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

Explanation:

Any vector in the plane $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ is

$\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{c}}$

$\overrightarrow{\mathrm{r}}=2\mathrm{i}-\mathrm{j}+\mathrm{k}+\lambda (\mathrm{i}+\mathrm{j}-2\mathrm{k})$

$\overrightarrow{\mathrm{r}}=(2+\lambda )\mathrm{i}+(\lambda -1)\mathrm{j}+(1-2\lambda )\mathrm{k}$...(i)

Projection of $\overrightarrow{\mathrm{r}}$ on $\overrightarrow{\mathrm{a}}$ is 

$\frac{\mathrm{r}.\mathrm{a}}{|\mathrm{a}|}$ = $\frac{\{(2+\lambda )\mathrm{i}+(\lambda -1)\mathrm{j}+(1-2\lambda )\mathrm{k}\}.(2\mathrm{i}-\mathrm{j}+\mathrm{k})}{\sqrt{4+1+1}}$

     = $\frac{2(2+\lambda )-(\lambda -1)+(1-2\lambda )}{\sqrt{4+1+1}}$

     = $\frac{6-\lambda }{\sqrt{4+1+1}}$

By given condition

$\frac{6-\lambda }{\sqrt{4+1+1}}$ = 0

⇒ λ = 6

Therefore the required vector is given by putting λ = 6 in (i) and is

r = 

Concept Used:

for ${\overrightarrow{d}}_1$ and ${\overrightarrow{d}}_2$ as diagonals of parallelogram, area  parallelogram is $\frac{1}{2}|{\overrightarrow{d}}_1\times {\overrightarrow{d}}_2|$

Calculation:

$\overrightarrow{a}+\overrightarrow{b}=2\hat{i}-3\hat{j}-\hat{k}-\hat{i}+\hat{k}$
$\Rightarrow \overrightarrow{a}+\overrightarrow{b}=\hat{i}-3\hat{j}$
 and  $\overrightarrow{b}+\overrightarrow{c}=-\hat{i}+\hat{k}+2\hat{j}-\hat{k}$
$\Rightarrow \overrightarrow{b}+\overrightarrow{c}=-\hat{i}+2\hat{j}$
 area of parallelogram $=\frac{1}{2}|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{b}+\overrightarrow{c})|$
$\Rightarrow \text{ area }=\frac{1}{2}\left|\begin{array}{ccc}\hat{2}& \hat{j}& \hat{k}\\ 1& -3& 0\\ -1& 2& 0\end{array}\right|$ 
$\Rightarrow \text{ area }=\frac{1}{2}|\hat{i}(0-0)-j(0-0)+\hat{k}(2-3)|$
$\Rightarrow \text{ area }=\frac{1}{2}|-\hat{k}|=\frac{1}{2}$

Concept - 

Use the cross product  of vectors.

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$

Solution -

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|$

=$2\hat{i}-2\hat{j}+\hat{k}$

$|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{4+4+1}=3$

so $|\overrightarrow{c}-\overrightarrow{a}|=2\sqrt{2}$

$\Rightarrow |\overrightarrow{c}-\overrightarrow{a}{|}^2=8$8

$\Rightarrow |\overrightarrow{c}{|}^2-2\overrightarrow{a}.\overrightarrow{c}+|\overrightarrow{a}{|}^2=8$

$\Rightarrow |\overrightarrow{c}{|}^2-2|\overrightarrow{c}|+9=8$

$\Rightarrow |\overrightarrow{c}{|}^2-2|\overrightarrow{c}|+1=0$

$\Rightarrow |\overrightarrow{c}|=1$

$|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=|(\overrightarrow{a}\times \overrightarrow{b})\overrightarrow{c}|sin{30}^{\circ }=\frac{3}{2}$

Hence the final answer is option is 2.

Calculation:

Given, $(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})$

$\Rightarrow (\overrightarrow{a}\times \overrightarrow{a})+(\overrightarrow{a}\times \overrightarrow{b})-(\overrightarrow{b}\times \overrightarrow{a})-(\overrightarrow{b}\times \overrightarrow{b})$

we know that, $\overrightarrow{a}\times \overrightarrow{a}=0,\overrightarrow{b}\times \overrightarrow{b}=0and,\overrightarrow{a}\times \overrightarrow{b}=-(\overrightarrow{b}\times \overrightarrow{a})$

$\Rightarrow (\overrightarrow{a}\times \overrightarrow{b})-(\overrightarrow{b}\times \overrightarrow{a})$

$\Rightarrow (\overrightarrow{a}\times \overrightarrow{b})-[-(\overrightarrow{a}\times \overrightarrow{b})]$

