Mathematics - Vector Question with Solution | TestHub

As$\left|\widehat{w}\right|=\left|\widehat{u}\right|=1=\left|\widehat{u}\times \overrightarrow{v}\right|$
Let$\varphi$is angle between$\widehat{w} \& (\widehat{u}\times \overrightarrow{v})$
$\left|\widehat{w}\right| \left|\widehat{u}\times \overrightarrow{v}\right|\cos \varphi =1 \Rightarrow \varphi =0$
$\Rightarrow \widehat{u}\times \overrightarrow{v}=\widehat{w}$
$\Rightarrow$There may be infinite vectors$\overrightarrow{v}$Satisfying this condition
If$\widehat{u}$his in$xy$plane :$\widehat{u}\times \overrightarrow{v}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ u_1& u_2& 0\\ v_1& v_2& v_3\end{array}\right|$$=\widehat{w}$
$\Rightarrow u_3=0$
Let$\overrightarrow{v}=v_1\widehat{i}+v_2\widehat{j}+v_3\widehat{k}$
$\widehat{w}=\left(u_2{v}_3\right)\widehat{i}-\left(u_1{v}_3\right)\widehat{j}+\left(u_1{v}_2-u_2{v}_1\right)\widehat{k}$$=\frac{1}{\sqrt{6}}\widehat{i}+\frac{1}{\sqrt{6}}\widehat{j}+\frac{2}{\sqrt{6}}\widehat{k}$
$u_2{v}_3=\frac{1}{\sqrt{6}}, -u{}_1{v}_3=\frac{1}{\sqrt{6}} \Rightarrow \left|u_1\right|=\left|u_2\right|$
If$\widehat{u}$his in$xz$plane$\Rightarrow u_2=0$:$\widehat{u}\times \overrightarrow{v}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ u_1& 0& u_3\\ v_1& v_2& v_3\end{array}\right|$
$\widehat{w}=\left(-v_2{u}_3\right)\widehat{i}-\left(u_1{v}_3-u_3{v}_1\right)\widehat{j}+\left(u_1{v}_2\right)\widehat{k}$$=\frac{1}{\sqrt{6}}\widehat{i}+\frac{1}{\sqrt{6}}\widehat{j}+\frac{2}{\sqrt{6}}\widehat{k}$
$-v_2{u}_3=\frac{1}{\sqrt{6}}, u_1{v}_2=\frac{2}{\sqrt{6}}$
$\Rightarrow$$2\left|u_3\right|=\left|u_1\right|$