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 $\overrightarrow{B}=B_x{\hat{a}}_x+B_y{\hat{a}}_y+B_z{\hat{a}}_z$ is given by:

$\mathrm{\nabla }\cdot \overrightarrow{B}=\frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_z}{\partial z}$

Given:

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

Calculation:

Since $\overrightarrow{B}$ 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:

$\mathrm{\nabla }\cdot \overrightarrow{B}=16+48=64$

Correct Answer: 4) 64

Explanation:

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

(This is always true for any three given vectors)

(b) We know that $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}$ is always true but $\overrightarrow{a}\times \overrightarrow{b}\ne \overrightarrow{b}\times \overrightarrow{a}$ because $\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}$

This can be true only when $\overrightarrow{a}\times \overrightarrow{b}=0$

So, $\overrightarrow{r}\cdot (\overrightarrow{p}\times \overrightarrow{q})=\overrightarrow{r}.(\overrightarrow{q}\times \overrightarrow{p})$

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

This can be true if

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

$\overrightarrow{p}\times \overrightarrow{q}=\hat{i}-2\hat{j}+j$

$(\overrightarrow{r}\cdot \overrightarrow{p}\times \overrightarrow{q})=0$ 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 $\overrightarrow{p}\times (\overrightarrow{q}\times \overrightarrow{r})\perp \overrightarrow{p}$ and $(\overrightarrow{p}\times \overrightarrow{q})\times \overrightarrow{r})\perp \overrightarrow{r}$

So, $(\overrightarrow{p}\times (\overrightarrow{q}\times \overrightarrow{r})\ne (\overrightarrow{p}\times \overrightarrow{q}))\times \overrightarrow{r}$

(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

$\because \overrightarrow{p}\cdot \overrightarrow{r}=\overrightarrow{r}\cdot \overrightarrow{p}$

$\overrightarrow{p}\cdot \overrightarrow{q}=\overrightarrow{q}\cdot \overrightarrow{p}$

$\overrightarrow{q}\cdot \overrightarrow{r}=\overrightarrow{r}\cdot \overrightarrow{q}$

(Hence this is proved) 

Explanation:

To evaluate, ∯_s z dx dy

where S(sphere) = x² + y² + z² = R² 

= ∯z dxdy
${\int }_S\overrightarrow{F}\cdot \hat{n}ds={\iint }_R(F_{1}dydz+F_{2}dxdz+F_{3}dxdy)$

$F_1=0,F_2=0,F_3=z$

$\overrightarrow{F}=Z\hat{K}$

$\mathrm{\nabla }\cdot \overrightarrow{F}=1$

Using Gauss Divergence Theorem:

${\int }_S\overrightarrow{F}\cdot \hat{n}ds={\int }_V\mathrm{\nabla }\cdot \overrightarrow{F}dV={\int }_{V}dV$

= 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

= $\int {\int }_S\surd 1+z_x^2+z_y^{2}dxdy$ 

= $\int {\int }_S\surd 1+5^2+3^{2}dxdy$ 

= $\int {\int }_S\surd 35dxdy$ 

= $\surd 35\int {\int }_{S}dxdy$

Now, the cylinder is 𝑥2 + 𝑦2 = 3

then surface area of S

=$\surd 35\int {\int }_{S}dxdy$  = √35 × π × ${\surd 3}^2$ = 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

$V=8\times \underset{2}{\overset{3}{\int }}xdx\underset{1}{\overset{2}{\int }}ydy\underset{0}{\overset{1}{\int }}zdz$

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

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 $\overline{a}=ai+bj+ck$ and $\overline{b}=di+ej+fk$ the cross, the product is given by: $\overline{a}\times \overline{b}=\left|\begin{array}{ccc}i& j& k\\ a& b& c\\ d& e& f\end{array}\right|=pi+qj+rk$

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 $\overline{a}\text{ and }\overline{b}$ form adjacent sides of the triangle then the area of the triangle is given by: $A={\displaystyle \frac{1}{2}}|\overline{a}\times \overline{b}|$

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)  $\overline{a}=3i+4j$ , and $\overline{b}=5i+7j+k$.

First, we will calculate the cross product as follow:

$\begin{array}{rl}\overline{a}\times \overline{b}& =\left|\begin{array}{ccc}i& j& k\\ 3& 4& 0\\ 5& 7& 1\end{array}\right|\\ & =i(4-0)-j(3-0)+k(21-20)\\ & =4i-3j+1k\end{array}$

Therefore, the magnitude of the cross product is:

$\begin{array}{rl}|\overline{a}\times \overline{b}|& =\sqrt{16+9+1}\\ & =\sqrt{26}\end{array}$

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

$A=\frac{1}{2}|a\times b|=\frac{1}{2}\times \sqrt{26}=\frac{\sqrt{26}}{2}$

Explanation:

Position vector $\overrightarrow{r}=x\hat{i}+y\hat{j}+zk$

$\left|\overrightarrow{r}\right|=\sqrt{x^2+y^2+z^2}$

$\varphi =\ln \left|\overrightarrow{r}\right|$

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

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

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

$\mathrm{\nabla }\varphi =\left\{\frac{xi+yj+zk}{x^2+y^2+z^2}\right\}$

$Gradient\mathrm{\nabla }\varphi =\frac{\overrightarrow{r}}{\overrightarrow{r}.\overrightarrow{r}}$

Concept:

