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Ab, AC and AD are three adjacent edges o...

Ab, AC and AD are three adjacent edges of a parallelpiped. The diagonal of the praallelepiped passing through A and direqcted away from it is vector `veca`. The vector of the faces containing vertices A, B , C and A, B, D are `vecb and vecc`, respectively , i.e. `vec(AB) xx vec(AC) and vec(AD) xx vec(AB) = vecc` the projection of each edge AB and AC on diagonal vector `veca is |veca|/3`
vector `vec(AD)` is

A

`1/3 veca+ (vecaxx(vecb-vecc))/|veca|^(2)`

B

`1/3 veca+ (vecaxx(vecb-vecc))/|veca|^(2) + (3(vecbxxveca))/|veca|^(2)`

C

`1/3 veca+ (vecaxx(vecb-vecc))/|veca|^(2) -(3(vecbxxveca))/|veca|^(2)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
c
`veca=vec(AP)=vec(AB)+vec(AC)+vec(AD)`
`vec(AB)xxvec(AC)=vecb`
`vec(AD)xxvec(AB)=vecc`
`vec(AB).veca/(|veca|)=|veca|/3 Rightarrowvec(AB).veca= (|veca|^(2))/3`
`vec(AB).veca/(|veca|)=|veca|/3 Rightarrowvec(AC).veca= (|veca|^(2))/3`
` (vec(AB) xx vec(AC))xxveca = vecb xxveca`
`vec(AC)-vec(AB)=3(vecbxxveca)/(|veca|^(2))`
`|veca|^(2)=vec(AB).veca+vec(AC).veca+vec(AD).veca`
`(|veca|^(2))/3=vec(AD).veca`
`(vec(AD)xxvec(AB))xxveca=veccxxveca`
`vec(AB)- vec(AD) = 3 (vecc xx veca)/(|veca|^(2))`
Now from (ii) and (iii), we get `vec(AC) and vec(AD)` as
`vec(AC)=1/3veca+ (vecaxx(vecb xx vecc))/(|veca|^(2))+(3(vecbxxveca))/(|veca|^(2))`
` vec(AD)= 1/3veca+ (vecaxx(vecb-vecc))/(|veca|^(2))- (3(vec cxxveca))/(|veca|^(2))`
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