Vectors Class 12 Notes Maths Chapter 10 - CBSE

Chapter:10

What are Vectors ?

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    Vectors come in handy when we study those physical objects that have direction as well as magnitude. Let’s look at some of the terms associated with vectors :

    Scalar

    It is a quantity which has magnitude but no direction. Examples : Mass, length, distance, speed etc.

    Vector

    It is a quantity which has a magnitude as well as a specific direction. Examples : Weight, displacement, velocity etc. Denoted as

    $$\overrightarrow{AB}$$

    Magnitude

    The distance between initial and terminal points of a vector is called the magnitude of the vector.

    $$\text{Denoted as}\space|\overrightarrow{AB}| \space \text{which is read as modulus of}\space \overrightarrow{AB}.$$

    Direction Cosines

    Direction cosines of a vector are the cosines of the angle between the vector and the three coordinate axes.

    • As you see in the figure the angle α, β,γ made by the vector A with the positive directions of x,y and z-axis respectively, are called its direction angles. The cosine values of these angles ,i.e., cos α, cos β and cos γ are called direction cosines.
    • Denoted by l,m and n.
    DIRECTION COSINES

    Direction Ratios

    Numbers that are proportional to the direction cosines of the line are called direction ratios of the line. 

    • Ax = A cos α, Ay = A cos β and Az = A cos γ
    • Denoted as a, b and c.
    • Let (l, m, n) be the direction cosines of a line and direction ratios of the line be (a, b, c). Then

    $$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{a^2+b^2+c^2}}=\pm \space \frac{1}{\sqrt{a^2+b^2+c^2}}\qquad\\l=\pm\frac{a}{\sqrt{a^2+b^2+c^2}}(\because l^2+m^2+n^2=1)\\m=\pm\frac{b}{\sqrt{a^2+b^2+c^2}}\space\space\space\space n=\pm\frac{c}{\sqrt{a^2+b^2+c^2}}$$

    Vectors

    Equal Vectors

    Two vectors are said to be equal vectors if they have same magnitude and direction.

    $$\overrightarrow{AB}=\overrightarrow{CD}$$

    Unit Vectors

    $$•\space\hat {a} \text{ is a unit vector in the direction of}\space \vec{a} \space\text{and}\space \hat a =\frac{\vec{a}}{|\vec{a}|}\\ \text{• Unit vectors in the directions of the positive x-axis, y-axis and z-axis are i, j, k.}$$

    Zero Vectors

    If the magnitude of the vector is one , then it is called unit vector.

    If the magnitude of the vector is zero , then it is called null vector or zero vector. It can have any arbitrary direction.

    Parallel Vectors

    Vectors which have same direction or exactly the opposite direction are called parallel vectors . The angle between them is 0° or 180°. Parallel vectors
    are of two types :

    Like Parallel Vectors

    If the angle between the vectors is 0°, then they are called like parallel vectors.

    Unlike Parallel Vectors

    If the angle between the vectors is 180°, then they are called unlike parallel vectors.

    Collinear Vectors

    Two or more vectors are said to be collinear vectors if they are parallel in same line irrespective of their magnitudes.

    Position Vector Of A Point

    The vector OA is said to be a position vector of A with respect to Origin O, if A be any terminal point and O is the origin which is fixed.

    POSITION VECTOR OF A POINT

    Negative Of A Vector

    Two vectors are called negative vectors of each other if they have same magnitude but opposite direction. Denoted as

