NEET Physics Solutions For Class 11 Chapter 10 Mathematical Tools – Vector

NEET Physics Solutions For Class 11 Chapter 10  Vector

In physics we deal with two types of physical quantity one is scalar and the other is a vector. Each scalar quantity has magnitude.

The magnitude of a physical quantity means a product of numerical value and unit of that physical quantity. For example mass = 4 kg

Magnitude of mass = 4 kg

and unit of mass = kg

Examples of scalar quantities: are mass, speed, distance, etc.

Scalar quantities can be added, subtracted, and multiplied by simple laws of algebra.

1. Definition Of Vector

If a physical quantity in addition to magnitude –

1. Has a specified direction.
2. It should obey the commutative law of additions
   1. \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}}\)
3. Obeys the law of parallelogram of addition, then and then only it is said to be a vector. If any of the above conditions is not satisfied the physical quantity cannot be a vector.

If a physical quantity is a vector it has a direction, but the converse may or may not be true, i.e. if a physical quantity has a direction, it may or may not a be vector. e.g. time, pressure, surface tension current, etc. have directions but are not vectors because they do not obey the parallelogram law of addition.

The magnitude of a vector \((\overrightarrow{\mathrm{A}})\) is the absolute value of a vector and is indicated by \(|\vec{A}|\) or A.

Examples of vector quantity are displacement, velocity, acceleration, force, etc.

Representation Of Vector: Geometrically, the vector is represented by a line with an arrow indicating the direction of the vector as

Mathematically, vector is represented by \(\overrightarrow{\mathrm{A}}\)

Sometimes it is represented by the bold letter A.

Important Points: If a vector is displaced parallel to itself it does not change

If a vector is rotated through an angle other than a multiple of 2π (or 360º) it changes.

If the frame of reference is translated or rotated the vector does not change (though its components may change).

Two vectors are called equal if their magnitudes and directions are the same,
and they represent values of the same physical quantity.

The angle between two vectors means the smaller of the two angles between the vectors when they are placed tail to tail by displacing either of the vectors parallel to itself (i.e. 0 ≤ θ ≤ π).

2. Unit Vector

A unit vector is a vector that has a unit magnitude and points in a particular direction. Any vector \((\overrightarrow{\mathrm{A}})\) can be written as the product of unit vector \((\hat{A})\) in that direction and magnitude of the given vector.

⇒ \(\overrightarrow{\mathrm{A}}=\mathrm{A} \hat{\mathrm{A}} \text { or } \hat{\mathrm{A}}=\frac{\vec{A}}{\mathrm{~A}}\)

A unit vector has no dimensions or units. Unit vectors along the positive x-, y- and z-axes of a rectangular coordinate system are denoted by

⇒ \(\hat{\mathrm{i}}, \hat{\mathrm{j}} \text { and } \hat{\mathrm{k}}\) and respectively such that

⇒ \(|\hat{\mathrm{i}}|=|\hat{\mathrm{j}}|=|\hat{\mathrm{k}}|\) = 1

Question 1. Three vectors \(\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}}, \overrightarrow{\mathrm{C}}\) are shown in the figure. Find angle between

1. \(\overrightarrow{\mathrm{A}} \text { and } \vec{B}\),
2. \(\overrightarrow{\mathrm{B}} \text { and } \vec{C}\)
3. \(\overrightarrow{\mathrm{A}} \text { and } \vec{C}\).

Answer:

To find the angle between two vectors we connect the tails of the two vectors. We can shift \(\overrightarrow{\mathrm{B}}\) such that tails of \(\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} \text { and } \overrightarrow{\mathrm{C}}\) are connected as shown in figure.

Now we can easily observe that angle between \(\vec{A} \text { and } \vec{B}\) is 60º, \(\vec{B} \text { and } \vec{C}\) is 15º and between \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{C}}\)is 75º.

