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Question Number 56460 by cesar.marval.larez@gmail.com last updated on 16/Mar/19

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19

w^→ =3i+4j+5k  u^→ =ai+2j−3k  v^→ =i+4j+ck  w^→ .u^→ =0  (3i+4j+5k).(ai+2j−3k)=0  3a+8−15=0  3a=7→a=(7/3)  w^→ .v^→ =0  (3i+4j+5k).(i+4j+ck)=0  3+16+5c=0→5c=−19  c=((−19)/5)  b)u^− .v^→ =∣u^→ ∣∣v^→ ∣cosθ  cosθ=((u^→ .v^→ )/(∣u^→ ∣∣v^→ ∣))=(((ai+2j−3k).(i+4j+ck))/(√(a^2 +4^ +(−3)^2 ×(√(1^2 +16+c^2 )))))  cosθ=((3a+8−3c)/((√(25+a^2 ))}))×(1/(√(17+c^2 )))=  now put value of and calculate

$$\overset{\rightarrow} {{w}}=\mathrm{3}{i}+\mathrm{4}{j}+\mathrm{5}{k} \\ $$$$\overset{\rightarrow} {{u}}={ai}+\mathrm{2}{j}−\mathrm{3}{k} \\ $$$$\overset{\rightarrow} {{v}}={i}+\mathrm{4}{j}+{ck} \\ $$$$\overset{\rightarrow} {{w}}.\overset{\rightarrow} {{u}}=\mathrm{0} \\ $$$$\left(\mathrm{3}{i}+\mathrm{4}{j}+\mathrm{5}{k}\right).\left({ai}+\mathrm{2}{j}−\mathrm{3}{k}\right)=\mathrm{0} \\ $$$$\mathrm{3}{a}+\mathrm{8}−\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{3}{a}=\mathrm{7}\rightarrow{a}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\overset{\rightarrow} {{w}}.\overset{\rightarrow} {{v}}=\mathrm{0} \\ $$$$\left(\mathrm{3}{i}+\mathrm{4}{j}+\mathrm{5}{k}\right).\left({i}+\mathrm{4}{j}+{ck}\right)=\mathrm{0} \\ $$$$\mathrm{3}+\mathrm{16}+\mathrm{5}{c}=\mathrm{0}\rightarrow\mathrm{5}{c}=−\mathrm{19} \\ $$$${c}=\frac{−\mathrm{19}}{\mathrm{5}} \\ $$$$\left.{b}\right)\overset{−} {{u}}.\overset{\rightarrow} {{v}}=\mid\overset{\rightarrow} {{u}}\mid\mid\overset{\rightarrow} {{v}}\mid{cos}\theta \\ $$$${cos}\theta=\frac{\overset{\rightarrow} {{u}}.\overset{\rightarrow} {{v}}}{\mid\overset{\rightarrow} {{u}}\mid\mid\overset{\rightarrow} {{v}}\mid}=\frac{\left({ai}+\mathrm{2}{j}−\mathrm{3}{k}\right).\left({i}+\mathrm{4}{j}+{ck}\right)}{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}^{} +\left(−\mathrm{3}\right)^{\mathrm{2}} ×\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{16}+{c}^{\mathrm{2}} }}} \\ $$$${cos}\theta=\frac{\mathrm{3}{a}+\mathrm{8}−\mathrm{3}{c}}{\left.\sqrt{\mathrm{25}+{a}^{\mathrm{2}} }\right\}}×\frac{\mathrm{1}}{\sqrt{\mathrm{17}+{c}^{\mathrm{2}} }}= \\ $$$${now}\:{put}\:{value}\:{of}\:{and}\:{calculate} \\ $$

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