    Components Of A Vector

    $$-\vec{a}$$

    $$\overrightarrow{\text{OA}},\overrightarrow{\text{OB}}\space\text{AND}\space \overrightarrow{\text{OC}}\text{are the unit vectors along the axes OX, OY, and OZ respectively,}\text{and denoted by}\space \hat{i}, \hat{j}, \hat{k}.$$

    components

    $$\text{Now, consider the position vector}\space\overrightarrow{\text{OP}}\text{of a point P (x,y,z) as in fig}.\\\text{Let}\space \text{P}_1\space \text{be the foot of the perpendicular from P on the plane XOY. We, thus, See that}\\\text{P}_1 \text{P is parallel to z-}\hat{axis}.\hat{}\space \hat{\text{As}}\text{i, j, k are the unit vectors along the x,y and z-axes respectively, and by the definition of the coordinates of P,}\text{we have}\overrightarrow{P_1P}=\overrightarrow{OR}=Z\hat{k}. \space\text{Similarly,}\\\overrightarrow{QP}_1=\overrightarrow{OS}=y\hat{j}\space\text{and}\space\overrightarrow{OQ}=x\hat{i}.$$

    component1

    $$\text{Therefore, it follows that}\space\overrightarrow{\text{OP}_1}={\overrightarrow{\text{OQ}}}+{\overrightarrow{\text{OP}}_1}=x\hat{i}+y\hat{j}\\\text{And}\overrightarrow{\text\space{OP}}=\overrightarrow{\text{OP}}_1+\overrightarrow{\text{P}_1P}=x\hat{i}+y\hat{j}+z\hat{k}\\\text{Hence , the position vector of P with reference to O is given by}\\\overrightarrow{\text{OP}}=r=x\hat{i}+y\hat{j}+z\hat{k}\\\text{This form of any vector is called its component form. Here x,y,z are called as the scalar components of }\text{are called the vector components of r along the respective axes.}$$

    Addition Of Vectors

    Triangle Law

    $$\overrightarrow{\text{AC}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\implies \vec{c}=\vec{a}+\vec{b}\\•\space \vec{\text{c}}\space \text{is called the resultant of\space }\vec{\text{a}}\space\text{and}\space\vec{\text{b}}$$

    The addition of vectors is achieved by ‘tail to nose’ placing of the directed line segments in a triangle.

    Triangle Law

    Parallelogram Law

    The result of adding two co-initial vectors is the vector represented by the diagonal of the parallelogram formed with the component vectors as adjacent sides. eg: in the fig. below,

    Parallelogram Law

    Multiplication Of A Vector By A Scalar

    $$\text{\text{The product of}}\space \text{by a scalar ‘m’ is denoted as m.}\vec{a}\space \text{such that :}\\\text{• The support of ma is same or parallel to that of }\vec{a},\\ • |\text{m}\vec{\text{a}}|=|\text{m}||\vec{\text{a}}|\\ \text{• m}\vec{\text{a}}\space \text{has the direction same or opposite to that of a, according as m is +ve or -ve.}$$

    Position Vector Of A Point Dividing A Line Segment In A Given Ratio

    For Internal Division

    $$\text{\text{Let}}\space\vec{\text{a}}\space \text{and}\space \vec{\text{b}}\space \text{be the position vectors of two points A and B respectively and let C be a point with position vector}\space\vec{\text{c}}\space \text{dividing AB internally in the ratio m : n. Then the position vector of C is}\\\vec{\text{c}}=\overrightarrow{\text{OC}}=\frac{\text{m}\vec{\text{b}}+\text{n}\vec{\text{a}}}{\text{m+n}}$$

    For External Division

    $$\text{\text{Let}}\space\vec{\text{a}}\space \text{and}\space \vec{\text{b}}\space \text{be the position vectors of two points A and B respectively and let C be the point with position vector}\space \vec{\text{c}} \text{dividing AB externally in the ratio m : n. Then the position vector of C is}\\ \vec{\text{c}}=\overrightarrow{\text{OC}}=\frac{\text{m}\vec{\text{b}}-\text{n}\vec{\text{a}}}{\text{m+n}}$$

