Topics

Angle between two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is defined as the smaller angle between the tail's or the head's of the two vectors.

Laws of addition and subtraction of vectors:

Triangle rule of addition:

Steps for adding two vectors representing same physical quantity by triangle law :
(i) Keep vectors such that tail of one vector coincides with head of other.
(ii) Join tail of first to head of the other by a line with arrow at head of the second.
(iii) This new vector is the sum of two vectors. (also called resultant)

For Example :

To add \(\overrightarrow{CD}\) to \(\overrightarrow{AB}\), place tail of \(\overrightarrow{CD}\) at the head of \(\overrightarrow{AB}\). The sum is the vector from the tail of \(\overrightarrow{AB}\) to head of \(\overrightarrow{CD}\) i.e. \(\overrightarrow{AD}\).

\[\overrightarrow{AB} + \overrightarrow{CD} = \overrightarrow{AD}\]

Graphical Representation :

(i)

(ii)

Polygon Law of addition:

This law is used for adding more than two vectors. This is extension of triangle law of addition. We keep on arranging vectors s.t. tail of next vector lies on head of former.
When we connect the tail of first vector to head of last we get resultant of all the vectors.

\[\overrightarrow{P} = (((\overrightarrow{a} + \overrightarrow{b}) + \overrightarrow{c}) + \overrightarrow{d}) = ((\overrightarrow{c} + \overrightarrow{a}) + \overrightarrow{b}) + \overrightarrow{d}\]

[Associative Law]

Parallelogram law of addition:

Steps:

(i) Keep two vectors such that their tails coincide.
(ii) Draw parallel vectors to both of them considering both of them as sides of a parallelogram.
(iii) Then the diagonal drawn from the point where tails coincide represents the sum of two
vectors, with its tail at point of coincidence of the two vectors.

\[\underset{\overrightarrow{a}}{\overset{/\overrightarrow{b}}{︸}}\]

Steps :

(i)

(ii)

(iii)

\[\overrightarrow{AC} = \overrightarrow{a} + \overrightarrow{b}\]

\(\overrightarrow{AC} = \overrightarrow{a} + \overrightarrow{b}\) and \(\overrightarrow{AC} = \overrightarrow{b} + \overrightarrow{a}\) thus \(\overrightarrow{c} = \overrightarrow{a} + \overrightarrow{b} = \overrightarrow{b} + \overrightarrow{a}\) [Commutative Law]

So, the full parallelogram would look like this

\[\overrightarrow{c} = \overrightarrow{a} + \overrightarrow{b};\overrightarrow{d} = \overrightarrow{a} - \overrightarrow{b}\]

Derivation of general formula :

\[\begin{matrix} \overrightarrow{R} & \ = \overrightarrow{a} + \overrightarrow{b},|\overrightarrow{R}| = R,|\overrightarrow{a}| = a,|\overrightarrow{\text{ }b}| = b \\ R & \ = \sqrt{(a + bcos\theta)^{2} + (bsin\theta)^{2}} \\ & \ = \sqrt{a^{2} + b^{2} + 2abcos\theta} \end{matrix}\]

as in figure angle between \(\overrightarrow{a}\) and \(\overrightarrow{R}\) is \(\alpha\).
\[tan\alpha = \frac{bsin\theta}{a + bcos\theta}\]

The angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is \(\theta\) and resultant \(\overrightarrow{R} = \overrightarrow{a} + \overrightarrow{b}\).
(i) If \(\theta = 0^{\circ}\ \Rightarrow \overrightarrow{a}\|\overrightarrow{b}\)
then, \(|\overrightarrow{R}| = |\overrightarrow{a}| + |\overrightarrow{b}|\)
& \(|\overrightarrow{R}|\) is maximum
(ii) If \(\theta = \pi\ \Rightarrow \overrightarrow{a}\) anti \(\|\overrightarrow{b}\)
then, \(|\overrightarrow{R}| = |\overrightarrow{a}| - |\overrightarrow{b}| \mid\)
\(\&|\overrightarrow{R}|\) is minimum
(iii) If \(\theta = \pi/2\ \Rightarrow \overrightarrow{a}\bot\overrightarrow{b}\)
\[R = \sqrt{a^{2} + b^{2}} \]& \(tan\alpha = b/a(\alpha\) is angle with \(\overrightarrow{a})\)

(iv) If \(|\overrightarrow{a}| = |\overrightarrow{b}| = a\)
\[|\overrightarrow{R}| = 2acos\theta/2 \]& \(\alpha = \theta/2\)

(v) \(\ \) If \(|\overrightarrow{a}| = |\overrightarrow{b}| = a\&\theta = 2\pi/3\)
then \(|\overrightarrow{R}| = a\)

