JEE MAIN - Mathematics (2005 - No. 32)
If $$C$$ is the mid point of $$AB$$ and $$P$$ is any point outside $$AB,$$ then :
$$\overrightarrow {PA} + \overrightarrow {PB} = 2\overrightarrow {PC} $$
$$\overrightarrow {PA} + \overrightarrow {PB} = \overrightarrow {PC} $$
$$\overrightarrow {PA} + \overrightarrow {PB} = 2\overrightarrow {PC} = \overrightarrow 0 $$
$$\overrightarrow {PA} + \overrightarrow {PB} = \overrightarrow {PC} = \overrightarrow 0 $$
Explanation
$$\overrightarrow {PA} + \overrightarrow {AP = 0} $$ and $$\overrightarrow {PC} + \overrightarrow {CP} = 0$$
$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {AC} + \overrightarrow {CP} = 0$$
and
$$\overrightarrow {PB} + \overrightarrow {BC} + \overrightarrow {CP} = 0$$
Adding, we get
$$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {AC} + \overrightarrow {BC} + 2\overrightarrow {CP} = 0.$$
Since
$$\overrightarrow {AC} = - \overrightarrow {BC} $$ $$\,\,\,\,\,\,$$ & $$\,\,\,\,\,\,$$ $$\overrightarrow {CP} = - \overrightarrow {PC} $$
$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {PB} - 2\overrightarrow {PC} = 0.$$
$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {AC} + \overrightarrow {CP} = 0$$
and
$$\overrightarrow {PB} + \overrightarrow {BC} + \overrightarrow {CP} = 0$$
Adding, we get
$$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {AC} + \overrightarrow {BC} + 2\overrightarrow {CP} = 0.$$
Since
$$\overrightarrow {AC} = - \overrightarrow {BC} $$ $$\,\,\,\,\,\,$$ & $$\,\,\,\,\,\,$$ $$\overrightarrow {CP} = - \overrightarrow {PC} $$
$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {PB} - 2\overrightarrow {PC} = 0.$$
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