    Properties And Application Of Scalar(Dot) Product Of Vectors

    $$\text{(i)}\space \vec{a}.\vec{a}=|\vec{a}|^2=a^2\\\qquad\text{(ii)}\space \vec{a}.\vec{b}=\vec{b}.\vec{a} \space \text{(commutativity)}\\\qquad\text{(iii)} \space \vec{a}.\vec{0}=0\\\qquad\text{(iv)}\space \vec{a}.(\vec{b}+\vec{c})=\vec{a}.\vec{b}+\vec{a}.\vec{c}\space\text{(Distributivity)}\\\qquad\text{(v)} \space\vec{a}.(\vec{b}-\vec{c})=\vec{a}.\vec{b}-\vec{a}.\vec{c}\\\qquad\text{(vi)}\space \vec{a}.\vec{b}=|\vec{a}|.|\vec{b}|.\text{cos} \space \theta\\\qquad \text{(vii) If x is any scalar, then}\space (x\vec{a}).\vec{b}=x(\vec{a}.\vec{b})=\vec{a}.(x\vec{b})\\\qquad\text{(viii)}\space(x\vec{a}).(y\vec{b})=\text{xy}(\vec{a}.\vec{b})\text{where x, y are scalars.}\\\qquad\text{(ix)}\space (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=|\vec{a}|^2-|\vec{b}|^2\\\qquad\text{(x)}\space \text{(a + b)}^2=|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}\\\qquad\text{(xi)}\space (\vec{\text{a}}-\vec{\text{b}})^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}.\vec{b}\\\qquad\text{(xii) If}\space \hat i, \hat j, \hat k\space\text{are three-unit vectors along three mutually perpendicular lines, then}\\ \qquad •\space\space \hat i .\hat i=\hat j.\hat j=\hat k.\hat k=1\\•\space \hat{i} .\hat{j} = \hat{j} . \hat{k} = \hat{k} . \hat{i} = \hat{j} . \hat{i} = \hat{k} . \hat{j} = \hat{i} . \hat{k} = 0.\\\qquad\text{(xiii) If OX, OY, OZ are three mutually perpendicular axes and}\space\hat{i},\hat{j},\hat{k}\space \text{are unit vectors along these axes. If co-ordinates of any point P, are (x, y, z) then its position vector is}\\\text{x}\hat{i}+\text{y}\hat{j}+\text{z}\hat{k}.$$

    Properties Of Vector (Cross) Product Of Vectors

    $$\text{(i)}\space \vec{a}×\vec{b}=-(\vec{b}×\vec{a})\\\qquad\text{(ii)}\space\vec{a}×\vec{a}=\vec{0}\\\qquad\text{(iii)}\space x(\vec{a}×\vec{b})=(x\vec{a})×\vec{b}=\vec{a}×(x\vec{b})\\\qquad\text{(iv)}\space(x\vec{a})×(y\vec{b})=\text{xy}(\vec{a}×\vec{b}).\\\qquad\text{(v)}\space|\vec{a}×\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}.\vec{b})^2.\\\qquad\text{(vi)}\space \hat{i}×\hat{i}=\hat{j}×\hat{j}=\hat{k}×\hat{k}=0.\\\qquad\text{(vii)}\space \hat{i}×\hat{j}=\hat{k},\hat{j}×\hat{k}=\hat{i},\hat{k}×\hat{i}=\hat{j}\\\qquad\text{(viii) Two non-zero vectors are collinear if their cross product is equal to zero vector.}\\\qquad\text{(ix)} \space\text{If} \space \vec{a}_1\hat{i}+\vec{a_2}\hat{j}+\vec{a_3}\hat{k}\\\text{And} \space\vec{b}=\vec{b}_1\hat{i}+\vec{b_2}\hat{j}+\vec{b_3}\hat{k}\\\text{Then}\space\vec{a}×\vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}=(a_2b_3-a_3b_2)\hat{i}-(a_1b_3-a_3b_1)\hat{j}+(a_1b_2-a_2b_1)\hat{k}\\\qquad\text{(x) If a, b, c are three vectors, then}\\\vec{a}×(\vec{b}+\vec{c})=\vec{a}×\vec{b}+\vec{a}×\vec{c}\\(\vec{b}+\vec{c})×\vec{a}=\vec{b}×\vec{a}+\vec{c}×\vec{a}$$

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