Category : 12th Class
Vector Algebra
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On this basis, all physical quantities are divided into two groups:
(i) Scalar quantities.
(ii) Vector quantities.
where \[\hat{i}\],\[\hat{j}\] and \[\hat{k}\]are said to be the unit vectors.
If \[P(x,y,z)\] be any point in space then position vector \[\overrightarrow{OP}=x\vec{i}+y\vec{j}+z\vec{k}\]
\[\therefore \,\,|\,\,\overrightarrow{OP}\,\,|\]\[=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\]
Internal Division: Let A and B be the two points with the position vectors \[\vec{a}\] & \[\vec{b}\] respectively and let P be any point dividing AB internally in the ratio m : n. Then the position vector of P can be written
as \[\overrightarrow{OP}=\frac{m.\vec{b}+n\vec{a}}{m+n}\]
When P divides AB externally, then the position vector of P is given by,
\[\overrightarrow{OP}=\frac{m.\vec{b}-n\vec{a}}{m-n}\]
If P is the midpoint of AB, then position vector of P is given by, \[\overrightarrow{OP}=\frac{\vec{b}+\vec{a}}{2}\]
Let \[\overrightarrow{OA}=\vec{a}\] and \[\overrightarrow{AB}=\vec{b}\] be the two vectors.
Two non-zero vectors can be added geometrically by placing the initial point of \[\overrightarrow{AB}\]at the terminal point of \[\overrightarrow{OA}\] and then finding the vector that has the same initial point as \[\overrightarrow{OA}\] and the same terminal point \[\overrightarrow{AB}\],as shown in the figure.
\[\therefore \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\]
It is known as triangle law of addition.
\[\therefore \vec{a}+\vec{b}=\vec{b}+\vec{a}\]
For any three vectors \[\vec{a},\vec{b}\]and \[\vec{c}\], we have \[\vec{a}+(\vec{b}+\vec{c})\]\[=(\vec{a}+\vec{b})+\vec{c}\]
For every vector \[\vec{a}\], we have \[\vec{0}+\vec{a}=\vec{a}+\vec{0}=\vec{a}\]
Here \[\vec{0}\] is said to be null or zero vector.
For a given vector \[\vec{a}\], there exists a negative vector \[-\vec{a}\], such that
\[\vec{a}+(-\vec{a})=0\]
The vector \[-\vec{a}\] is said to be the additive inverse of\[\vec{a}\].
Here \[\overrightarrow{AO}\] is the component of \[\overrightarrow{OP}\] along x -axis.
Definition: A vector \[\vec{\ell }\]is said to be the linear combination of vectors \[\overrightarrow{a},\vec{b},\vec{c}\]...etc., if there exists scalar quantities x, y, z, ... etc. such that \[\vec{r}=x\vec{a}+y\vec{b}+z\vec{c}+\]…
Sol. 1st Method: Let the given points be A, B and C related to any origin O. Now if we can find the three scalars\[\ell \], m and n such that\[\ell .\overrightarrow{OA}+m.\overrightarrow{OB}+n.\overrightarrow{OC}=0\], provided that \[\ell +m+n=0\]then A, B, C will be collinear
Now, \[\ell .\overrightarrow{OA}+m.\overrightarrow{OB}+n.\overrightarrow{OC}=0\]
\[=\ell \,(\vec{a}-2\vec{b}+3\vec{c})+m(2\vec{a}+3\vec{b}-4\vec{c})\]
\[+n(-7\vec{b}+10\vec{c})=0\]
\[=(\ell +2m)\vec{a}+(-2\ell +3m-7n)\vec{b}\]
\[+(3\ell -4m+10n)\vec{c}=0\]
Equating the coefficient, we get
\[\ell +2m=0\] ........ (i)
\[-2\ell +3m-7n=0\] ........ (ii)
and \[3\ell -4m+10n=0\] ........ (iii)
On solving equations, (i), (ii) and (iii), we get
\[\ell =2,\]\[m=-1\]and\[n=-1\]
\[\because \] \[\ell +m+n=2-1-1=0\]
Hence the given points are collinear.
2nd Method:
Let the given points be A, B, C respectively, we have \[\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}\]
\[=(-7\vec{b}+10\vec{c})-(\vec{a}-2\vec{b}+3\vec{c})=\overrightarrow{-a}-5\vec{b}+7\vec{c}\]
\[\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}=(\vec{a}+2\vec{b}+3\vec{c})-(2\vec{a}+3\vec{b}-4\vec{c})=\overrightarrow{-a}-5\vec{b}+7\vec{c}\]
\[\overrightarrow{BA}\] \[=-\vec{a}-5\vec{b}+7\vec{c}\]
Hence \[\overrightarrow{AC}=\overrightarrow{BA}\]
\[\therefore \] Points A, B and C are collinear.
