Question Number 175279 by naka3546 last updated on 26/Aug/22
$$\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{angle}\:\:\mathrm{between}\:\:\mathrm{the}\:\:\mathrm{lines} \\ $$$${r}\:=\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{4}}\end{pmatrix}\:\:+\:\lambda\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$${r}\:=\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{4}}\end{pmatrix}\:\:+\:\lambda\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$ \\ $$
Answered by mahdipoor last updated on 26/Aug/22
$${point}\:{A}:\: \\ $$$${r}_{\mathrm{1}} \left({a}\right)={r}_{\mathrm{2}} \left({b}\right)\:\Rightarrow\begin{pmatrix}{\mathrm{1}+\mathrm{2}{a}}\\{\mathrm{0}−{a}}\\{\mathrm{4}−{a}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}+\mathrm{3}{b}}\\{\mathrm{0}}\\{\mathrm{4}+{b}}\end{pmatrix} \\ $$$${a}=\mathrm{0}\:,\:{b}=\mathrm{0}\:\Rightarrow\Rightarrow{A}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{4}}\end{pmatrix} \\ $$$${point}\:{B}\::\:{r}_{\mathrm{1}} \left(\mathrm{1}\right)=\begin{pmatrix}{\mathrm{3}}\\{−\mathrm{1}}\\{\mathrm{3}}\end{pmatrix} \\ $$$${point}\:{C}\::\:{r}_{\mathrm{2}} \left(\mathrm{1}\right)=\begin{pmatrix}{\mathrm{4}}\\{\mathrm{0}}\\{\mathrm{5}}\end{pmatrix} \\ $$$$\Delta{ABC}\::\: \\ $$$${a}={BC}=\sqrt{\left(\mathrm{3}−\mathrm{4}\right)^{\mathrm{2}} +\left(−\mathrm{1}−\mathrm{0}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{5}\right)^{\mathrm{2}} }=\sqrt{\mathrm{6}} \\ $$$${b}={AC}=\sqrt{\mathrm{10}} \\ $$$${c}={AB}=\sqrt{\mathrm{6}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}.{cosA}={a}^{\mathrm{2}} \Rightarrow \\ $$$${cosA}=\frac{\mathrm{6}−\mathrm{10}−\mathrm{6}}{−\mathrm{2}.\sqrt{\mathrm{10}}.\sqrt{\mathrm{6}}}=\frac{\mathrm{5}}{\:\mathrm{2}\sqrt{\mathrm{15}}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{6}} \\ $$$$\Rightarrow\Rightarrow{A}={cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{15}}}{\mathrm{6}}\right) \\ $$
Commented by naka3546 last updated on 26/Aug/22
$$\mathrm{Thank}\:\:\mathrm{you} \\ $$
Answered by MJS_new last updated on 26/Aug/22
$$\mathrm{cos}\:\alpha\:=\frac{\mid\overset{\rightarrow} {{p}}\ast\overset{\rightarrow} {{q}}\mid}{\mid\overset{\rightarrow} {{p}}\mid×\mid\overset{\rightarrow} {{q}}\mid} \\ $$$$\mathrm{with}\:\overset{\rightarrow} {{p}}\ast\overset{\rightarrow} {{q}}=\begin{pmatrix}{{p}_{\mathrm{1}} }\\{{p}_{\mathrm{2}} }\\{…}\\{{p}_{{n}} }\end{pmatrix}\:\:\ast\begin{pmatrix}{{q}_{\mathrm{1}} }\\{{q}_{\mathrm{2}} }\\{…}\\{{w}_{{n}} }\end{pmatrix}\:\:=\:{p}_{\mathrm{1}} {q}_{\mathrm{1}} +{p}_{\mathrm{2}} {q}_{\mathrm{2}} +…+{p}_{{n}} {q}_{{n}} \\ $$
Answered by MikeH last updated on 26/Aug/22
$${d}_{\mathrm{1}} \:=\:\mathrm{2}{i}−{j}−{k}\:\mathrm{and}\:{d}_{\mathrm{2}} \:=\:\mathrm{3}{i}+{k} \\ $$$$\Rightarrow\:\mathrm{cos}\:\theta\:=\:\frac{\left(\mathrm{2}{i}−{j}−{k}\right).\left(\mathrm{3}{i}+{k}\right)}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }×\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{6}×\mathrm{10}}}\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{60}}} \\ $$$$\Rightarrow\:\theta\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{60}}}\right) \\ $$