Class 10 Optional Mathematics Note

Vector

Cartesian form of vector in two dimensions

Vectors i and j are vectors of length 1 in the directions OX and OY respectively.

The vectorOR⃗is xi. The vectorOS⃗is yj.. The vectorOP⃗is the sum ofOR⃗and OS⃗, that is

 

OP⃗=xi+ yj

 The modulus of the vector is positive number that measures the length of the vector in terms of components

OP ⃗= $\sqrt {{x^2} + {y^2}} $

 

 

Addition of vectors

if two vectors such as AB and BC are representing the two sides of a triangle ABC, then the third side AC closing the other side of the triangle in opposite direction represents the sum of two vectors both in magnitude and vectors.

AC⃗= AB⃗+ BC⃗

 

 

Polygon law of vector addition

For the addition of more than three vectors polygon law of vectors is used.

Figure:

R⃗ =A⃗+ B⃗+ C⃗+ D⃗

 The resultant is the vector sum of two or more vectors. It is the result of adding two or more vectors together. If displacement vectors A, B, C,D are added together, the result will be vector R as shown in the diagram, vector R can be determined.

Scalar product of two vectors

Let a⃗ = ( a1,a2) andb⃗ = ( b1 , b2) be any two vectors thenthe scalar product of two vectors ordot product of two vectors is given by

a⃗ . b⃗ = a1.b1 + a2 b2

It can be also written as

a⃗ .b⃗ = |a⃗| |b⃗|cosθ = abcosθ

 

Geometrical interpretation of Scalar product of two vectors

 

 

Here, OA ⃗ = a⃗ , OB⃗= b⃗ and θ is the angle between

a⃗. b⃗ = |a⃗| |b⃗|cosθ = abcosθ=OA OB cosθ= OA*OE= magnitude of a⃗ × projection of b⃗ on a⃗

 

Condition of perpendicularity

a⃗. b⃗ = a1.b1 + a2 b2+ a3b3=abcos 90° = 0

 

Angle between the vectors

If a⃗ = ( a1,a2) andb⃗ = ( b1 , b2) be any two vectors then,

cosθ=$\frac{{{a_1}.{b_1}\:  + \: {a_2}\: {b_2}\: }}{{ab}}$=$\frac{{{\rm{\vec a}}.{\rm{\: \vec b}}}}{{ab}}$

 

If $\overrightarrow {{\rm{a }}} $ = x1${\rm{\veci}}$ + y1${\rm{\vec j}}$ and $\overrightarrow {{\rm{b }}} $ = x2${\rm{\veci}}$ + y2${\rm{\vec j}}$ then, $\overrightarrow {{\rm{a }}} .\overrightarrow {{\rm{b }}} $ = x1.x2 + y1.y2. where, $\overrightarrow {{\rm{i }}} .\overrightarrow {{\rm{i }}} $ =$\overrightarrow {{\rm{ j }}} .\overrightarrow {{\rm{j }}} $ = 1 and $\overrightarrow {{\rm{i }}} {\rm{ }}.\overrightarrow {{\rm{j }}} $ = $\overrightarrow {{\rm{j }}} .\overrightarrow {{\rm{i }}} $ = 0

-|$\overrightarrow {{\rm{a }}} $| = $\sqrt {{\rm{x}}_1^2 + {\rm{y}}_1^2} $ and |$\overrightarrow {{\rm{b }}} $|= $\sqrt {{\rm{x}}_2^2 + {\rm{y}}_2^2} $ ${\rm{ }}$ and cos? = $\frac{{{{\rm{x}}_{1.}}{{\rm{x}}_2} + {{\rm{y}}_1}.{{\rm{y}}_2}}}{{\left| {\overrightarrow {{\rm{a }}} } \right|\left| {\overrightarrow {{\rm{b }}} } \right|}}{\rm{ }}$

 

 

Vector Geometry

 

If the vector of A is $\overrightarrow {{\rm{a }}} $ and the vector of B is $\overrightarrow {{\rm{b }}} $. Then the mid-point of AB is M so, $\overrightarrow {{\rm{m }}} $= $\frac{{\overrightarrow {{\rm{a }}} + \overrightarrow {{\rm{b }}} }}{2}$

If P divides line AB internally in ratio m:n then $\overrightarrow {{\rm{p }}} = \frac{{{\rm{n}}\overrightarrow {{\rm{a }}} + {\rm{m}}\overrightarrow {{\rm{b }}} }}{{{\rm{m}} + {\rm{n}}}}$ and when it divides the line externally $\overrightarrow {{\rm{p }}} = \frac{{{\rm{m}}\overrightarrow {{\rm{b }}} - {\rm{n}}\overrightarrow {{\rm{a }}} }}{{{\rm{m}} - {\rm{n}}}}$

 

Position vector of the centroid of the given triangle is given byr⃗$\frac{{\vec a + \: \vec b + \: \vec c}}{3}$

 

Example 1

 Find the value ofKif p ⃗= k i⃗+3 j⃗= x1i⃗ + y1 j⃗, q⃗ = 9 i⃗+ 27 j⃗=x2i⃗ + y2 j⃗

 

p⃗= k i⃗+3 j⃗= x1i⃗ + y1 j⃗ = (k, 3 )

q⃗ = 9 i⃗+ 27 j⃗=x2i⃗ + y2 j⃗ = (9, -27)

if the p⃗ and q⃗parallelto each other p⃗ = a q⃗

$\left( {\begin{array}{*{20}{c}}k\\3\end{array}} \right) = a$$\left( {\begin{array}{*{20}{c}}9\\{27}\end{array}} \right)$

Equating the corresponding term

 3 =27 a

∴ a = $\frac{1}{9}$

Andk = 9 a

 Or, k = 1

 

Example 2

 

Find the position vector of the centroid of the triangle ABC

Letp⃗be the position vector of the centroid of the triangle and a⃗, b⃗ and c⃗ bet the position vector of the A, B and C respectively.

a⃗ = $\left( {\begin{array}{*{20}{c}}{ - 1}\\0\end{array}} \right)$b⃗ =$\left({\begin{array}{*{20}{c}}2\\3\end{array}} \right)$c⃗=$\left( {\begin{array}{*{20}{c}}{ - 2}\\{ - 7}\end{array}} \right)$

 Then,

p⃗= $\frac{1}{3}\: \left( {\vec a + \: \vec b + \: \vec c} \right)$

= $\frac{1}{3}\: $[$\left( {\begin{array}{*{20}{c}}{ - 1}\\0\end{array}} \right)$+$\left({\begin{array}{*{20}{c}}2\\3\end{array}} \right)$+ $\left( {\begin{array}{*{20}{c}}{ - 2}\\{ - 7}\end{array}} \right)$

 =$\frac{1}{3}\left( {\begin{array}{*{20}{c}}{ - 1 + 2 - 2\: }\\{0 + 3 - 7\: }\end{array}} \right)$

 = $\frac{1}{3}\: \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 4}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{ - \frac{1}{3}}\\{ - \frac{4}{3}}\end{array}} \right)$= $ - \frac{1}{3}i + \:  - \frac{4}{3}j$

 

 


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