Question 2. A unit vector along East is defined as \(\hat{\mathbf{i}}\). A force of 10⁵ dynes acts westwards. Represent the force in terms of \(\hat{\mathbf{i}}\)

Answer:

⇒ \(\overrightarrow{\mathrm{F}}=-10^5 \hat{\mathrm{i}}\) dynes

3. Multiplication Of A Vector By A Scalar

Multiplying a vector \(\overrightarrow{\mathrm{A}}\) with a positive number λ gives a vector \(\vec{B}(=\lambda \overrightarrow{\mathrm{A}})\) whose magnitude is changed by the factor λ but the direction is the same as that of \(\overrightarrow{\mathrm{A}}\).

Multiplying a vector \(\overrightarrow{\mathrm{A}}\) by a negative number λ gives a vector \(\overrightarrow{\mathrm{B}}\) whose direction is opposite to the direction of \(\overrightarrow{\mathrm{A}}\) and whose magnitude is − λ times \(|\vec{A}|\)

Question 3. A physical quantity (m = 3kg) is multiplied by a vector \(\overrightarrow{\mathrm{a}}\) such that \(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}\). Find the magnitude and direction of \(\overrightarrow{\mathrm{F}}\) if

1. \(\overrightarrow{\mathrm{a}}=3 \mathrm{~m} / \mathrm{s}^2\) Eastwards
2. \(\overrightarrow{\mathrm{a}}=-4 \mathrm{~m} / \mathrm{s}^2\) Northwards

Answer:

1. \(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}=3 \times 3 \mathrm{~ms}^{-2}\) East wards = 9 N Eastwards
2. \(\overrightarrow{\mathrm{F}}=\mathrm{m} \vec{a}=3 \times(-4)\) Northwards

= – 12 N North wards = 12 N Southwards

4. Addition Of Vectors

The addition of vectors is done by parallelogram law or the triangle law :

Parallelogram Law Of Addition Of Vectors: If two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) are represented by two adjacent sides of a parallelogram both pointing outwards (and their tails coinciding) as shown. Then the diagonal drawn through the intersection of the two vectors represents the resultant (i.e., the vector sum of \(\vec{A} \text { and } \vec{B}\)).

R = \(\sqrt{A^2+B^2+2 A B \cos \theta}\)

The direction of resultant vector \(\overrightarrow{\mathrm{R}} \text { from } \overrightarrow{\mathrm{A}}\) given by

⇒ \(\tan \phi=\frac{\mathrm{MN}}{\mathrm{PN}}\)

⇒ \(\frac{\mathrm{MN}}{\mathrm{PQ}+\mathrm{QN}}\)

⇒ \(\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}\)

⇒ \(\varphi=\tan ^{-1}\left(\frac{B \sin \theta}{A+B \cos \theta}\right)\)

Triangle law of addition of vectors: To add two vectors \(\vec{A} \text { and } \vec{B}\) shift any of the two vectors parallel to itself until the tail of \(\vec{B}\) is at the head of \(\vec{A}\).

The sum \(\vec{A}+\vec{B}\) is a vector \(\vec{R}\) drawn from the tail of \(\vec{A}\)to the head of \(\vec{B}\), i.e.,\(\vec{A}+\vec{B}=\vec{R}\). As the figure formed is a triangle, this method is called the ‘ triangle method’ of the addition of vectors.

If the ‘triangle method’ is extended to add any number of vectors in one operation as shown. Then the figure formed is a polygon and hence the name Polygon Law of addition of vectors is given to such type of addition.

Important Points:

To a vector only a vector of the same type can be added that represents the same physical quantity and the resultant is a vector of the same type.

As R = [A² + B² + 2AB cosθ]^1/2 so R will be maximum when, cos θ = max = 1, i.e., θ = 0º, i.e. vectors are like or parallel and R_max= A + B.

The resultant will be minimum if, cos θ = min = -1, i.e., θ = 180º, i.e. vectors are antiparallel, and R_min = A ~ B.

If the vectors A and B are orthogonal, i.e., θ = 90º, R = \(\sqrt{A^2+B^2}\)

As previously mentioned the resultant of two vectors can have any value from (A ~ B) to (A + B) depending on the angle between them and the magnitude of the resultant decreases as θ increases from 0º to 180º

The minimum number of unequal coplanar vectors whose sum can be zero is three.