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

If $\overrightarrow{f}=f_1\hat{i}+f_2\hat{j}+f_3\hat{k}$

Then

$\mathrm{\nabla }\times \overrightarrow{f}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ f_1& f_2& f_3\end{array}\right|$

Calculation:

Given vector is

$\overrightarrow{u}=yz\hat{i}+3zx\hat{j}+z\hat{k}$

Than $\mathrm{\nabla }\times \overrightarrow{u}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ yz& 3zx& z\end{array}\right|$

$\mathrm{\nabla }\times \overrightarrow{u}=\hat{i}(\frac{\partial (z)}{\partial y}-\frac{\partial (3zx)}{\partial z})-\hat{j}(\frac{\partial (z)}{\partial x}-\frac{\partial (yz)}{\partial z})+\hat{k}(\frac{\partial (3zx)}{\partial x}-\frac{\partial (yz)}{\partial y})$

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:

${\displaystyle \frac{x-x_1}{x_2-x_1}}={\displaystyle \frac{y-y_1}{y_2-y_1}}={\displaystyle \frac{z-z_1}{z_2-z_1}}=t$

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

Calculation:

Given:

$\mathrm{I}=\int \left(2\mathrm{x}{\mathrm{y}}^2\mathrm{d}\mathrm{x}+2{\mathrm{x}}^2\mathrm{y}\mathrm{d}\mathrm{y}+\mathrm{d}\mathrm{z}\right)$, A (0, 0, 0) and B(1, 1, 1).

Equation of line i.e. path

$\frac{\mathrm{x}-0}{1-0}=\frac{\mathrm{y}-0}{1-0}=\frac{\mathrm{z}-0}{1-0}=\mathrm{t}$

$\therefore \mathrm{x}=\mathrm{y}=\mathrm{z}=\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{t}:0\to 1$

$\therefore \mathrm{I}=\underset{0}{\overset{1}{\int }}\left(2{\mathrm{t}}^3\mathrm{d}\mathrm{t}+2{\mathrm{t}}^3\mathrm{d}\mathrm{t}+\mathrm{d}\mathrm{t}\right)$

$=4\cdot {\left[\frac{{\mathrm{t}}^4}{4}\right]}_0^1+{\left[\mathrm{t}\right]}_0^1=1+1=2$

$\therefore \int \left(2\mathrm{x}{\mathrm{y}}^2\mathrm{d}\mathrm{x}+2{\mathrm{x}}^2\mathrm{y}\mathrm{d}\mathrm{y}+\mathrm{d}\mathrm{z}\right)=2$

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

$V={\int }_a^b\pi y^{2}dx$

similarly for revolution about y-axis:

$V=\int \pi x^{2}dy$

Calculation:

Given:

$V=\underset{1}{\overset{2}{\int }}\pi y^{2}dx$

$V=\frac{3\pi }{2}$

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

Concept:

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are any non-zero vector then

$[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})....(1)$

Calculation: 

Given, $[\overrightarrow{a},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}]$

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

$=\overrightarrow{a}.[\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}+\overrightarrow{c}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{c}]$

$=\overrightarrow{a}.[\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}+\overrightarrow{c}\times \overrightarrow{b}]$

$=[\overrightarrow{a}\overrightarrow{b}\overrightarrow{a}]+[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]+[\overrightarrow{a}\overrightarrow{c}\overrightarrow{a}]+[\overrightarrow{a}\overrightarrow{c}\overrightarrow{b}]$

$(since[\overrightarrow{a}\overrightarrow{b}\overrightarrow{a}]=0)$

$[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]-[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]$

= 0

Concept:

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

Calculation:

Given:

$|\overrightarrow{a}|=2,|\overrightarrow{b}|=3,|\overrightarrow{c}|=5$

As  $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$

∴$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=0$

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

$\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{c}=0$

$\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}+2\overrightarrow{a}.\overrightarrow{b}+2\overrightarrow{b}.\overrightarrow{c}+2\overrightarrow{c}.\overrightarrow{a}=0$

$\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

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

$2^2+3^2+5^2+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

$2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=-38$

Hence $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=-19$

Given that,

F = a_x y - a_y x

y = x²

By differentiating on both sides,

⇒ dy = 2x dx

$\int F.dl={\int }_C\left(a_{x}y-a_{y}x\right).\left(dx{a}_x+dy{a}_y\right)$

$={\int }_C\left(ydx-xdy\right)$

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

$\int F.dl={\int }_{-1}^2{x}^{2}dx-2x^{2}dx$

$=-{\int }_{-1}^2{x}^{2}dx=-\frac{1}{3}{\left[x^3\right]}_{-1}^2=-3$

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:

${\int }_C\overrightarrow{F}\cdot dr={\int }_0^{{45}^{\circ }}\left[\left(-r\cos \theta \right)\left(-r\sin \theta \right)d\theta +\left(r\sin \theta \right)\left(r\cos \theta \right)d\theta \right]$

${\int }_0^{{45}^{\circ }}\left(r^2\cos \theta \sin \theta +r^2\sin \theta \cos \theta \right)d\theta$

With r = 1, the above integral becomes:

$=\frac{1}{2}{\int }_0^{{45}^{\circ }}\left(\sin 2\theta +\sin 2\theta \right)d\theta$

r = 1

$={\int }_0^{{45}^{\circ }}\sin 2\theta d\theta$

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

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