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Question Number 98018 by hardylanes last updated on 11/Jun/20

t h e v e c t o r e q u a t i o n s o f t w o l i n e s l_{1} a n d l_{2} a r e g i v e n b y

l_{1} : r = 5 i - j + k + λ (- 3 i + 2 j)

l_{2} : r = 2 i + 3 j + 2 k + μ (2 j + k)

f i n d

t h d p o s i t i o n v d c t o r o f i n t e r s e c t i o n o f t h f l i n d s l_{1} a n d l_{2}

t h d c a r t e s i a n e q u a t k o n o f t h e p l a n e π, c o n t a i n i n g t h e l i n e s l_{1} a n d l_{2}

t h e s i n e o f t h e a n h l e b e t w e e n t h e p l a n e π a n d t h e l i n e, l_{3} : r = i - 5 j - 2 k + s (2 i + 2 j - k)

Answered by Rio Michael last updated on 11/Jun/20

(a) l_{1} : r = (5 - 3 λ) i + (- 1 + 2 λ) j + k = (\begin{matrix} 5 - 3 λ \\ - 1 + 2 λ \\ 1 \end{matrix})

l_{2} : r = 2 i + (3 + 2 μ) j + (2 + μ) k = (\begin{matrix} 2 \\ 3 + 2 μ \\ 2 + μ \end{matrix})

at intersection, l_{1} = l_{2} \Rightarrow 5 - 3 λ = 2 \Leftrightarrow λ = 1

and 2 + μ = 1 \Leftrightarrow μ = - 1

point of intersection r_{0} = 2 i + j + k

(b) equation of plane r . n = r_{0} . n

\bar{n} = d_{1} \times d_{2} = | \begin{matrix} i & j & k \\ - 3 & 2 & 0 \\ 0 & 2 & 1 \end{matrix} | = i | \begin{matrix} 2 & 0 \\ 2 & 1 \end{matrix} | - j | \begin{matrix} - 3 & 0 \\ 0 & 1 \end{matrix} | + k | \begin{matrix} - 3 & 2 \\ 0 & 2 \end{matrix} | = 2 i + 3 j - 6 k

\Rightarrow r . (2 i + 3 j - 6 k) = (2 i + j + k) . (2 i + 3 j - 6 k)

2 x + 3 y - 6 z = 1

(c) \sin θ = \frac{n . d}{∣ n ∣∣ d ∣} = \frac{(2 i + 3 j - 6 k) . (2 i + 2 j - k)}{\sqrt{49} . \sqrt{9}} = \frac{16}{21}

\Rightarrow \sin θ = \frac{16}{21}

Commented by hardylanes last updated on 11/Jun/20

thanks again

Commented by peter frank last updated on 11/Jun/20

good