$=(\overrightarrow{a}\times \overrightarrow{b})+(\overrightarrow{a}\times \overrightarrow{b})=2(\overrightarrow{a}\times \overrightarrow{b})$

Explanation:

The magnitude $|\mathbf{v}|$ of a vector $\mathbf{v}=a\hat{i}+b\hat{j}+c\hat{k}$ is given by:

$|\mathbf{v}|=\sqrt{a^2+b^2+c^2}$

In this case, the vector is $3\hat{j}+4\hat{k}$ , so $a=0$ , b = 3 , and c = 4 . The magnitude

is $|\mathbf{v}|=\sqrt{0^2+3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5.$

The unit vector in the direction of a given vector is obtained by dividing the

vector by its magnitude. So, the unit vector $\hat{u}along3\hat{j}+4\hat{k}$ is

$\hat{u}=\frac{1}{5}(3\hat{j}+4\hat{k}).$

The correct option is option 1.

Concept Used:

for ${\overrightarrow{d}}_1$ and ${\overrightarrow{d}}_2$ as diagonals of parallelogram, area  parallelogram is $\frac{1}{2}|{\overrightarrow{d}}_1\times {\overrightarrow{d}}_2|$

Calculation:

$\overrightarrow{a}+\overrightarrow{b}=2\hat{i}-3\hat{j}-\hat{k}-\hat{i}+\hat{k}$
$\Rightarrow \overrightarrow{a}+\overrightarrow{b}=\hat{i}-3\hat{j}$
 and  $\overrightarrow{b}+\overrightarrow{c}=-\hat{i}+\hat{k}+2\hat{j}-\hat{k}$
$\Rightarrow \overrightarrow{b}+\overrightarrow{c}=-\hat{i}+2\hat{j}$
 area of parallelogram $=\frac{1}{2}|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{b}+\overrightarrow{c})|$
$\Rightarrow \text{ area }=\frac{1}{2}\left|\begin{array}{ccc}\hat{2}& \hat{j}& \hat{k}\\ 1& -3& 0\\ -1& 2& 0\end{array}\right|$ 
$\Rightarrow \text{ area }=\frac{1}{2}|\hat{i}(0-0)-j(0-0)+\hat{k}(2-3)|$
$\Rightarrow \text{ area }=\frac{1}{2}|-\hat{k}|=\frac{1}{2}$

Scalar Quantity: The physical quantities which require only magnitude to express are called a scalar quantity.

• Examples: Mass, Distance, time, speed, volume, temperature, density, volume, magnetic flux, magnetic potential, electric current, work, power, relative permeability, etc.

Vector Quantity: The physical quantities which require both magnitude and direction to express are called vector quantities.

• Examples: Displacement, Weight, velocity, acceleration, force, momentum, Impulse, electric field, magnetic field density, magnetic field intensity, etc.

Additional Information

• Magnetic field intensity: The ratio of the MMF needed to create a certain Flux Density (B) within a particular material per unit length of that material is called magnetic field intensity.
• Intensity of magnetization (I): It is the degree to which a substance is magnetized when placed in a magnetic field. 
• It can also be defined as the pole strength per unit cross-sectional area of the substance or the induced dipole moment per unit volume

Hence, $I=\frac{m}{A}=\frac{M}{V}$

• It is a vector quantity,
• Its S.I. unit is Ampere/meter.

Concept:

The dot product of the two vectors is given by:

a.b = |a||b| cosθ 

Calculation:

Given, |a| = 9 and |b| = 5$\sqrt{2}$ and θ = 45°.

a.b = 9 × 5$\sqrt{2}$ × cos 45° 

a.b = 45

Additional InformationThe cross product of the two vectors is given by:

a x b = |a||b| sinθ 

Concept:

Dot product:

The dot product of two vectors is given by:

$\overrightarrow{A}\overrightarrow{B}=|A||B|cos\theta$

where, cos θ = Angle between vectors A and B

|A| = Magnitude of vector A

|B| = Magnitude of vector B

$\overrightarrow{B}=|A|cos\theta$

Hence, the dot product is also the projection of one vector on another vector.