Sol. \[\because \]\[(\vec{x}-\vec{a}).(\vec{x}+\vec{a})=12\]
\[\Rightarrow \vec{x}.\vec{x}-\vec{a}.\vec{x}+\vec{x}.\vec{a}-\vec{a}.\vec{a}=12\]
\[\Rightarrow \,\,|\vec{x}{{|}^{2}}-1=12\]
\[\Rightarrow \,\,|\vec{x}{{|}^{2}}=12+1=13\]
\[\Rightarrow \,\,|\vec{x}|=\sqrt{13}\]
Sol. \[\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=2\hat{i}-\hat{j}+\hat{k}-(3\hat{i}-4\hat{j}-4\hat{k})=-\hat{i}+3\hat{j}+5\hat{k}\]
\[\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\hat{i}-3\hat{j}+5\hat{k}-(2\hat{i}-\hat{j}-\hat{k})=-\hat{i}+2\hat{j}+6\hat{k}\]
\[\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=3\hat{i}-4\hat{j}+4\hat{k}-(\hat{i}-3\hat{j}-5\hat{k})=2\hat{i}-\hat{j}+\hat{k}\]
Now \[|\overrightarrow{BC}|\,\,=\,\,\sqrt{{{(-1)}^{2}}+{{(-2)}^{2}}+{{(-6)}^{2}}}={{(41)}^{2}}\]
\[|\overrightarrow{AB}|\,\,=\,\,\sqrt{{{1}^{2}}+{{3}^{2}}+{{5}^{2}}}\]\[=\sqrt{35}\]and\[|\overrightarrow{CA}|\,\,=\,\,\sqrt{{{2}^{2}}+{{1}^{2}}+{{1}^{2}}}=\sqrt{6}\]
Thus, \[|\overrightarrow{BC}{{|}^{2}}\,=\]\[|\overrightarrow{AB}{{|}^{2}}+|\overrightarrow{CA}{{|}^{2}}=35+6=41\]
Hence, the point A, B and C are the vertices of a right angled triangle.
(i) If the vectors \[\vec{a}\] and \[\vec{b}\] are perpendicular then:
\[\cos \theta =0\]and hence \[\vec{a}\,.\,\vec{b}=0\]
(ii) Unit vectors: Let i,] and k be unit vectors
Then \[\hat{i}\,.\,\,\hat{i}={{i}^{2}}=1\]
\[j.\hat{j}={{j}^{2}}=1\]
\[\hat{k}.\hat{k}={{k}^{2}}=1\]
Scalar product of two like unit vectors is 1 whereas the scalar product of two unlike unit vector is zero i.e.
\[\hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}=0.\]
(iii) Scalar product is \[\vec{a}.\vec{b}=\vec{b}.\vec{a}\] commutative i.e.
(iv) \[m\vec{a}.n\vec{b}=mn(\vec{a}.\vec{b})\]
(v) \[\vec{a}.(\vec{b}+\vec{c})=\vec{a}\,.\,\vec{b}+\vec{a}\,.\,\vec{c}\] (distributive law)
(vi) Let \[\vec{a}={{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k}\]\[\vec{b}={{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k}\]
Then \[\vec{a}.\vec{b}={{a}_{1}}.{{a}_{2}}+{{b}_{1}}.{{b}_{2}}+{{c}_{1}}.{{c}_{2}}\]
(vii) Rules of sign:
\[\vec{a}.(-\vec{b})=-\vec{a}.\,\vec{b}\]
\[(-\vec{a}).\vec{b}=-\vec{a}.\,\vec{b}\]
\[(-\vec{a}).(-\vec{b})=\vec{a}.\,\vec{b}\]
(viii) Geometrical Interpretation: The scalar product (dot product) \[\vec{a}\,.\,\vec{b}\] may be defined as the product of \[|\vec{a}|\] and the projection of \[\vec{b}\] on \[\vec{a}\] or product of \[\vec{b}\]and the projection of \[\vec{a}\]on\[\vec{b}\].
