The position vector of the point P is π→=xi^+yj^ +zk^, where x,y,z∈N. If a→=i^+j^+k^ and π→.a→=10, then the number of possible positions of point P is…

Question

The position vector of the point P is π→=xi^+yj^ +zk^, where x,y,z∈N. If a→=i^+j^+k^ and π→.a→=10, then the number of possible positions of point P is equal to

MARKS Content Team · Accepted Answer

Correct answer: (A) 36. π→.a→=10 ⇒x+y+z=10 The required number of positions = The number of positive integral solutions of x+y+z=10 =C3-110-1 =(9×8)/2=36

The position vector of the point $P$ is $\vec{π} = x \hat{i} + y \hat{j}$ $+ z \hat{k}$ , where $x, y, z \in N.$ If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{π} . \vec{a} = 10,$ then the number of possible positions of point $P$ is equal to

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