Concept-

Length of vector $a\hat{i}+b\hat{j}+c\hat{k}=\sqrt{(a{)}^2+(b{)}^2+(c{)}^2}$

Let Initial point A(x₁, y₁, z₁) and terminal point B(x₂, y2, z2) of a vector than 

$|\overrightarrow{AB}|=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{i}+(z_2-z_1)\hat{i}$

Calculation-

Initial point is P(2,-3,4) and terminal point Q(-2, 1, 1)

Vector $|\overrightarrow{PQ}|$ = (-2 - 2)$\hat{i}$ + (1 - (-3))$\hat{j}$+ (1 - 4)$\hat{k}$

Vector $|\overrightarrow{PQ}|$ = -4$\hat{i}$ + 4$\hat{j}$ - 3$\hat{k}$

Now the length of a vector  $|\overrightarrow{PQ}|$ is given by 

$|\overrightarrow{PQ}|=\sqrt{(-4{)}^2+(4{)}^2+(-3{)}^2}$

$|\overrightarrow{PQ}|=\sqrt{16+16+9}$

$|\overrightarrow{PQ}|=\sqrt{41}$

∴ length of vector $|\overrightarrow{PQ}|is\sqrt{41}$

Concept:

The vector product is given by:

a x b = $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$

Calculation:

a x b = $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 2\\ 1& -2& 3\end{array}\right|$

a x b = $1\hat{i}-7\hat{j}-5\hat{k}$

Additional Information The scalar product is given by:

a.b = $(a_1{b}_1)\hat{i}+(a_2{b}_2)\hat{j}+(a_3{b}_3)\hat{k}$

Concept:

The cross product of two vectors A and B is given by:

$\overrightarrow{A}\times \overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|sin\theta \hat{n}$

The cross product of $\hat{i},\hat{j}and\hat{k}$

$\hat{i}\times \hat{j}=\hat{k}$ and $\hat{j}\times \hat{i}=-\hat{k}$

$\hat{j}\times \hat{k}=\hat{i}$ and $\hat{k}\times \hat{j}=-\hat{i}$

$\hat{k}\times \hat{i}=\hat{j}$ and $\hat{i}\times \hat{k}=-\hat{j}$

Explanation:

The vector product is anti- commutative.

$\overrightarrow{A}\times \overrightarrow{B}=-(\overrightarrow{B}\times \overrightarrow{A})$

$(\hat{i}\times \hat{j})=-(\hat{j}\times \hat{i})$

$\hat{k}=-(-\hat{k})$

$\hat{k}=\hat{k}$

Concept:

Zero vector/Null vector:

• The vector having magnitude equal to zero is called a null vector. It is generally represented by O.
• In zero vector the initial and terminal points coincide with each other. Hence length is zero
• A point is generally taken as a null vector.
• $\left|\overrightarrow{A}\right|=0$

Additional InformationThe dot product or scaler of two vectors is given as:

$D=\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}\right|.\left|\overrightarrow{B}\right|cos\theta$

The cross product or the vector product of two vectors is given as:

$\overrightarrow{C}=\overrightarrow{A}\times \overrightarrow{B}$

Calculation:

Given, $(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})$

$\Rightarrow (\overrightarrow{a}\times \overrightarrow{a})+(\overrightarrow{a}\times \overrightarrow{b})-(\overrightarrow{b}\times \overrightarrow{a})-(\overrightarrow{b}\times \overrightarrow{b})$

we know that, $\overrightarrow{a}\times \overrightarrow{a}=0,\overrightarrow{b}\times \overrightarrow{b}=0and,\overrightarrow{a}\times \overrightarrow{b}=-(\overrightarrow{b}\times \overrightarrow{a})$

$\Rightarrow (\overrightarrow{a}\times \overrightarrow{b})-(\overrightarrow{b}\times \overrightarrow{a})$

$\Rightarrow (\overrightarrow{a}\times \overrightarrow{b})-[-(\overrightarrow{a}\times \overrightarrow{b})]$

$=(\overrightarrow{a}\times \overrightarrow{b})+(\overrightarrow{a}\times \overrightarrow{b})=2(\overrightarrow{a}\times \overrightarrow{b})$

Concept:

• Scalar quantities: The physical quantities which have only magnitude and no direction are called scalar quantities or scalars.
  • A scalar quantity can be specified by a single number, along with the proper unit.
  • Examples: Mass, volume, density, time, temperature, electric current, luminous intensity, voltage etc.
• Vector quantities: The physical quantities which have both magnitude and direction and obey the laws of vector addition are called vector quantities or vectors.
  • A vector quantity is specified by a number with a unit and its direction.
  • Examples Displacement, velocity, force, momentum, etc.

Analysis:

1. Since the electric current (I) doesn't follow the vector addition rule. The current can be added or divided into several components arithmetically. So the electric current is a scalar quantity.
2. An electric potential is the amount of work needed to move a unit of positive charge from a reference point to a specific point inside the field.
   It is a scalar quantity because work is not a vector quantity. However, scalars are allowed to be negative. The minus sign on the potential does not indicate the vector direction. A negative potential can be attracted by a positive potential and repulsed by another negative potential.
3. Resistance is the opposition delivered by the conductor in which current flows through it. SI unit of the electricity is Ampere(A) and it is a scalar quantity.
4. The electric field is the region around the electric charge in which another charge can feel the force. It is a vector quantity.