Question-78652

Question Number 78652 by peter frank last updated on 19/Jan/20

Answered by mr W last updated on 08/Feb/20

$${three}\:{vectors}\:{are}\:{non}−{coplanar}\:{if} \\ $$$${and}\:{only}\:{if}\:{their}\:{scalar}\:{triple}\:{product} \\ $$$${is}\:{not}\:{equal}\:\mathrm{0}. \\ $$$$ \\ $$$$\left(\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}\right)\centerdot\left[\left(\lambda\boldsymbol{{b}}+\mathrm{4}\boldsymbol{{c}}\right)×\left(\mathrm{2}\lambda−\mathrm{1}\right)\boldsymbol{{c}}\right] \\ $$$$=\left(\mathrm{2}\lambda−\mathrm{1}\right)\left(\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}\right)\centerdot\left(\lambda\boldsymbol{{b}}×\boldsymbol{{c}}+\mathrm{4}\boldsymbol{{c}}×\boldsymbol{{c}}\right) \\ $$$$=\left(\mathrm{2}\lambda−\mathrm{1}\right)\left[\lambda\boldsymbol{{a}}\centerdot\boldsymbol{{b}}×\boldsymbol{{c}}+\mathrm{4}\boldsymbol{{a}}\centerdot\boldsymbol{{c}}×\boldsymbol{{c}}+\mathrm{2}\lambda\boldsymbol{{b}}\centerdot\boldsymbol{{b}}×\boldsymbol{{c}}+\mathrm{8}\boldsymbol{{b}}\centerdot\boldsymbol{{c}}×\boldsymbol{{c}}+\mathrm{3}\lambda\boldsymbol{{c}}\centerdot\boldsymbol{{b}}×\boldsymbol{{c}}+\mathrm{12}\boldsymbol{{c}}\centerdot\boldsymbol{{c}}×\boldsymbol{{c}}\right] \\ $$$$=\left(\mathrm{2}\lambda−\mathrm{1}\right)\lambda\left(\boldsymbol{{a}}\centerdot\boldsymbol{{b}}×\boldsymbol{{c}}\right)\neq\mathrm{0} \\ $$$${since}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\:{are}\:{non}−{coplanar},\:{i}.{e}.\:\boldsymbol{{a}}\centerdot\boldsymbol{{b}}×\boldsymbol{{c}}\neq\mathrm{0} \\ $$$$\left(\mathrm{2}\lambda−\mathrm{1}\right)\lambda\neq\mathrm{0} \\ $$$$\Rightarrow\lambda\neq\mathrm{0}\:{and}\:\lambda\neq\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{answer}\:\left({c}\right)\:{is}\:{correct}. \\ $$

Commented by peter frank last updated on 08/Feb/20

$${thank}\:{you}\:{very}\:{much} \\ $$

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