Practice Exercise

Q. 1 Two vectors \(\overrightarrow{P}\) & \(\overrightarrow{Q}\) are arranged in such a way that they form adjacent sides of a parallelogram as shown in figure

Which of the following options have correct relationship
(A) \(\overrightarrow{Q} = \overrightarrow{R} + \overrightarrow{S}\)
(B) \(\overrightarrow{R} = \overrightarrow{P} + \overrightarrow{Q}\)
(C) \(\overrightarrow{R} = \overrightarrow{P} + \overrightarrow{S}\)
(D) \(\overrightarrow{S} = \overrightarrow{Q} - \overrightarrow{P}\)
Q. 2 Two vectors of \(\mathbf{10}\) units & \(\mathbf{5}\) units make an angle of \(\mathbf{120}^{\mathbf{\circ}}\) with each other. Find the magnitude & angle of resultant with vector of \(\mathbf{10}\) unit magnitude.

Lets say we have a vector \(\overrightarrow{a}\) and \(k\) is a Scalar. Vector \(\overrightarrow{b} = k\overrightarrow{a}\) is defined as a vector of magnitude \(|ka|\).
If k is a positive then direction of \(\overrightarrow{b}\) is along \(\overrightarrow{a}\).
If \(k\) is negative then direction of \(\overrightarrow{b}\) is opposite to \(\overrightarrow{a}\).
If

\[\begin{matrix} & \ |k| > 1 \Rightarrow |\overrightarrow{\text{ }b}| > |\overrightarrow{a}| \\ & \ |k| < 1 \Rightarrow |\overrightarrow{\text{ }b}| < |\overrightarrow{a}| \end{matrix}\ \overset{\overset{\phantom{\overrightarrow{a}\ }}{\rightarrow}3\overrightarrow{a}}{\underset{- 1.5\overrightarrow{a}}{\longrightarrow}}\]

A vector may be multiplied by a pure number or by a scalar.
When a vector is multiplied by a scalar, the new vector may become a different physical quantity for example, when velocity (a vector) is multiplied by time (a scalar) we obtain a displacement (a vector).

Subtraction of vectors :

To subtract two vectors, reverse the direction of the vector being subtracted and add the inverted vector to the vector from which you are subtracting.
Let say we want to obtain \(\overrightarrow{a} - \overrightarrow{b}\)

\[\overrightarrow{s} = \overrightarrow{a} - \overrightarrow{b} = \overrightarrow{a} + ( - \overrightarrow{b})\]

i.e. \(\overrightarrow{a} - \overrightarrow{b}\) can be understood as summation of \(\overrightarrow{a} + ( - \overrightarrow{b})\)

For \(\overrightarrow{s} = \overrightarrow{a} - \overrightarrow{b}\)

Steps :

(i) Put tail of \(\overrightarrow{b}\) at head of \(\overrightarrow{a}\)
(ii) Take- \(\overrightarrow{b}\)
(iii) Resultaul vector \(\overrightarrow{s}\) from tail of \(\overrightarrow{a}\) to head of \(- \overrightarrow{b}\).
(iv) Angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is \(\theta\), then angle between \(\overrightarrow{a}\) and \(- \overrightarrow{b}\) or between \(- \overrightarrow{a}\) and \(\overrightarrow{b}\) is \(\left( 180^{\circ} - \theta \right)\).
the angle between \(\overrightarrow{a}\) and \(- \overrightarrow{b}\) becomes \(\pi - \theta\)
so we get \(s = \sqrt{(a + bcos(\pi - \theta))^{2} + (bsin(\pi - \theta))^{2}}\)

\[= \sqrt{a^{2} + b^{2} - 2abcos\theta}\]

and \(tan\alpha = \frac{bsin(\pi - \theta)}{a + bcos(\pi - \theta)} = \frac{bsin\theta}{a - bcos\theta}\)

(i) Aunit vector is vector of unit length used to specify direction.
(ii) A unit vector is a dimensionless vector with a magnitude of exactly 1 .
(iii) Unit vectors are used to specify a direction and have no other physical significance

A unit vector in direction of vector \(\overrightarrow{A}\) is represented as \(\widehat{A}\)
& \(\widehat{A} = \frac{\overrightarrow{A}}{|\overrightarrow{A}|}\)

or \(\overrightarrow{A}\) can be expressed in terms of a unit vector in its direction i.e. \(\overrightarrow{A} = |\overrightarrow{A}|\widehat{A}\)

Unit Vectors along three coordinates axes:

unit vector along \(x\)-axis is \(\widehat{i}\)
unit vector along \(y\)-axis is \(\widehat{j}\)
unit vector along \(z\)-axis is \(\widehat{k}\)

Co-ordinate systems are used to describe the position of a point in space.
Coordinate system consists of:
(i) A fixed reference point called the origin
(ii) Specific axes with scales and labels
(iii) Instructions on how to label a point relative to the origin and the axes

Cartesian Coordinate System :

Also called rectangular coordinate system
x - and y - axes intersect at the origin
Points are labeled ( \(x,y\) )
In figure, points \(P\) and \(Q\) are shown as \((5,3)\) and \(( - 3,4)\) respectively.