The resultant of three non-coplanar vectors can never be zero, or a minimum number of noncoplanar vectors whose sum can be zero is four.

Subtraction of a vector from a vector is the addition of a negative vector, i.e.,

⇒ \(\vec{A}-\vec{B}=\vec{A}+(-\vec{B})\)

From figure it is clear that \(\vec{A}-\vec{B}\) is equal to addition of \(\overrightarrow{\mathrm{A}}\) with reverse of \(\overrightarrow{\mathrm{B}}\)

⇒ \(|\vec{A}-\vec{B}|=\left[(A)^2+\left(B^2\right)+2 A B \cos \left(180^{\circ}-\theta\right)\right]^{1 / 2}\)

⇒ \(|\vec{A}-\vec{B}|=\sqrt{A^2+B^2-2 A B \cos \theta}\)

Change in a vector’s physical quantity means subtraction of the initial vector from the final vector.

Question 4. Find the resultant of two forces each having magnitude F₀, and the angle between them is θ.

Answer:

⇒ \(F_{\text {Resultant }}^2=F_0^2+F_0^2+2 F_0^2 \cos \theta\)

⇒ \(2 F_0^2(1+\cos \theta)\)

⇒ \(2 F_0^2\left(1+2 \cos ^2 \frac{\theta}{2}-1\right)\)

⇒ \(2 \times 2 F_0^2 \cos ^2 \frac{\theta}{2}\)

⇒ \(\mathrm{F}_{\text {resultant }}=2 \mathrm{~F}_0 \cos \frac{\theta}{2}\)

Question 5. Two non zero vectors \(\vec{A} \text { and } \vec{B}\) are such that \(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|\). Find angle between \(\overrightarrow{\mathrm{A}}\)and \(\overrightarrow{\mathrm{B}}\)

Answer:

⇒ \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\)

⇒ A² + B² + 2AB cos θ

= A² + B² – 2AB cos θ

⇒ 4AB cos θ = 0

⇒ cos θ = 0

⇒ θ = \(\frac{\pi}{2}\)

Question 6. If the sum of two unit vectors is also a unit vector. Find the magnitude of their difference.

Answer:

Let \(\hat{A} \text { and } \hat{B}\) are the given unit vectors and \(\hat{R}\) is their resultant then

⇒ \(|\hat{R}|=|\hat{A}+\hat{B}|\)

⇒ \(1=\sqrt{(\hat{A})^2+(\hat{B})^2+2|\hat{A} \| \hat{B}| \cos \theta}\)

1 = 1 + 1 + 2 cos θ

⇒ cos θ = \(-\frac{1}{2}\)

⇒ \(|\hat{A}-\hat{B}|=\sqrt{(\hat{A})^2+(\hat{B})^2-2|\hat{A}||\hat{B}| \cos \theta}\)

⇒ \(\sqrt{1+1-2 \times 1 \times 1\left(-\frac{1}{2}\right)}\)

⇒ \(\sqrt{3}\)

5. Resolution Of Vectors

If \(\vec{a} \text { and } \vec{b}\) be any two nonzero vectors in a plane with different directions and \(\overrightarrow{\mathrm{A}}\)be another vector in the same plane.

⇒ \(\overrightarrow{\mathrm{A}}\) can be expressed as a sum of two vectors – one obtained by multiplying \(\overrightarrow{\mathrm{a}}\) by a real number and the other obtained by multiplying

⇒ \(\overrightarrow{\mathrm{b}}\) by another real number.

⇒ \(\overrightarrow{\mathrm{A}}=\lambda \overrightarrow{\mathrm{a}}+\mu \overrightarrow{\mathrm{b}}\) (where λ and μ are real numbers)

We say that \(\overrightarrow{\mathrm{A}}\) has been resolved into two component vectors namely

⇒ \(\lambda \vec{a} \text { and } \mu \vec{b}\)

λ \(\vec{a} \text { and } \mu \vec{b}\) along \(\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}}\) respectively. Hence one can resolve a given vector into two component vectors along a set of two vectors − all three lie in the same plane.

