Question

position vector of x-z plane is given by
a.r=yi + zk
b.r=xi+yk​

Answer 1

Answer

r =x i^ +y j^ +z k^u = i^ + j^ + k^r ⋅

u =10

(x i^ +y j^ +z k^ )⋅( i^+j+k^ )=10

x+y+z=10

HERE x,y,z are natural number So x,y,z cannot be negative and zero as well

so atleast x,y,z should be 1

so values of x,y,z vary from 2 to 8

So combination of values becomes ( 37 )+1

possible values= 4!3! 7! +1

possible values= 3×2×1

7×6×5+1

possible values=35+1

possible values=36

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