Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 49468 by munnabhai455111@gmail.com last updated on 07/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

O(0,0,0) A(x_1 ,y_1 ,z_1 ) B(x_2 ,y_2 ,z_2 ) OA^→ =ix_1 +jy_1 +kz_1 OB^→ =ix_2 +jy_2 +kz_2 area of OAB=(1/2)∣OA^→ ×OC^→ ∣ OA^→ ×OB^→ =i(y_1 z_2 −y_2 z_1 )−i(x_1 z_2 −x_2 z_1 )+k(x_1 y_2 −x_2 y_1 ) so (1/2)∣OA^→ ×OB^→ ∣ =(1/2)(√((y_1 z_2 −y_2 z_1 )^2 +(x_1 z_2 −x_2 z_1 )^2 +(x_1 y_2 −x_2 y_1 )^2 ))

$${O}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:\:{A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)\:{B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$${O}\overset{\rightarrow} {{A}}={ix}_{\mathrm{1}} +{jy}_{\mathrm{1}} +{kz}_{\mathrm{1}} \\ $$$${O}\overset{\rightarrow} {{B}}={ix}_{\mathrm{2}} +{jy}_{\mathrm{2}} +{kz}_{\mathrm{2}} \\ $$$${area}\:{of}\:{OAB}=\frac{\mathrm{1}}{\mathrm{2}}\mid{O}\overset{\rightarrow} {{A}}×{O}\overset{\rightarrow} {{C}}\mid \\ $$$${O}\overset{\rightarrow} {{A}}×{O}\overset{\rightarrow} {{B}}={i}\left({y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} \right)−{i}\left({x}_{\mathrm{1}} {z}_{\mathrm{2}} −{x}_{\mathrm{2}} {z}_{\mathrm{1}} \right)+{k}\left({x}_{\mathrm{1}} {y}_{\mathrm{2}} −{x}_{\mathrm{2}} {y}_{\mathrm{1}} \right) \\ $$$${so}\:\frac{\mathrm{1}}{\mathrm{2}}\mid{O}\overset{\rightarrow} {{A}}×{O}\overset{\rightarrow} {{B}}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({x}_{\mathrm{1}} {z}_{\mathrm{2}} −{x}_{\mathrm{2}} {z}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({x}_{\mathrm{1}} {y}_{\mathrm{2}} −{x}_{\mathrm{2}} {y}_{\mathrm{1}} \right)^{\mathrm{2}} }\: \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com