(components means parts)
We can move from tail of \(\overrightarrow{a}\) to its head via various paths. But, if we move with \(\overrightarrow{a}\) as the diameter of circle as shown, then the two vectors ( \(\overrightarrow{c}\&\overrightarrow{\text{ }d}\) and \({\overrightarrow{a}}_{x}\&{\overrightarrow{a}}_{y}\) ) would be perpendicular to each other. Such perpendicular vectors are called rectangular components of \(\overrightarrow{a}\). But, if

we choose \({\overrightarrow{a}}_{x}\) & \({\overrightarrow{a}}_{y}\). Then we can write them in terms of standard unit vectors \(\widehat{i}\&\widehat{j}\) respectively. Then we would say that \(a_{x}\) is the component of vector along the x -axis & \(a_{y}\) is the component of vector along y -axis. Now according to triangle law of addition:
\[\overrightarrow{a} = {\overrightarrow{a}}_{x} + {\overrightarrow{a}}_{y} \]So, we write \(\overrightarrow{a} = a_{x}(\widehat{i}) + a_{y}(\widehat{j}) = acos\theta(\widehat{i}) + asin\theta(\widehat{j})\)
This is a very convenient form of representing vectors.
\(acos\theta\) is known as component of \(\overrightarrow{a}\) along x -axis \(\left( a_{x} \right)\)

\(asin\theta\) is known as component of \(\overrightarrow{a}\) along \(y\)-axis ( \(a_{y}\) )
This is best form of representing vectors because we can do exact addition and subtraction using simple laws of algebra without needing to draw vectors

Results :

1. Unit vector along \(\overrightarrow{A}\) is \(\widehat{A}\).

Since, \(\overrightarrow{A} = |\overrightarrow{A}|\widehat{A}\)
& \(\overrightarrow{A} = |\overrightarrow{A}|(cos\theta\widehat{i} + sin\theta\widehat{j})\)
\[\widehat{A} = \widehat{i}cos\theta + \widehat{j}sin\theta \]2. If components of a vector along \(x\& y\)-axis are known, then that vector can be completely represented as
\[\overrightarrow{A} = A_{x}\widehat{i} + A_{y}\widehat{j} \]3. \(|\overrightarrow{A}| = \sqrt{A_{x}^{2} + A_{y}^{2}}\)
4. \(tan\theta = \left( \frac{A_{y}}{A_{x}} \right)\)

\(\theta\) is angle with x -axis
5. The components can be positive or negative and will have the same units as the original vector
6. The signs of the components will depend on the angle

Alternate

\[{{\overrightarrow{r}}_{i} = o\widehat{i} + o\widehat{j},{\overrightarrow{r}}_{f} = 8\widehat{i} - 6\widehat{j} }{\overrightarrow{d} = {\overrightarrow{r}}_{f} - {\overrightarrow{r}}_{i} = (8\widehat{i} - 6\widehat{j}) - (0\widehat{i} + 0\widehat{j}) }{\overrightarrow{d} = 8\widehat{i} - 6\widehat{j}}\]

Equilibrium means net force acting on body is zero.
i.e. \(\ \Sigma\overrightarrow{F} = \overrightarrow{0}\) (for translational equilibrium)
State of rest or moving with constant velocity are the condition of translational equilibrium.

Suppose that a particle displaces from position \({\overrightarrow{\mathbf{r}}}_{1}\) to \({\overrightarrow{\mathbf{r}}}_{2}\), then the particle's displacement is given by:

28
\[{{\overrightarrow{F}}_{3} = - \left( {\overrightarrow{F}}_{I} + {\overrightarrow{F}}_{2} \right) }{= - ( - 6\widehat{i} - 2\widehat{j}) }{{\overrightarrow{F}}_{3} = 6\widehat{i} + 2\widehat{j}N}\]

Subtraction of vectors (applications):

To find change in velocity.

A general way of locating a particle is with a position vector \(\overrightarrow{\mathbf{r}}\), which is a vector that extends from a reference point to the particle. This reference point is usually the origin.