Resolution Along Rectangular Component:

It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude, which we call unit vectors. \(\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}}\) are unit vector along x,y and z-axis as shown in figure below:

Resolution in two Dimension

Consider a vector \(\overrightarrow{\mathrm{A}}\) that lies in xy plane as shown in figure,

⇒\(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{A}}_1+\overrightarrow{\mathrm{A}}_2\)

⇒ \(\vec{A}_1=A_x \hat{i}, \quad \vec{A}_2=A_y \hat{j}\)

⇒ \(\vec{A}=A_x \hat{i}+A_y \hat{j}\)

The quantities A_x and A_y are called x- and y- components of the vector \(\overrightarrow{\mathrm{A}}\)

A_x is itself not a vector but \(A_x \hat{i}\) is a vector and so is \(A_y \hat{j}\)

A_x= A cos θ and A_y= A sin θ

It’s clear from the above equation that a component of a vector can be positive, negative, or zero depending on the value of θ. A vector \(\overrightarrow{\mathrm{A}}\) can be specified in a plane by two ways:

Its magnitude A and the direction θ it makes with the x-axis; or

Its components are A_x and A_y.

A = \(A=\sqrt{A_x^2+A_y^2}, \theta=\tan ^{-1} \frac{A_y}{A_x}\)

Note: If A = Ax ⇒ Ay = 0 and if A = Ay ⇒ Ax= 0 i.e. components of a vector perpendicular to itself is always zero.

The rectangular components of each vector and those of the

sum

⇒\(\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) are shown in the figure. We saw that

⇒\(\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) is equivalent to both

C_x = A_x+ B_x

and C_y = A_y+ B_y

Resolution In Three Dimensions: A vector \(\overrightarrow{\mathrm{A}}\) in components along x-, y- and z-axis can be written as:

⇒\(\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{BP}}=\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{BP}}\)

⇒\(\vec{A}=\vec{A}_z+\vec{A}_x+\vec{A}_y=\vec{A}_x+\vec{A}_y+\vec{A}_z\)

⇒ \(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\)

A = \(\sqrt{A_x^2+A_y^2+A_z^2}\)

⇒ \(A_x=A \cos \alpha, A_y=A \cos \beta, A_z=A \cos \gamma\)

where cos α, cos β, and cos γ are termed as Direction Cosines of a given vector \(\overrightarrow{\mathrm{A}}\).

cos² α + cos² β + cos² γ = 1

Question 7. A mass of 2 kg lies on an inclined plane as shown in the figure. Resolve its weight along and perpendicular to the plane.(Assumeg=10m/s²)

Answer:

Component along the plane = 20 sin 30 = 10 N

component perpendicular to the plane = 20 cos 30 = \(10 \sqrt{3} \mathrm{~N}\)

Question 8. A vector makes an angle of 30º with the horizontal. If a horizontal component of the vector is 250. Find the magnitude of a vector and its vertical component.

Answer: Let vector is \(\overrightarrow{\mathrm{A}}\)

⇒ \(A_x=A \cos 30^{\circ}\) = 250

⇒ \(\frac{A \sqrt{3}}{2}\)

A= \(\frac{500}{\sqrt{3}}\)

⇒ \(A_y=A \sin 30^{\circ}\)

⇒ \(\frac{500}{\sqrt{3}} \times \frac{1}{2}\)

⇒ \(\frac{250}{\sqrt{3}}\)

Question 9. A = \(\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{k}\), when a vector \(\overrightarrow{\mathrm{B}}\) is added to \(\overrightarrow{\mathrm{A}}\), we get a unit vector along x-axis. Find the value of \(\overrightarrow{\mathrm{B}}\)? Also, find its magnitude

Answer:

⇒\(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}\)

⇒ \(\vec{B}=\hat{i}-\vec{A}=\hat{i}-(\hat{i}+2 \hat{j}-3 \hat{k})=-2 \hat{j}+3 \hat{k}\)

⇒ \(|\vec{B}|=\sqrt{(2)^2+(3)^2}=\sqrt{13}\)

Question 10. In the above question find a unit vector along \(\vec{B}\).

