The position vector of a point P is r-x-y-zk- where x- N-y - N-z- N-and a-i-j-k- If r-a-10- the number of possible positions of P is

The position vector of a point P is $\to r = x^i + y^j + z^k$ , when $x, y, z ϵ$ N and $\to a =^i +^j +^k . I f \to r . \to a = 10$ , the number of possible position of P is

Q. Let

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k,

and

\to C = 4 \to i - 3 \to j + 7 \to k

Determine a vector

\to R

satisfying

\to R \times \to B = \to C \times \to B

and

\to R . \to A = 0

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The position vector of a point P is →r=x^i→+y^j+z^k, where xϵ N,yϵ N,zϵ N,and→a=→i+→j+→k→. If →r.→a=10, the number of possible positions of P is

The correct option is A 36The no. of positive integral solution of the equation x1+x2+x3+...xr=n is n−1Cr−1.We have, →r=xi+yj+zkAlso, →r⋅→a=10⇒x+y+z=10Thus number of positive integral solution is =10−1C3−1=36

The position vector of a point P is $\to r = x^i \to + y^j + z^k,$ where $x ϵ N, y ϵ N, z ϵ N,$ and $\to a = \to i + \to j + \to k \to$ . If $\to r . \to a = 10$ , the number of possible positions of P is