In unit vector notation: \(\overrightarrow{r} = x\widehat{i} + y\widehat{j} + z\widehat{k}\)

\(\overrightarrow{s} = p.v\). of \({\overrightarrow{r}}_{2} - p.v\). of \({\overrightarrow{r}}_{1} = \left( x_{2}\widehat{i} + y_{2}\widehat{j} + z_{2}\widehat{k} \right) - \left( x_{1}\widehat{i} + y_{1}\widehat{j} + z_{1}\widehat{k} \right)\)
\[\therefore\overrightarrow{s} = (\Delta x\widehat{i} + \Delta y\widehat{j} + \Delta z\widehat{k})\]

There are two ways in which vectors can be multiplied :
(1) Scalar product or dot product.
(2) Vector product or cross product
(1) Scalar product
\(\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}|.|\overrightarrow{b}|cos\theta\), where \(\theta\) is angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\) It is called a scalar product because its product is a scalar quantity.

(i) \(\widehat{i} \cdot \widehat{i} = |\widehat{i}| \cdot |\widehat{i}|cos0^{\circ} = 1 = \widehat{j} \cdot \widehat{j} = \widehat{k} \cdot \widehat{k}\)
\[\widehat{i} \cdot \widehat{j} = |\widehat{i}| \cdot |\widehat{j}|cos90^{\circ} = 0 = \widehat{j} \cdot \widehat{k} = \widehat{k} \cdot \widehat{i} \](ii) \(\overrightarrow{a} \cdot \overrightarrow{b} = 0 \Rightarrow \overrightarrow{a}\bot\overrightarrow{b}\) : used to test orthogonality. (means two vectors are mutually perpendicular to each other).
(iii) If \(\overrightarrow{a} = a_{x}\widehat{i} + a_{y}\widehat{j} + a_{z}\widehat{k}\ \overrightarrow{b} = b_{x}\widehat{i} + b_{y}\widehat{j} + b_{z}\widehat{k}\)

\[\begin{matrix} & \overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}\|\overrightarrow{b}|cos\theta \\ & a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z} = \sqrt{a_{x}^{2} + a_{y}^{2} + a_{z}^{2}}\sqrt{b_{x}^{2} + b_{y}^{2} + b_{z}^{2}}cos\theta \end{matrix}\]

\(cos\theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}\|\overrightarrow{b}|}\) (this is used to find the angle between two vectors).
(iv) Work done by vector \(\overrightarrow{F}\) is defined as \(w = \overrightarrow{F} \cdot \overrightarrow{d}\)

Where \(\overrightarrow{d}\) displacement vector.

Find the vector component of (i) \(\overrightarrow{a}\) along \(\overrightarrow{b}\) (ii) \(\overrightarrow{a}\bot\) to \(\overrightarrow{b}\)

(i) Magnitude of Projection of \(\overrightarrow{a}\) on \(\overrightarrow{b} = |\overrightarrow{a}|cos\theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|} = \overrightarrow{a} \cdot \widehat{b}\)

Thus, vector Component of vector \(\overrightarrow{a}\) in the direction of \(\overrightarrow{b}\) is \((\overrightarrow{a}.\widehat{b})\widehat{b}\)
(ii) Vector component of vector \(\overrightarrow{a}\) in the perpendicular direction of \(\overrightarrow{b}\) is \((\overrightarrow{a} - (\overrightarrow{a},\widehat{b})\widehat{b})\).

The vector or cross product of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is defined as

\[\overrightarrow{a} \times \overrightarrow{b} = absin\theta\widehat{n}\]

The product is defined to be a vector of magnitude ab \(sin\theta\) that points in the direction of the unit vector \(\widehat{n}\) normal (perpendicular) to the plane of \(\overrightarrow{a}\) and \(\overrightarrow{b}\). The angle \(\theta\) is the smaller angle between the vector.

The direction of \(\widehat{n}\) is still ambiguous. This ambiguity is removed by using a convention called the righthand rule. Curl the fingers of your right hand and stick out your thumb as if you were hitch-hiking as in figure. The sense of rotation of the fingers should be from the first vector \(\overrightarrow{a}\) to the second vector \(\overrightarrow{b}\) through the smaller angle between them. The thumb indicates the direction \(\widehat{n}\).

