}{2}\)

\( = \frac{{ - 0 - \left( { - 1} \right)}}{2} = \frac{1}{2}\)