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The position vector of a point $P$ is given by $\overrightarrow r=x\hat i+y\hat j+z\hat k$, where $x,y,z\in N$. Position vector of point A is given by $\overrightarrow a=\hat i+\hat j+\hat k$. How many points P are possible so that $\overrightarrow r.\overrightarrow a=5$ ?

$\begin {array} {1 1} (A)\;27 & \quad (B)\;12 \\ (C)\;6 & \quad (D)\;2 \end {array}$

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Since $\overrightarrow r.\overrightarrow a=5$, $x+y+z=5$, where $x,y,z\in N$
$\therefore$ The no. of possible points p is taken from no. of combinations of 3 natural numbers whose sum is 5 along with their arrangements.
$(1,1,3)..........3\: arrangements.$
$(2,2,1)..........3\: arrangements.$
$\therefore$ The number of possible points = 6
answered Sep 3, 2013 by rvidyagovindarajan_1
edited Jul 2, 2014 by balaji.thirumalai
 

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