Question Number 132856 by bramlexs22 last updated on 17/Feb/21
$$\mathrm{Given}\:\mathrm{vector}\:\overset{\rightarrow} {{a}}\:=\:\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}}\:,\:\overset{\rightarrow} {\mathrm{c}}=\hat {\mathrm{j}}−\hat {\mathrm{k}}\:; \\ $$$$\:\overset{\rightarrow} {\mathrm{a}}\:×\:\overset{\rightarrow} {\mathrm{b}}\:=\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{3}\:\mathrm{then}\:\mid\overset{\rightarrow} {\mathrm{b}}\mid\:=\:? \\ $$
Answered by EDWIN88 last updated on 17/Feb/21
$$\mathrm{from}\:\overset{\rightarrow} {\mathrm{a}}\:×\:\overset{\rightarrow} {\mathrm{b}}\:=\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{we}\:\mathrm{get}\:\overset{\rightarrow} {\mathrm{a}}×\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)\:=\:\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{c}} \\ $$$$\:\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{c}}=\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\mathrm{1}\:\:−\mathrm{1}}\end{vmatrix}=\:−\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {\mathrm{a}}×\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)=\left(\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\right)\overset{\rightarrow} {\mathrm{a}}−\left(\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{a}}\right)\overset{\rightarrow} {\mathrm{b}}=−\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}} \\ $$$$\Rightarrow\mathrm{3}\overset{\rightarrow} {\mathrm{a}}−\mathrm{3}\overset{\rightarrow} {\mathrm{b}}=\:−\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}}\: \\ $$$$\:\overset{\rightarrow} {\mathrm{b}}=\frac{\left(\mathrm{3},\mathrm{3},\mathrm{3}\right)−\left(−\mathrm{2},\mathrm{1},\mathrm{1}\right)}{\mathrm{3}}\:=\:\frac{\mathrm{5}\hat {\mathrm{i}}+\mathrm{2}\hat {\mathrm{j}}+\mathrm{2}\hat {\mathrm{k}}}{\mathrm{3}} \\ $$$$\mathrm{Therefore}\:\mid\overset{\rightarrow} {\mathrm{b}}\mid\:=\:\frac{\sqrt{\mathrm{25}+\mathrm{4}+\mathrm{4}}}{\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{33}}}{\mathrm{3}}\:=\:\sqrt{\frac{\mathrm{11}}{\mathrm{3}}} \\ $$