Answer:

⇒ \(\vec{B}=\frac{\vec{B}}{B}=\frac{-2 \hat{j}+3 \hat{k}}{\sqrt{13}}\)

Question 11. Vector \(\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} \text { and } \overrightarrow{\mathrm{C}}\) have magnitude 5, \(5 \sqrt{2}\) and 5 respectively, direction of \(\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} \text { and } \overrightarrow{\mathrm{C}}\) are towards east, North-East and North respectively. If \(\hat{i} \text { and } \hat{j}\) and are unit vectors along East and North respectively. Express the sum \(\vec{A}+\vec{B}+\vec{C}\) in terms of \(\hat{i}, \hat{j}\). Also, Find the magnitude and direction of the resultant.

Answer:

⇒ \(\overrightarrow{\mathrm{A}}=5 \hat{i}\)

⇒ \(\overrightarrow{\mathrm{C}}=5 \hat{j}\)

⇒ \(\vec{B}=5 \sqrt{2} \cos 45 \hat{i}+5 \sqrt{2} \sin 45 \hat{j} \quad=5 \hat{i}+5 \hat{j}\)

⇒ \(\vec{A}+\vec{B}+\vec{C}=5 \hat{i}+5 \hat{i}+5 \hat{j}+5 \hat{j} \quad=10 \hat{i}+10 \hat{j}\)

⇒ \(|\vec{A}+\vec{B}+\vec{C}|\)

⇒ \(\sqrt{(10)^2+(10)^2}=10 \sqrt{2}\)

⇒ \(\tan \theta=\frac{10}{10}=1\)

θ = 45º from East

Question 12. You walk 3 Km west and then 4 Km headed 60° north of east. Find your resultant displacement

1. Graphically and
2. Using vector components.

Answer:

Picture the Problem: The triangle formed by the three vectors is not a right triangle, so the magnitudes of the vectors are not related by the Pythagoras theorem. We find the resultant graphically by drawing each of the displacements to scale and measuring the resultant displacement.

If we draw the first displacement vector 3 cm long and the second one 4 cm long, we find the resultant vector to be about 3.5 cm long.

Thus the magnitude of the resultant displacement is 3.5 Km. The angle θ made between the resultant displacement and the west direction can then be measured with a protractor. It is about 75°.

Let \(\overrightarrow{\mathrm{A}}\) be the first displacement and choose the x-axis to be in the easterly direction. Compute \(A_x \text { and } A_y, A_x=-3, A_y=0\)

Similarly, compute the components of the second displacement \(\vec{B}, B_x=4 \cos 60^{\circ}=2, B_y=4\) sin 60° = \(2 \sqrt{3}\)

The components of the resultant displacement = \(\vec{A}+\vec{B}\) are found by addition,

⇒ \(\overrightarrow{\mathrm{C}}=(-3+2) \hat{i}+(2 \sqrt{3}) \hat{j}=-\hat{i}+2 \sqrt{3} \hat{j}\)

The Pythagorean theorem gives the magnitude of \(\overrightarrow{\mathrm{C}}\)

⇒ \(C=\sqrt{1^2+(2 \sqrt{3})^2}\)

⇒ \(\sqrt{13}\)

⇒ 3.6

The ratio of C_y to C_x gives the tangent of the angle θ between \(\overrightarrow{\mathrm{C}}\) and the x-axis.

⇒ tan θ = \(2 \sqrt{3}\)/(-1)

⇒ \(\theta=-74^{\circ}\)

Remark: Since the displacement (which is a vector) was asked for, the answer must include either the magnitude and direction, or both components. In (2) we could have stopped at step 3 because the x and y components completely define the displacement vector.

• We converted the magnitude and direction to compare with the answer to part (1). Note that in step 5 of (2), a calculator gives the angle as –74°. However, the calculator can’t distinguish whether the x or y components are negative.
• We noted in the figure that the resultant displacement makes an angle of about 75° with the negative x-axis and an angle of about 105° with the positive x-axis. This agrees with the results in (a) within the accuracy of our measurement.