Because of the right-hand rule, the order of the vectors in the cross product is important. The vector product is noncommutative :

\[\overrightarrow{b} \times \overrightarrow{a} = - \overrightarrow{a} \times \overrightarrow{b}\]

(i) Cross product non-commutative :

\[\overrightarrow{a} \times \overrightarrow{b} \neq \overrightarrow{b} \times \overrightarrow{a}\text{~}\text{i.e.}\text{~}\overrightarrow{b} \times \overrightarrow{a} = - \overrightarrow{a} \times \overrightarrow{b}\]

(ii) Follows distributive law:

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

(iii) If \(a\) and \(b\) are any vectors, and \(m\) is any real number (positive or negative) then \((m\overrightarrow{a}) \times \overrightarrow{b} = m(\overrightarrow{a} \times \overrightarrow{b}) = \overrightarrow{a} \times (m\overrightarrow{b})\)
(iv) Does not follow associative law :

\[\overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \neq (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}\]

(v) \(\widehat{i} \times \widehat{j} = \widehat{k},\widehat{j} \times \widehat{k} = \widehat{i},\widehat{k} \times \widehat{i} = \widehat{j}\)
and \(\widehat{i} \times \widehat{i} = 0,\widehat{j} \times \widehat{j} = 0,\widehat{k} \times \widehat{k} = 0\)
(vi) \(\overrightarrow{a} = a_{x}\widehat{i} + a_{y}\widehat{j} + a_{z}\widehat{k}\)

\[\begin{matrix} & \overrightarrow{b} = b_{x}\widehat{i} + b_{y}\widehat{j} + b_{z}\widehat{k} \\ & \overrightarrow{a} \times \overrightarrow{b} = \left( a_{x}\widehat{i} + a_{y}\widehat{j} + a_{z}\widehat{k} \right) \times \left( b_{x}\widehat{i} + b_{y}\widehat{j} + b_{z}\widehat{k} \right) \\ & \ = \left( a_{y}b_{z} - a_{z}b_{y} \right)\widehat{i} + \left( a_{z}b_{x} - a_{x}b_{z} \right)\widehat{j} + \left( a_{x}b_{y} - a_{y}b_{x} \right)\widehat{k} \end{matrix}\]

(vii) Angle between two vectors

\[\begin{matrix} & \ |\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}|sin\theta \\ & sin\theta = |\overrightarrow{a} \times \overrightarrow{b}|/|\overrightarrow{a}||\overrightarrow{b}| \end{matrix}\]

Let
\(\overrightarrow{a} = a_{1}\widehat{i} + a_{2}\widehat{j} + a_{3}\widehat{k}\) and \(\overrightarrow{b} = b_{1}\widehat{i} + b_{2}\widehat{j} + b_{3}\widehat{k}\)
\[{\overrightarrow{a} \times \overrightarrow{b} = \left( a_{2}{\text{ }b}_{3} - a_{3}{\text{ }b}_{2} \right)\widehat{i} - \left( a_{1}{\text{ }b}_{3} - a_{3}{\text{ }b}_{1} \right)\widehat{j} + \left( a_{1}{\text{ }b}_{2} - a_{2}{\text{ }b}_{1} \right)\widehat{k} }{\Rightarrow |\overrightarrow{a} \times \overrightarrow{b}| = \sqrt{\left( a_{2}b_{3} - a_{3}b_{2} \right)^{2} + \left( a_{1}b_{3} - a_{3}b_{1} \right)^{2} + \left( a_{1}b_{2} - a_{2}b_{1} \right)^{2}} }{|\overrightarrow{a}| = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}},|\overrightarrow{\text{ }b}| = \sqrt{b_{1}^{2} + b_{2}^{2} + b_{3}^{2}} }{sin\theta = \frac{\sqrt{\left( a_{2}b_{3} - a_{3}b_{2} \right)^{2} + \left( a_{3}b_{1} - a_{1}b_{3} \right)^{2} + \left( a_{1}b_{2} - a_{2}b_{1} \right)^{2}}}{\sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}\sqrt{b_{1}^{2} + b_{2}^{2} + b_{3}^{2}}} }\](ix) The cross product of a vector with itself is a NULL vector i.e.,

\[\overrightarrow{a} \times \overrightarrow{a} = |\overrightarrow{a}||\overrightarrow{a}|sin0^{\circ}\widehat{n} = \overrightarrow{0}\]

(x) The cross product of two vectors represents the area of the parallelogram formed by them, (Figure., shows a parallelogram \(PQRS\) whose adjacent sides \(\overrightarrow{PQ}\) and \(\overrightarrow{PS}\) are represented by vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) respectively.