Solved Example
Sol. \[\because \]\[\vec{a}\,.\,\vec{b}=(2\hat{i}+2\hat{j}-\hat{k}).(6\hat{i}-3\hat{j}+2\hat{k})=2\times 6+2\times (-3)+(-1).2\]
\[=12-6-2=4\]
\[\because \]We know that \[\vec{a}\,.\,\vec{b}=\,\,|\vec{a}|.|\vec{b}|\cos \theta \]
Now, \[|\vec{a}|\,\,=\sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}}=\sqrt{9}=3\]
\[|\vec{b}|\,\,=\sqrt{{{6}^{2}}+{{3}^{2}}+{{2}^{2}}}=\sqrt{49}=7\]
\[\therefore \vec{a}\,.\,\vec{b}=\,\,|\vec{a}||\vec{b}|cos\theta \]
\[\cos \theta =\frac{\vec{a}\,.\,\vec{b}}{|\vec{a}||\overrightarrow{b|}}=\frac{4}{3\times 7}=\frac{4}{21}\]
Thus if \[\hat{n}\] be the unit vector normal to the plane of \[\vec{a}\] and \[\vec{b}\] then,
\[\vec{a}\times \vec{b}=\,\,|\vec{a}|.|\vec{b}|.\sin \theta .\hat{n}\]
Corollary: \[\vec{a}\times \vec{b}=0\]
(i) The vector product does not satisfy the commutative law i.e. \[\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}\]
(ii) Rules of sign:
\[(-\vec{a})\times \vec{b}=-\vec{a}\times \vec{b}\]
\[\vec{a}\times (-\vec{b})=-\vec{a}\times \vec{b}\]
\[(-\vec{a})\times (-\vec{b})=\vec{a}\times \vec{b}\]
(iii) Distributive law:
\[\vec{a}\times (\vec{b}+\vec{c})=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}\]
(iv) Unit vectors:
\[\vec{i}\times \vec{i}=\vec{j}\times \vec{j}=\vec{k}\times \vec{k}=0\]
\[\vec{i}\times \vec{j}=\vec{k}\]\[\Rightarrow \vec{j}\times \vec{i}=-\vec{k}\]
\[\vec{j}\times \vec{k}=\vec{i}\]\[\Rightarrow \vec{k}\times \vec{j}=-\vec{i}\]
\[\vec{k}\times \vec{i}=\vec{j}\]\[\Rightarrow \vec{i}\times \vec{k}=-\vec{j}\]
(v) Let \[\vec{a}={{a}_{1}}\hat{i}+{{b}_{1}}j+{{c}_{1}}\hat{k}\]
\[\vec{b}={{a}_{2}}\hat{i}+{{b}_{2}}j+{{c}_{2}}\hat{k}\]
\[+\,\,\hat{k}({{a}_{1}}.{{b}_{2}}-{{a}_{2}}.{{b}_{1}})\]
(vi) Geometrical Interpretation:
\[\vec{a}\times \vec{b}\]represents a vectors whose magnitude is equal to the area of the parallelogram of which \[\vec{a}\] and \[\vec{b}\] are adjacent sides.
Note: \[\frac{1}{2}\vec{a}\times \vec{b}\]is the vectorial area of the triangle whose adjacent sides are \[\vec{a}\]and\[\vec{b}\].
Solved Examples
\[{{(\vec{a}.\vec{b})}^{2}}+{{(\vec{a}\times \vec{b})}^{2}}={{a}^{2}}.{{b}^{2}}\]
Sol. \[\because {{(\vec{a}.\vec{b})}^{2}}={{(|\vec{a}|.|\vec{b}|.\cos \theta )}^{2}}={{a}^{2}}.{{b}^{2}}.{{\cos }^{2}}\theta \]
\[{{(\vec{a}\times \vec{b})}^{2}}={{[|\vec{a}|.|\vec{b}|.sin\theta .\hat{n}]}^{2}}={{a}^{2}}.{{b}^{2}}.si{{n}^{2}}\theta \]
where \[\hat{n}\]is the unit vector.
L.H.S.\[={{(\vec{a}.\vec{b})}^{2}}+{{(\vec{a}\times \vec{b})}^{2}}={{a}^{2}}{{b}^{2}}.{{\cos }^{2}}\theta +{{a}^{2}}.{{b}^{2}}.{{\sin }^{2}}\theta \]
\[={{a}^{2}}{{b}^{2}}(co{{s}^{2}}\theta +{{\sin }^{2}}\theta )={{a}^{2}}{{b}^{2}}=\]R.H.S. Hence proved
Sol.
\[=\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)=0\hat{i}=19\vec{j}-19\vec{k}\]
\[\therefore \]\[|\vec{a}\times \vec{b}|\,\,=\sqrt{{{(-19)}^{2}}+{{(19)}^{2}}}=\sqrt{361+361}=\sqrt{2\times 361}=19\sqrt{2}\]
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