Vector Solutions, Shortcuts, Formulas

Observation

  • \vec{a}\cdot \vec{b} is real number
  • Let \vec{a}\:and\:\vec{b}, be two nonzero vector , then \vec{a}\cdot \vec{b}=0 if and only if \vec{a}\:and\:\vec{b}, are perpendicular to each other i.e \vec{a}\cdot \vec{b}=0 ,
  • If θ = 0, then , \vec{a}.\vec{b}=|\vec{a}||\vec{b}| in particular \vec{a}\cdot \vec{a}=|\vec{a}^{2}| as θ in this case is 0.
  • If θ = π , then \vec{a}\cdot \vec{b}=-|\vec{a}||\vec{b}| in particular \vec{a}\cdot \vec{a}=|\vec{a}^{2}| as θ in this case is  \Pi
  • In view of the Observations 2 and 3, for mutually perpendicular unit vector  \hat{i},\hat{j},\hat{k} we have , \hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\: or\: \hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0
  • The angle between two non zero vector \vec{a}\:and\:\vec{b} is given by \cos\:\:\frac{\vec{a}\:\:\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\:\:or\:\:\theta
    =\cos^{-1}\:\left ( \frac{\vec{a}\:\:\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |} \right )
  • The scalar product is commutative i.e \vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}
  • Cross product of two vector : he three dimensional right handed rectangular coordinate system. In this system, when the positive x -axis is rotated counter clockwise nto the positive y -axis, a right handed (standard) screw would advance in the direction of the positive z -axis (Fig 10.22(i)). In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z -axis when the fingers are curled in the direction away from the positive x-axis toward the positive y -axis (Fig 10.22(ii).

 

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