Mathematics - Vector Question with Solution | TestHub

$\overrightarrow{\text{x}}\equiv \widehat{\text{i}}+\widehat{\text{j}}+\widehat{\text{k}}$
$\overrightarrow{\text{y}}\equiv \widehat{\text{i}}+\widehat{\text{j}}-\widehat{\text{k}}$
$\overrightarrow{\text{z}}\equiv \widehat{\text{i}}+\widehat{\text{j}}+\widehat{\text{k}}$
$\overrightarrow{\omega }=\widehat{\text{i}}-\widehat{\text{j}}-\widehat{\text{k}}$
${\overline{\text{x}}}^{\prime }=-\widehat{\text{i}}-\widehat{\text{j}}-\widehat{\text{k}}$
${\overrightarrow{\text{y}}}^{\prime }=-\widehat{\text{i}}-\widehat{\text{j}}+\widehat{\text{k}}$
${\overrightarrow{\text{z}}}^{\prime }=-\widehat{\text{i}}+\widehat{\text{j}}-\widehat{\text{k}}$
${\overrightarrow{\omega }}^{\prime }=-\widehat{\text{i}}+\widehat{\text{j}}+\widehat{\text{k}}$
If we take$\overrightarrow{\text{x}}\text{, }{\overrightarrow{\text{x}}}^{\prime }$and any one of remaining six, vectors will always be coplanar
$\therefore$No. of coplanar vectors = 6
Similarly on taking$\overrightarrow{\text{y}}\text{,}{\overrightarrow{\text{y}}}^{\prime }=6$
$\text{z}\text{,}{\overrightarrow{\text{z}}}^{\prime }=6$
$\omega \text{,}{\overrightarrow{\omega }}^{\prime }=6$
$\therefore$No. of set coplanar vectors = 24
${}^8{\text{C}}_3-24=32$