Learn examples of scalar and vector quantities as a physics quantities, concept of unit and displacement vector. Practice analytical method of vector addition using triangle law.
1. Scalars
2. Vectors
For example:-
(1) What is your height?
(2) What is your mass?
(3) How many apples are there in the bag?
Examples:-
(1) For complete specification of a force, we need to know the magnitude as well as the direction of the force.
(2) For complete description of a point in space, both its distance and its direction from a fixed point (reference point) are needed.
Note:-
Assigning a direction to a scalar quantity will not convert it into a vector quantity.
e.g. Current
A 20 m towards east
B 10 m towards right
C Position of particle from origin
D Your height
(1) Magnitude
(2) Direction
e.g. : Let A is a vector quantity
\(\underbrace{20\,m}_{magnitude}\,\,\,\,\, \underbrace{\text{towards east}}_{Unit \,\,vector}\)
Vector = magnitude × unit vector
\(\hat A\) means towards \(\vec A\)
\(\hat B\) means towards \(\vec B\)
Hence,
\(\underbrace{\vec A}_{Vector} =\,\,\,\underbrace{|\vec A|}_{Magnitude}\,\, \underbrace{\hat A}_{direction} \,\,\text{or}\,\, A.\hat A\)
\(\text {towards (+) }x- axis \,\,\,\,\hat i\)
\(\text {towards (+) }y- axis \,\,\,\,\hat j\)
\(\text {towards (+) }z- axis \,\,\,\,\hat k\)
A East to West
B West to East
C North to South
D South to North
\(\vec{PQ} = \vec {\Delta r} = \text {Displacement vector}\)
A \(\sqrt{13}\;km,\, tan^{-1}(1.5)\) south of west
B \(5\,km,\, tan^{-1}(1.5) \) south of west
C \(2\,km,\, tan^{-1}(2) \) south of west
D \(6\,km,\, tan^{-1}(3) \) south of west
Displacement of man \(= (4\,\hat i+3 \, \hat j) \,m\)
Here (+) sign doesn't mean addition; rather it represents continuity. We can interpret it as: -
\(4\,m \,\hat i \text{(towards }x- axis) + 3\,m\,\hat j \,\,(\text {towards } y- axis)\)
Graphically,
We can say that \(\vec{OA} = (4\,\hat i) \,m\)
\(\vec{AB} = (3\,\hat j) \,m\)
Hence, displacement = \(\vec{OA} +\vec{AB}= (4\,\hat i+3\,\hat j) \,m\)
A \((\hat i - \hat j)\,m\)
B \((7\,\hat i - 2\,\hat j)\,m\)
C \((2\,\hat i + \,\hat j)\,m\)
D \((-2\,\hat i +\,\hat j)\,m\)
\(\text{ Result of Robin = Result of Sara}\\ \\\\\,\,\text{A to B = A to C then C to B} \\ \vec{AB} = \vec {AC} + \vec {CB}\)
A to C then C to B is equivalent to A to B
\(\vec{AC} + \vec {CB} = \vec {AB}\)
Here, B is turning point.
Similarly,
\(\vec {AC} + \vec {CD} = \vec {AD}\)
This is known as triangle law of vector addition.
Step 1 : Observe the arrows carefully.
Step 2 : Check which two sides are continuous or in same order.
Step 3 : Select the side which is in opposite order.
Step 4 : Selected side is the resultant of other two sides taken in same order.
Note:
Using triangle rule, we can conclude that: -
\(\vec{AB} =\vec{OB} -\vec{OA}\)
\(\vec{AB}=\) Position vector of B – Position vector of A
A \(\vec {AB} =\vec {BC} +\vec {BA}\)
B \(\vec {AC} =\vec {AB} +\vec {BC}\)
C \(\vec {BA} =\vec {BC} +\vec {AC}\)
D \(\vec {BC} =\vec {BA} +\vec {AC}\)
\(\vec{PQ} = \vec {\Delta r} = \text {Displacement vector}\)
\(\vec{PQ} = \vec{OQ}-\vec{OP}\)
Here, \(\Delta \vec r + \vec r_P = \vec r_Q\)
where \(\vec r_P \,\,\,and \,\,\, \vec r_Q\) are two vectors
\(\Delta r \to\) displacement vector
\(\vec {PQ} = \vec r _Q - \vec r _P\)
\(\vec {PQ} = \) Position vector of Q – Position vector of P
A \(4\,\hat i - 3\,\hat j\)
B \(5\,\hat i - 6\,\hat j\)
C \(\,\hat i+ 3\,\hat j\)
D \(4\,\hat i - 4\,\hat j\)
Let's consider three points in space i.e. O, A and B.
\(x \) component of AB = Q \(cos\,\theta\)
\(y\) component of AB = Q \(sin\,\theta\)
where
\(R \,cos \,\alpha = P+Q \,cos \,\theta\)
\(R \,sin \,\alpha = Q \,sin \,\theta\)
Calculating resultant vector |R|
\(|R| = \sqrt{(R\,cos\,\alpha)^2 +(R\,sin\,\alpha)^2 }\)
\(= \sqrt{(P+Q\,cos\,\theta)^2 + (Q\,sin \,\theta)^2}\)
calculating angle \(\alpha\) which is form in between \(\vec P \,\,\,\text {and}\,\,\,\vec R\)
\(tan\,\alpha =\dfrac{Q\,sin\,\theta}{P+Q\,cos\,\theta}\)
\(\alpha = tan^{-1} \left(\dfrac{Q\,sin\,\theta}{P+Q\,cos\,\theta}\right)\)
A 30 km /hr, \(tan^{-1} \,2/3\) with the horizontal
B 40 km /hr, \(tan^{-1} \,5/3\) with the horizontal
C 20 km /hr, \(tan^{-1} \,1/3\) with the vertical
D 50 km /hr, \(tan^{-1} \,4/3\) with the horizontal