# Vector Solutions, Shortcuts, Formulas

## Components of a Vector

Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x -axis, y -axis and z -axis, respectively. Then, clearly .
$|\vec{OA}|=1,|\vec{OB}|=1\:and\:\:\vec{OC}=1$
The vector $|\vec{OA}|,|\vec{OB}|,\vec{OC}$ each having magnitude 1,are called unit vectors along the axes OX, OY and OZ respectively and denoted by $\hat{i},\hat{j}\: and\: \hat{k}$ respectively (Fig 10.13).
Now, consider the position vector OP of a point P (x , y, z) as in Fig 10.14. Let P1 be the foot of the perpendicular from P on the plane XOY. We, thus, see that P1 P is parallel to z – axis.
Therefore it follows that $\vec{OP_{1}}=\vec{OQ}+\vec{QP_{1}}=x\hat{i}+y\hat{j}$
$\vec{OP_{1}}(\vec{a})=x\hat{i}+y\hat{j}+z\hat{k}$
The length of any vector = , is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP1. (Fig 10.14) $\vec{op} = \ \sqrt{X^{2}+Y^{2}}$

$| \vec{r} |= \sqrt{x^{2}+y^{2}+z^{2}}$
Vector joining two point : $p_{1}(x_{1},y_{1},z_{1})\: and\: p_{2}(x_{2},y_{2},z_{1})$ are any two points $p_{1} and p_{2}$ with the origin O, and applying triangle law , from the triangle $OP_{1}P_{2}$ ,we have

$p_{1}p_{2}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

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