If the position vector of a point P is $\vec r=x\hat i+y\hat j+z\hat k$ where $x, y, z ∈N$ and $\vec a$ is a vector given by $\vec a = \hat i +\hat j+\hat k$, then the total number of possible positions of point P for which $\vec r.\vec a = 10$, is |
36 72 66 none of these |
36 |
The correct answer is Option (1) → 36 $\vec r.\vec a = 10⇒ x + y + z=10$ as $x,y,z∈N$ assuming 1 to each variables now we need to distribute $(10-3)=7$ among $x,y,z$ Bagger's Method No. of ways of doing So using 2 distribution partitions $=\frac{(7+2)!}{2!7!} =\frac{9!}{2!7!}= 36$. |