Vector Solutions, Shortcuts, Formulas

Multiplication of a Vector by a Scalar

Let be a given vector and λ a scalar. Then the product of the vector a by the scalar λ , denoted as \lambda \vec{a} , is called the multiplication of vector \vec{a} , by the scalar λ. Note that , \lambda \vec{a} , is also a vector, collinear to the vector \vec{a} , . The vector \lambda \vec{a} , has the direction same (or opposite) to that of vector \vec{a} , according as the value of λ is positive (or negative).Also, the magnitude of vector \lambda \vec{a} , Is |λ| times the magnitude of the vector \vec{a} , i.e , \left | \lambda \vec{a} \right |=\left | \lambda \right |\left | \vec{a} \right |
A geometric visualisation of multiplication of a vector by a scalar is given in( Fig 10.12)

When \lambda =-1,\: then\: \lambda \vec{a}=-\vec{a} which is a vector having magnitude equal to the magnitude of \vec{a} the vector -\vec{a} is called negative vector .(or additive inverse ) of vector a \vec{a} and we always have .
\vec{a}+(-\vec{a})=(-\vec{a})+(\vec{a}) . Also , if \lambda =\frac{1}{|\vec{a}|} , provide 0 , i.e \vec{a} is not a null vector , then \lambda \vec{a} represents the unit vector in the direction of. We write it as \hat{a}=\frac{1}{|\vec{a}|}

Note:

  • One may observe that whatever be the value of λ , the vector \lambda \vec{a} is always collinear to the vector \vec{a} . In fact, two vectors \vec{a}\: and\: \vec{b} are collinear if and only if there exists a nonzero scalar λ such that \vec{b}=\lambda \vec{a} . If the vectors \vec{a}\: and\: \vec{b} are given in component form , i.e , \vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}\: and\: \vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}
  • If two vectors are collinear if and only if \lambda (a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k})\:=\:b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}\Rightarrow \lambda a_{1},b_{2}=\lambda a_{2},b_{3}=\lambda a_{3}
  • \frac{b_{1}}{a_{1}}=\frac{b_{2}}{a_{2}}=\frac{b_{3}}{a_{3}}
  • If (a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}) ,this is called the direction of ratio \vec{a}
  • In case if it is given that l ,m ,n are direction cosine of a vector , then l\hat{i}+m\hat{j}+n\hat{k}=(\cos\alpha)\hat{i}+(\cos\beta)\hat{j}+(\cos \gamma )\hat{k} is the unit vector in the direction of that vector, where α , β and γ are the angles which the vector makes with x , y and z axes respectively .

Example: find the unit vector in the direction of vector \vec{a}=2\hat{i}+3\hat{j}+\hat{k}
Solution: The unit vector in the direction of a vector \vec{a} is given by \hat{a}=\frac{1}{|\vec{a}|}\vec{a}
Now |\vec{a}|=\sqrt{2^{2}+3^{2}+1^{2}}=\sqrt{14}

Therefore \hat{a}=\frac{1}{\sqrt{14}}(2\hat{i}+3\hat{j}+\hat{k})=\frac{2}{\sqrt{14}}\hat{i}+\frac{3}{\sqrt{14}}\hat{j}+\frac{1}{\sqrt{14}}\hat{k}

Example: Find the unit vector in the direction of the sum of the vector \vec{a}=2\hat{i}+2\hat{j}-5\hat{k}\: and\: \vec{b}=2\hat{i}+\hat{j}+3\hat{k}
Solution: The sum of the given vector \vec{a}+\vec{b}=4\hat{i}+3\hat{j}+2\hat{k} , Let \vec{a}+\vec{b}=\vec{c}
|\vec{c}|=\sqrt{4^{2}+3^{^{3}}+(-2)^{2}}=\sqrt{29} , Thus the require unit vector is

\hat{c}=\frac{1}{|\vec{c}|}\vec{c}=\frac{1}{\sqrt{29}}(4\hat{i}+3\hat{j}-2\hat{k})=\frac{4}{\sqrt{29}}\hat{i}+\frac{3}{\sqrt{29}\hat{j}}-\frac{2}{\sqrt{29}}\hat{k}

 

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