Find-the-angle-between-the-lines-r-1-0-4-2-1-1-r-1-0-4-3-0-1-

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

Find the angle between the lines

r = (\begin{matrix} 1 \\ 0 \\ 4 \end{matrix}) + λ (\begin{matrix} 2 \\ - 1 \\ - 1 \end{matrix})

r = (\begin{matrix} 1 \\ 0 \\ 4 \end{matrix}) + λ (\begin{matrix} 3 \\ 0 \\ 1 \end{matrix})

Answered by mahdipoor last updated on 26/Aug/22

p o i n t A :

r_{1} (a) = r_{2} (b) \Rightarrow (\begin{matrix} 1 + 2 a \\ 0 - a \\ 4 - a \end{matrix}) = (\begin{matrix} 1 + 3 b \\ 0 \\ 4 + b \end{matrix})

a = 0, b = 0 \Rightarrow\Rightarrow A = (\begin{matrix} 1 \\ 0 \\ 4 \end{matrix})

p o i n t B : r_{1} (1) = (\begin{matrix} 3 \\ - 1 \\ 3 \end{matrix})

p o i n t C : r_{2} (1) = (\begin{matrix} 4 \\ 0 \\ 5 \end{matrix})

Δ A B C :

a = B C = \sqrt{{(3 - 4)}^{2} + {(- 1 - 0)}^{2} + {(3 - 5)}^{2}} = \sqrt{6}

b = A C = \sqrt{10}

c = A B = \sqrt{6}

\Rightarrow b^{2} + c^{2} - 2 b c . c o s A = a^{2} \Rightarrow

c o s A = \frac{6 - 10 - 6}{- 2 . \sqrt{10} . \sqrt{6}} = \frac{5}{2 \sqrt{15}} = \frac{\sqrt{15}}{6}

\Rightarrow\Rightarrow A = {c o s}^{- 1} (\frac{\sqrt{15}}{6})

Commented by naka3546 last updated on 26/Aug/22

Thank you

Answered by MJS_new last updated on 26/Aug/22

\cos α = \frac{∣ \vec{p} * \vec{q} ∣}{∣ \vec{p} ∣ \times ∣ \vec{q} ∣}

with \vec{p} * \vec{q} = (\begin{matrix} p_{1} \\ p_{2} \\ \dots \\ p_{n} \end{matrix}) * (\begin{matrix} q_{1} \\ q_{2} \\ \dots \\ w_{n} \end{matrix}) = p_{1} q_{1} + p_{2} q_{2} + \dots + p_{n} q_{n}

Answered by MikeH last updated on 26/Aug/22

d_{1} = 2 i - j - k and d_{2} = 3 i + k

\Rightarrow \cos θ = \frac{(2 i - j - k) . (3 i + 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 θ = \cos^{- 1} (\frac{5}{\sqrt{60}})

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