Mathematics - Vector Question with Solution | TestHub

Let D is origin and position vectors of A, B, C are$\vec{a},\vec{b},\vec{c}$ respectively,

then$|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|=\ell$ and$|\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{c}}\cdot \overrightarrow{\mathrm{a}}|={\ell }^2\cos 60^{\circ }=\frac{{\ell }^2}{2}$ Now${\left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]}^2=\left|\begin{array}{lllcccc}\vec{a}\cdot \vec{a}& \vec{a}\cdot \vec{b}& \vec{c}\cdot \vec{c}\vec{b}\cdot \vec{a}& \vec{b}\cdot \vec{b}& \vec{b}\cdot \vec{c}\vec{c}\cdot \vec{a}& \vec{c}\cdot \vec{b}& \vec{c}\cdot \vec{c}\end{array}\right|=\frac{{\ell }^6}{2}$ now volume of tetrahedron$=\frac{1}{3}\times$ Area of base × height$\Rightarrow \frac{1}{6}\left[\begin{array}{lll}\overrightarrow{\mathrm{a}}& \overrightarrow{\mathrm{b}}& \overrightarrow{\mathrm{c}}\end{array}\right]=\frac{1}{3}\times \left(\frac{\sqrt{3}}{4}{\ell }^2\right)\times$ height$\Rightarrow$ height$=\frac{\frac{{\ell }^3}{\sqrt{2}}\times \frac{1}{2}}{\frac{\sqrt{3}}{4}{\ell }^2}=\sqrt{\frac{2}{3}}\ell$