Question Number 132972 by rexford last updated on 17/Feb/21
Answered by Ar Brandon last updated on 17/Feb/21
$$\begin{vmatrix}{\mathrm{i}}&{\mathrm{j}}&{\mathrm{k}}\\{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}=\mathrm{7i}+\mathrm{5j}+\mathrm{k} \\ $$
Commented by rexford last updated on 17/Feb/21
$${please},{is}\:{there}\:{any}\:{explanation}\:{behind} \\ $$$${the}\:{use}\:{of}\:{the}\:{determinant} \\ $$
Commented by mr W last updated on 17/Feb/21
$$\boldsymbol{{a}}×\boldsymbol{{b}}\:{is}\:{perpendicular}\:{both}\:{to}\:\boldsymbol{{a}}\:{and} \\ $$$${to}\:\boldsymbol{{b}}. \\ $$$$\boldsymbol{{a}}×\boldsymbol{{b}}=\begin{vmatrix}{{i}}&{{j}}&{{k}}\\{{a}_{{i}} }&{{a}_{{j}} }&{{a}_{{k}} }\\{{b}_{{i}} }&{{b}_{{j}} }&{{b}_{{k}} }\end{vmatrix} \\ $$
Answered by liberty last updated on 17/Feb/21
$$\mathrm{your}\:\mathrm{asked}\:\mathrm{unit}\:\mathrm{vector}\: \\ $$$$\mathrm{call}\:\mathrm{it}'\mathrm{s}\:\overset{\rightarrow} {{c}}\:\mathrm{where}\:\overset{\rightarrow} {{c}}=\:\frac{\begin{vmatrix}{\mathrm{1}\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{1}\:\:\:−\mathrm{1}\:\:−\mathrm{2}}\end{vmatrix}}{\mid\mid\overset{\rightarrow} {{c}}\mid\mid} \\ $$$$\overset{\rightarrow} {{c}}\:=\:\frac{\mathrm{7}\hat {\mathrm{i}}+\mathrm{5}\hat {\mathrm{j}}+\hat {\mathrm{k}}}{\:\sqrt{\mathrm{49}+\mathrm{25}+\mathrm{1}}}\:=\:\frac{\mathrm{7}\hat {\mathrm{i}}+\mathrm{5}\hat {\mathrm{j}}+\hat {\mathrm{k}}}{\mathrm{5}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$