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Question Number 43856 by MASANJA J last updated on 16/Sep/18

given that a×b=3i + j +k a×c=−i −2j +k find i)c×a ii)a×(b×c) iii)(a×b)•(a×c)

$${given}\:{that} \\ $$$${a}×{b}=\mathrm{3}{i}\:+\:{j}\:+{k} \\ $$$${a}×{c}=−{i}\:−\mathrm{2}{j}\:+{k} \\ $$$${find} \\ $$$$\left.{i}\right){c}×{a} \\ $$$$\left.{ii}\right){a}×\left({b}×{c}\right) \\ $$$$\left.{iii}\right)\left({a}×{b}\right)\bullet\left({a}×{c}\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18

i)c×a=−a×c c×a=−(−i−2j+k)=i+2j−k ii)a×(b×c)=(a.c)b−(a.b)c so a×(b×c)=(accosθ_1 )−(abcosθ_2 ) now a×c=acsinθ_1 n^→ ∣a×c∣=acsinθ_1 sinθ_1 =((√((−1)^2 +(−2)^2 +(1)^2 ))/(ac))=(((√6) )/(ac)) cosθ_1 =((√(a^2 c^2 −6))/(ac)) sinθ_2 =((√(3^2 +1^2 +1^2 ))/(ab))=((√(11))/(ab)) cosθ_2 =((√(a^2 b^2 −11))/(ab)) a×(b×c)=(a.c)b−(a.b)c =(accosθ_1 )b−(abcosθ_2 )c a×(b×c)={(√(a^2 c^2 −6)) }b^→ −{(√(a^2 b^2 −11)) }c^→

$$\left.{i}\right){c}×{a}=−{a}×{c} \\ $$$${c}×{a}=−\left(−{i}−\mathrm{2}{j}+{k}\right)={i}+\mathrm{2}{j}−{k} \\ $$$$ \\ $$$$\left.{ii}\right){a}×\left({b}×{c}\right)=\left({a}.{c}\right){b}−\left({a}.{b}\right){c} \\ $$$$ \\ $$$${so}\:{a}×\left({b}×{c}\right)=\left({accos}\theta_{\mathrm{1}} \right)−\left({abcos}\theta_{\mathrm{2}} \right) \\ $$$${now} \\ $$$${a}×{c}={acsin}\theta_{\mathrm{1}} \overset{\rightarrow} {{n}} \\ $$$$\mid{a}×{c}\mid={acsin}\theta_{\mathrm{1}} \\ $$$${sin}\theta_{\mathrm{1}} =\frac{\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} }}{{ac}}=\frac{\sqrt{\mathrm{6}}\:}{{ac}} \\ $$$${cos}\theta_{\mathrm{1}} =\frac{\sqrt{{a}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{6}}}{{ac}} \\ $$$${sin}\theta_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}{{ab}}=\frac{\sqrt{\mathrm{11}}}{{ab}} \\ $$$${cos}\theta_{\mathrm{2}} =\frac{\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{11}}}{{ab}} \\ $$$${a}×\left({b}×{c}\right)=\left({a}.{c}\right){b}−\left({a}.{b}\right){c} \\ $$$$\:\:=\left({accos}\theta_{\mathrm{1}} \right){b}−\left({abcos}\theta_{\mathrm{2}} \right){c} \\ $$$$ \\ $$$${a}×\left({b}×{c}\right)=\left\{\sqrt{{a}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{6}}\:\:\right\}\overset{\rightarrow} {{b}}\:−\left\{\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{11}}\:\:\right\}\overset{\rightarrow} {{c}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18

iii)(a×b).(a×c)={(3×−1)+(1×−2)+(1×1) ={−3−2+1}=−4

$$\left.{iii}\right)\left({a}×{b}\right).\left({a}×{c}\right)=\left\{\left(\mathrm{3}×−\mathrm{1}\right)+\left(\mathrm{1}×−\mathrm{2}\right)+\left(\mathrm{1}×\mathrm{1}\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left\{−\mathrm{3}−\mathrm{2}+\mathrm{1}\right\}=−\mathrm{4} \\ $$

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