Now, area of parallelogram \(= |\overrightarrow{PQ}| \cdot |\overrightarrow{SM}| = |\overrightarrow{PQ}| \cdot |\overrightarrow{PS}|sin\theta = |\overrightarrow{a}| \cdot |\overrightarrow{b}|sin\theta = |\overrightarrow{a} \times \overrightarrow{b}|\) hence cross product of two vectors represents the area of parallelogram formed by it. It is worth noting that area vector \(\overrightarrow{a} \times \overrightarrow{b}\) acts along the perpendicular to the plane of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

Unit Vector Perpendicular to two given vectors

Let be a unit vector perpendicular to two (non-zero) vectors \(a,b\) and positive for right handed rotation from a to b and \(\theta\) be the angle between the vectors \(a,b\) then

\[\begin{matrix} & \overrightarrow{a} \times \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}|sin\theta\widehat{n} \\ & \ |\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}|sin\theta \end{matrix}\]

Thus we get \(= \overrightarrow{a} \times \overrightarrow{b}/|\overrightarrow{a} \times \overrightarrow{b}| = \widehat{n}\).

Q. 1 Given that \(\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = \overrightarrow{0}\), but of three two are equal in magnitude and the magnitude of third vector is \(\sqrt{2}\) times that of either of the vectors two having equal magnitude. Then the angles between vectors are given by -
(A) \(30^{\circ},60^{\circ},90^{\circ}\)
(B) \(45^{\circ},45^{\circ},90^{\circ}\)
(C) \(45^{\circ},60^{\circ},90^{\circ}\)
(D) \(90^{\circ},135^{\circ},135^{\circ}\)

Sol. (D) From polygon law, there vectors having summation zero, should from a closed polygon (triangle). Since the two vectors are having same magnitude and the third vector is \(\sqrt{2}\) times that of either of two having equal magnitude. i.e. the triangle should be right angled triangle.

Angle between A and B is \(90^{\circ}\)
Angle between B and C is \(135^{\circ}\)
Angle between A and C is \(135^{\circ}\)
Q. 2 If a particle moves 5 m in +x -direction. Show the displacement of the particle-
(A) \(5\widehat{j}\)
(B) \(5\widehat{i}\)
(C) \(- 5\widehat{j}\)
(D) \(5\widehat{k}\)

Sol. \(\ \) Magnitude of vector \(= 5\)

\[\begin{matrix} & \text{~}\text{Unit vector in}\text{~} + x\text{~}\text{direction is}\text{~}\widehat{i} \\ & \text{~}\text{displacement}\text{~} = 5\widehat{i} \end{matrix}\]

Hence correct answer is (B).
Q. 3 A car travels 6 km towards north at an angle of \(45^{\circ}\) to the east then travels distnace of 4 km towards north at an angle of \(135^{\circ}\) to the east. How far is its final position due east and due north? How far is the point from the strating point? What angle does the straight line joining its initial and final position makes with the east? What is the total distnace travelled by the car?

Sol. Net movement along X - direction

\[\begin{matrix} & \ = (6 - 4)cos45^{\circ}\widehat{i} \\ & \ = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}\text{ }km \end{matrix}\]

Net movement along Y -direction

\[\begin{matrix} & \ = (6 + 4)sin45^{\circ}\widehat{j} \\ & \ = 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2}\text{ }km \end{matrix}\]

Net movement form starting point (Total distance travelled)

\[= 6 + 4 = 10\text{ }km\]

Angle which makes with the east direction

\[\begin{matrix} & tan\theta = \frac{Y - \text{~}\text{component}\text{~}}{X - \text{~}\text{component}\text{~}} \\ & \ = \frac{5\sqrt{2}}{\sqrt{2}} \\ & \theta = \tan^{- 1}(5) \end{matrix}\]

Q. 4 Abody is moving with uniform speed v on a horizontal circle in anticlockwise direction from A as shown in figure. What is the change in velocity in (a) half revolution (b) first quarter revolution.

Sol. Change in velocity in half revolution

\[\begin{matrix} & \Delta\overrightarrow{v} = {\overrightarrow{v}}_{c} - {\overrightarrow{v}}_{A} \\ & \ = v( - \widehat{j}) - v(\widehat{j}) \end{matrix}\]

\[\Delta\overrightarrow{v} = - 2v\widehat{j}\]

\(|\Delta\overrightarrow{v}| = 2v\) directed towards negative y -axis
change in velocity in first quarter revolution

\[\begin{matrix} & \Delta\overrightarrow{v} = {\overrightarrow{v}}_{B} - {\overrightarrow{v}}_{A} \\ & \ = v( - \widehat{i}) - v(\widehat{j}) \\ & \ = - v(\widehat{i} + \widehat{j}) \end{matrix}\]

\(|\Delta\overrightarrow{v}| = \sqrt{2}v\) and directed towards south-west.
Q. 5 The sum of the magnitudes of two forces acting at a point is 18 and the magnitude of their Resultant is 12. if the resultant is at \(90^{\circ}\) with the force of smaller magnitude, what are the magnitudes of forces?
(A) 12,5
(B) 14,4
(C) 5,13
(D) 10,8

