Question Number 62241 by vishnurajput8081@gmail.com last updated on 18/Jun/19
$$\boldsymbol{{if}}\:\boldsymbol{{the}}\:\boldsymbol{{point}}\:{A}\:{B}\:{C}\:{with}\:{position}\:{vector}\: \\ $$$$\left(\mathrm{20}\hat {{i}}+\lambda\hat {{j}}\right)\:\left(\mathrm{5}\hat {{i}}−\hat {{j}}\right)\:{and}\left(\mathrm{10}\hat {{i}}−\mathrm{13}\hat {{j}}\right)\:{are} \\ $$$${collinear}\:{then}\:{the}\:{value}\:{of}\:\lambda\:{is}: \\ $$
Answered by tanmay last updated on 18/Jun/19
![AB^→ =OB^→ −OA^→ =i(5−20)+j(−1−λ)=i(−15)+j(−1−λ) BC^→ =OC^→ −OB^→ =i(10−5)+j(−13+1)=i(5)+j(−12) AB^→ ×BC^→ =0 [(−15)i+j(−1−λ)]×[5i−12j]=0 −75(i×i)+180(i×j)+(−5−5λ)(j×i)+(12+12λ)(j×j)=0 (i×j)(180+5+5λ)=0 5λ+185=0 λ=−37](Q62248.png)
$${A}\overset{\rightarrow} {{B}}={O}\overset{\rightarrow} {{B}}−{O}\overset{\rightarrow} {{A}}={i}\left(\mathrm{5}−\mathrm{20}\right)+{j}\left(−\mathrm{1}−\lambda\right)={i}\left(−\mathrm{15}\right)+{j}\left(−\mathrm{1}−\lambda\right) \\ $$$${B}\overset{\rightarrow} {{C}}={O}\overset{\rightarrow} {{C}}−{O}\overset{\rightarrow} {{B}}={i}\left(\mathrm{10}−\mathrm{5}\right)+{j}\left(−\mathrm{13}+\mathrm{1}\right)={i}\left(\mathrm{5}\right)+{j}\left(−\mathrm{12}\right) \\ $$$${A}\overset{\rightarrow} {{B}}×{B}\overset{\rightarrow} {{C}}=\mathrm{0} \\ $$$$\left[\left(−\mathrm{15}\right){i}+{j}\left(−\mathrm{1}−\lambda\right)\right]×\left[\mathrm{5}{i}−\mathrm{12}{j}\right]=\mathrm{0} \\ $$$$−\mathrm{75}\left({i}×{i}\right)+\mathrm{180}\left({i}×{j}\right)+\left(−\mathrm{5}−\mathrm{5}\lambda\right)\left({j}×{i}\right)+\left(\mathrm{12}+\mathrm{12}\lambda\right)\left({j}×{j}\right)=\mathrm{0} \\ $$$$\left({i}×{j}\right)\left(\mathrm{180}+\mathrm{5}+\mathrm{5}\lambda\right)=\mathrm{0} \\ $$$$\mathrm{5}\lambda+\mathrm{185}=\mathrm{0}\:\:\:\: \\ $$$$\lambda=−\mathrm{37} \\ $$