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The vectors `veca and vecb ` are non collinear. Find for what value of x the vectors `vecc=(x-2)veca+vecb and vecd=(2x+1) veca-vecb` are collinear.? |
Answer» Both the vectors `vecc and vecd` are non-zero as the cofficients of `vecb` in the both are non-zero. Two vectors `vecc and vecd` are collinear if one of them is a linear multiple of the other. Therefore, `" "vecd=lamdavecc` or `" "(2x+1)veca-vecb=lamda{(x-2)veca+vecb}" "`(i) or `" "{(2x+1)-lamda(x-2)}veca-(1+lamda)vecb=0` The above expression is of the form `pveca+qvecb=0,` where `veca and vecb` are non-collinear, and hence we have `p=0 and q=0`. Therefore, `" "2x+1-lamda(x-2)=0" "`(ii) and `" "1+lamda=0" "`(iii) From (iii), `lamda=-1`, and putting this value in (i), we get `x=(1)/(3)` Alternate method : `" "vecc=(x-2)veca+vecb and vecd = (2x+1)veca-vecb` are collinear. If `(x-2)/(2x+1)=(1)/(-1)`, then `x=(1)/(3)` |
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