Sol. Let P be the smaller force then it is given that

\[\begin{array}{r} P + Q = 18\#(1) \end{array}\]

\[\begin{array}{r} R = \sqrt{P^{2} + Q^{2} + 2PQcos\theta} = 12\#(2) \end{array}\]

\[Qsin\theta/P + Qcos\theta = tan\phi = tan90^{\circ} = \infty\]

\[\begin{array}{r} P + Qcos\theta = 0\#(3) \end{array}\]

Substituting the value of P

\[\begin{array}{r} Q(1 - cos\theta) = 18\#(4) \end{array}\]

and subtracting square of equation (2) from (1)

\[\begin{array}{r} 2PQ\lbrack 1 - cos\theta\rbrack = 18^{2} - 12^{2} = 180\#(5) \end{array}\]

Dividing equation (5) by (4)
\(2P = 10\) i.e. \(P = 5\), so \(Q = 13\)
So the magnitude of forces are (5 and 13)
Hence correct answer is (C)
Q. 6 Let \(\overrightarrow{a} = 2\widehat{i} + 3\widehat{j} - \widehat{k};\overrightarrow{b} = - \widehat{i} + 3\widehat{j} + 4\widehat{k}\). Evaluate
(i) \(|\overrightarrow{a}|;|\overrightarrow{b}|\)
(ii) \(\overrightarrow{a} \cdot \overrightarrow{b}\)
(iii) the angle between the vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\)
(iv) the projection of \(\overrightarrow{a}\) on \(\overrightarrow{b}\)
(v) the projection of \(\overrightarrow{b}\) on \(\overrightarrow{a}\)
(vi) area of the \(\bigtriangleup AOB\) where O is origin

Sol. Give \(\overrightarrow{a} = 2\widehat{i} + 3\widehat{j} - \widehat{k},\overrightarrow{b} = - \widehat{i} + 3\widehat{j} + 4\widehat{k}\)
(i) \(|\overrightarrow{a}| = \sqrt{2^{2} + 3^{2} + ( - 1)^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}\)

\[|\overrightarrow{b}| = \sqrt{( - 1)^{2} + 3^{2} + 4^{2}} = \sqrt{1 + 9 + 16} = \sqrt{26}\]

(ii) \(\overrightarrow{a} \cdot \overrightarrow{b} = 2( - 1) + 3 \times 3 + ( - 1)(4) = 3\)
(iii) The angle \(\theta\) between the vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is given by

\[cos\theta = \frac{\overrightarrow{a} \cdot \overrightarrow{\text{ }b}}{|\overrightarrow{a}\|\overrightarrow{b}|} = \frac{3}{\sqrt{14}\sqrt{26}} = \frac{3}{2\sqrt{91}}\]

(iv) the projection of \(\overrightarrow{a}\) on \(\overrightarrow{b} = |\overrightarrow{a}|cos\theta\)

\[\sqrt{14} \times \frac{3}{\sqrt{14}\sqrt{26}} = \frac{3}{\sqrt{26}}\]

(v) The projection of \(\overrightarrow{b}\) on \(\overrightarrow{a} = |\overrightarrow{b}|cos\theta\)

\[\sqrt{26} \times \frac{3}{\sqrt{14}\sqrt{26}} = \frac{3}{\sqrt{14}}\]

(vi) Area of \(\bigtriangleup AOB = \frac{1}{2}|\overrightarrow{a}\|\overrightarrow{b}|sin\theta\)

Now \(\sin^{2}\theta = 1 - \cos^{2}\theta = 1 - \left( \frac{3}{2\sqrt{91}} \right)^{2}\)

\[= 1 - \frac{9}{364} = \frac{355}{364}\]

Area of \(\bigtriangleup AOB = \frac{1}{2}\sqrt{14}\sqrt{26} \cdot \sqrt{\frac{355}{364}}\)

\[= \frac{\sqrt{355}}{2} = 9.42\text{~}\text{sq. unit approx.}\text{~}\]

Q. 7 The torque of force \(\overrightarrow{F} = - 3\widehat{i} + \widehat{j} + 5\widehat{k}\) acting at the point \(\overrightarrow{r} = 7\widehat{i} + 3\widehat{j} + \widehat{k}\) is \(\_\_\_\_\) ?
(A) \(14\widehat{i} - 38\widehat{j} + 16\widehat{k}\)
(B) \(4\widehat{i} + 4\widehat{j} + 6\widehat{k}\)
(C) \(- 21\widehat{i} + 4\widehat{j} + 4\widehat{k}\)
(D) \(- 14\widehat{i} + 34\widehat{j} - 16\widehat{k}\)

