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Question Number 175279 by naka3546 last updated on 26/Aug/22

$$\mathrm{Find}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines}$$$$r=\left(\begin{array}{c}1\\ 0\\ 4\end{array}\right)+\lambda \left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)$$$$r=\left(\begin{array}{c}1\\ 0\\ 4\end{array}\right)+\lambda \left(\begin{array}{c}3\\ 0\\ 1\end{array}\right)$$$$$$

Answered by mahdipoor last updated on 26/Aug/22

$$pointA:$$$$r_1\left(a\right)=r_2\left(b\right)\Rightarrow \left(\begin{array}{c}1+2a\\ 0-a\\ 4-a\end{array}\right)=\left(\begin{array}{c}1+3b\\ 0\\ 4+b\end{array}\right)$$$$a=0,b=0\Rightarrow \Rightarrow A=\left(\begin{array}{c}1\\ 0\\ 4\end{array}\right)$$$$pointB:r_1\left(1\right)=\left(\begin{array}{c}3\\ -1\\ 3\end{array}\right)$$$$pointC:r_2\left(1\right)=\left(\begin{array}{c}4\\ 0\\ 5\end{array}\right)$$$$\mathrm{\Delta }ABC:$$$$a=BC=\sqrt{{(3-4)}^2+{(-1-0)}^2+{(3-5)}^2}=\sqrt{6}$$$$b=AC=\sqrt{10}$$$$c=AB=\sqrt{6}$$$$\Rightarrow b^2+c^2-2bc.cosA=a^2\Rightarrow$$$$cosA=\frac{6-10-6}{-2.\sqrt{10}.\sqrt{6}}=\frac{5}{2\sqrt{15}}=\frac{\sqrt{15}}{6}$$$$\Rightarrow \Rightarrow A={cos}^{-1}\left(\frac{\sqrt{15}}{6}\right)$$

Commented by naka3546 last updated on 26/Aug/22

$$\mathrm{Thank}\mathrm{you}$$

Answered by MJS_new last updated on 26/Aug/22

$$\cos \alpha =\frac{\mid \overrightarrow{p}\ast \overrightarrow{q}\mid }{\mid \overrightarrow{p}\mid \times \mid \overrightarrow{q}\mid }$$$$\mathrm{with}\overrightarrow{p}\ast \overrightarrow{q}=\left(\begin{array}{c}{p}_1\\ p_2\\ ...\\ p_n\end{array}\right)\ast \left(\begin{array}{c}{q}_1\\ q_2\\ ...\\ w_n\end{array}\right)=p_1{q}_1+p_2{q}_2+...+p_n{q}_n$$

Answered by MikeH last updated on 26/Aug/22

$$d_1=2i-j-k\mathrm{and}{d}_2=3i+k$$$$\Rightarrow \cos \theta =\frac{(2i-j-k).(3i+k)}{\sqrt{2^2+1^2+1^2}\times \sqrt{3^2+1^2}}=\frac{5}{\sqrt{6\times 10}}=\frac{5}{\sqrt{60}}$$$$\Rightarrow \theta ={\cos }^{-1}\left(\frac{5}{\sqrt{60}}\right)$$

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