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Prove that if `cos alpha ne 1, cos beta ne1 and cos gamma ne 1`, then the vectors `a=hati cos alpha+hatj+hatk,b=hati+hatj cos beta+hatk`. `c=hati+hatj+hatk cos gamma` can never be coplanar. |
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Answer» Suppose that, a,b and c are coplanar. `implies|(cosalpha,1,1),(1,cosbeta,1),(1,1cos gamma)|=0` On applying `R_(2) to R_(2)-R_(1) and R_(3) to R_(3)-R_(1)` `implies|(cos alpha,1,1),(1-cosalpha,cosbeta-1,0),(1-cosalpha,0,cosgamma-1)|=0` `implies cosalpha(cosbeta-1)(cosgamma-1)-(1-cosalpha)(cosgamma-1)-(1-cosalpha)(cosbeta-1)=0` On dividing throughout by `(1-cosalpha)(1-cosbeta)(1-cosgamma),` we get `(cosalpha)/(1-cosalpha)+(1)/(1-cosbeta)+(1)/(1-cosgamma)=0` `implies(-(1-cosalpha)+1)/(1-cosalpha)+(1)/(1-cosbeta)+(1)/(1-cosgamma)=0` `implies-1+(1)/((1-cosalpha))+(1)/((1-cosbeta))+(1)/((1-cosgamma))=0` `(1)/(1-cosalpha)+(1)/(1-cosbeta)+(1)/(1-cosgamma)=1` `implies"cosec"^(2)(alpha)/(2)+"cosec"^(2)(beta)/(2)+"cosec"^(2)(gamma)/(2)=2,` which is not possible As, `"cosec"^(2)(alpha)/(2)ge1,"cosec"^(2)(beta)/(2)ge1` and `"cosec"^(2)(gamma)/(2)ge1` `because`They cannot be coplanar. |
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