Sol. (A) The torque is defined as \(\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}\)
\[{\overrightarrow{r} \times \overrightarrow{F} = \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 7 & 3 & 1 \\ - 3 & 1 & 5 \end{matrix} \right| }{= \widehat{i}\left| \begin{matrix} 3 & 1 \\ 1 & 5 \end{matrix} \right| + \widehat{j}\left| \begin{matrix} 1 & 7 \\ 5 & - 3 \end{matrix} \right| + \widehat{k}\left| \begin{matrix} 7 & 3 \\ - 3 & 1 \end{matrix} \right| }{= \widehat{i}(15 - 1) + \widehat{j}( - 3 - 35) + \widehat{k}(7 - ( - 9)) }{= 14\widehat{i} - 38\widehat{j} + 16\widehat{k} }\]Thus the answer is (A)
Q. 8 The vectors from origin to the points \(A\) and \(B\) are \(\overrightarrow{a} = 3\widehat{i} - 6\widehat{j} + 2\widehat{k}\) and \(\overrightarrow{b} = 2\widehat{i} + \widehat{j} - 2\widehat{k}\) respectively. Find the area of:
(i) the triangle OAB
(ii) the parallelogram formed by OA and OB as adjacent sides.

Sol. Given \(\overrightarrow{OA} = \overrightarrow{a} = 3\widehat{i} - 6\widehat{j} + 2\widehat{k}\)
and \(\overrightarrow{OB} = \overrightarrow{b} = 2\widehat{i} + \widehat{j} - 2\widehat{k}\)
\[{\therefore\ (\overrightarrow{a} \times \overrightarrow{b}) = \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & - 6 & 2 \\ 2 & 1 & - 2 \end{matrix} \right| }{= (12 - 2)\widehat{i} - ( - 6 - 4)\widehat{j} - (3 + 12)\widehat{k} }{= 10\widehat{i} + 10\widehat{j} + 15\widehat{k} }{\Rightarrow |\overrightarrow{a} \times \overrightarrow{b}| = \sqrt{10^{2} + 10^{2} + 15^{2}} = \sqrt{425} = 5\sqrt{17} }\](i) Area of \(\bigtriangleup OAB = \frac{1}{2}|\overrightarrow{a} \times \overrightarrow{b}| = \frac{1}{2} \cdot 5\sqrt{17}\) sq. units
\(= \frac{5}{2}\sqrt{17}\) sq. units
(ii) Area of parallelogram fromed by OA and OB as
adjacent sides \(= |\overrightarrow{a} \times \overrightarrow{b}| = 5\sqrt{17}\) sq. units.
Q. 9 A buoy is attached to three tugboats by three ropes. The tugboats are engaged in a tug-of-war. One tugboat pulls west on the buoy with a force \({\overrightarrow{F}}_{1}\) of magnitude 1000 N . The second tugboat pulls south on the buoy with a force \({\overrightarrow{F}}_{2}\) of magnitude 2000 N . The third tugboat pulls northeast (that is, half way between north and east), with a force \({\overrightarrow{F}}_{3}\) of magnitude 2000 N .
(a) Draw a diagram of forces acting on the buoy to represent this situation.
(b) Express each force in unit vector form \((\widehat{i},\widehat{j})\).
(c) Calculate the magnitude of the resultant force.

Sol. (a)

(b)

\[\begin{matrix} & {\overrightarrow{F}}_{1} = - 1000\widehat{i}(\text{ }N), \\ & {\overrightarrow{F}}_{2} = - 2000\widehat{j}(\text{ }N), \\ & {\overrightarrow{F}}_{3} = 2000\left( cos45^{\circ}\widehat{i} + sin45^{\circ}\widehat{j} \right)(N) = 1000\sqrt{2}\widehat{i} + 1000\sqrt{2}\widehat{j} \end{matrix}\]

(c) \(\ {\overrightarrow{F}}_{\text{resultant}\text{~}} = {\overrightarrow{F}}_{1} + {\overrightarrow{F}}_{2} + {\overrightarrow{F}}_{3} = 1000(\sqrt{2} - 1)\widehat{i} - 1000(2 - \sqrt{2})\widehat{j}N\)

\[\begin{matrix} & F_{x} = 1000(\sqrt{2} - 1)N \\ & F_{y} = - 1000(2 - \sqrt{2})N \end{matrix}\]

\[|\overrightarrow{F}| = \sqrt{F_{x}^{2} + F_{y}^{2}} = 1000(\sqrt{2} - 1) \times \sqrt{}3 = 1000(\sqrt{2} - 1)\sqrt{3}\text{ }N\]