# If $P$ is point whose position vector is $\overrightarrow r=x\hat i+y\hat j+z\hat k$ where $x,y,z \in N$ and $\overrightarrow a=\hat i+\hat j+\hat k$, then the number of possible points $P$ for which $\overrightarrow r.\overrightarrow a=10$ is ?

Since given that $\overrightarrow r.\overrightarrow a= 10,$
$x+y+z=10$
That is sum of three natural numbers is 10.
$\therefore$ The no. of points $P$ satisfying the given condition is
no. of set of 3 natural numbers whose sum is $10$
which is $^{10-1}C_{3-1}=^9C_2=36$
answered Jan 7, 2014