Question Number 129591 by mohammad17 last updated on 16/Jan/21
Answered by mr W last updated on 16/Jan/21
$${AC}=\left(\mathrm{2},\mathrm{4}\right) \\ $$$${BD}=\left(\mathrm{4},−\mathrm{2}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{{AC}\centerdot{BD}}{\mid{AC}\mid×\mid{BD}\mid}=\frac{\mathrm{2}×\mathrm{4}−\mathrm{4}×\mathrm{2}}{\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\frac{\mathrm{0}}{\mathrm{20}}=\mathrm{0} \\ $$$$\Rightarrow\theta=\mathrm{90}° \\ $$
Commented by mohammad17 last updated on 16/Jan/21
$${sir}\:{what}\:{is}\:{the}\:{value}\:{of}\:\mid{AC}\mid×\mid{BD}\mid \\ $$
Commented by mr W last updated on 16/Jan/21
$${i}\:{thought}\:{you}\:{can}\:{get}\:\mid{AC}\mid×\mid{BD}\mid \\ $$$${by}\:{yourself}. \\ $$