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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...