the vector equations of two lines L_1 and L_2 is given by L_1 :r= i−j+3k + λ(i−j +k) L_2 : r= 2i+aj + 6k + μ(2i + j + 3k) where a,λ,μ are real constants. given that L_1 and L_2 intersect find a. the value of the constant a. b. the position vector of the point of intersection between L_1 and L_2 c. the cosine of the acute angle between L_1 and L


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Question Number 75101 by Rio Michael last updated on 07/Dec/19

$t h e v e c t o r e q u a t i o n s o f t w o l i n e s L_{1} a n d L_{2} i s g i v e n b y$ $L_{1} : r = i - j + 3 k + λ (i - j + k)$ $L_{2} : r = 2 i + a j + 6 k + μ (2 i + j + 3 k)$ $w h e r e a, λ, μ a r e r e a l c o n s t a n t s .$ $g i v e n t h a t L_{1} a n d L_{2} i n t e r s e c t f i n d$ $a . t h e v a l u e o f t h e c o n s t a n t a .$ $b . t h e p o s i t i o n v e c t o r o f t h e p o i n t o f$ $i n t e r s e c t i o n b e t w e e n L_{1} a n d L_{2}$ $c . t h e c o s i n e o f t h e a c u t e a n g l e b e t w e e n L_{1} a n d L_{2}$ $p l e a s e h e l p$
Commented by Kunal12588 last updated on 07/Dec/19

$a r e t h e s e y o u r h w q u e s t i o n c l a s s 12$
Commented by Rio Michael last updated on 07/Dec/19

$h a h a n o s i r, m y r e v i s i o n q u e s t i o n f o r m y$ ${c o u n t r y}^{'} s A l e v e l e x a m i n a t i o n, i w a n n a c o m f i r m m y$ $a n s w e r s$
	Answered by Kunal12588 last updated on 07/Dec/19

	$L_{1} a n d L_{2} i n t e r s e c t$ $∴ s h o r t e s t d i s t b / w L_{1} & L_{2} = 0$ $∴∣ \frac{({\vec{b}}_{1} \times {\vec{b}}_{2}) \cdot ({\vec{a}}_{2} - {\vec{a}}_{1})}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣} ∣= 0$ $\Rightarrow \| \begin{matrix} 2 - 1 & a + 1 & 6 - 3 \\ 1 & - 1 & 1 \\ 2 & 1 & 3 \end{matrix} \| = 0$ $\Rightarrow \| \begin{matrix} 1 & a + 1 & 3 \\ 1 & - 1 & 1 \\ 2 & 1 & 3 \end{matrix} \| = 0$ $\Rightarrow 1 (- 3 - 1) - (a + 1) (3 - 2) + 3 (1 + 2) = 0$ $\Rightarrow - 4 - a - 1 + 9 = 0$ $\Rightarrow a = 4$
	Commented by peter frank last updated on 07/Dec/19

	$t h a n k y o u$
	Answered by Kunal12588 last updated on 07/Dec/19

	$A n o t h e r w a y$ $r o f L_{1} = r o f L_{2}$ $(1 + λ) i + (- 1 - λ) j + (3 + λ) k = (2 + 2 μ) i + (a + μ) j + (6 + 3 μ) k$ $1 + λ = 2 + 2 μ - 1 - λ = a + μ 3 + λ = 6 + 3 μ$ $\Rightarrow λ - 2 μ = 1 \Rightarrow λ + μ = - 1 - a \Rightarrow λ - 3 μ = 3$ $\Rightarrow - 2 μ + 3 μ = 1 - 3$ $\Rightarrow μ = - 2 \Rightarrow λ + 4 = 1 \Rightarrow λ = - 3$ $\Rightarrow - 2 - 3 = - 1 - a$ $\Rightarrow a = 4$
	Answered by Kunal12588 last updated on 07/Dec/19
	$r=i−j+3k+λ(i−j+k) a^→ =i−j+3k−3(i−j+k) {λ=−3 for intersection} =(1−3)i−(1−3)j+(3−3)k =−2i+2j$
	$r = i - j + 3 k + λ (i - j + k)$ $\vec{a} = i - j + 3 k - 3 (i - j + k) {λ = - 3 f o r i n t e r s e c t i o n}$ $= (1 - 3) i - (1 - 3) j + (3 - 3) k$ $= - 2 i + 2 j$
	Commented by Rio Michael last updated on 07/Dec/19

	$t h a n k y o u s i r$
	Answered by Kunal12588 last updated on 07/Dec/19

	$a n g l e b e t w e e n L_{1} & L_{2}$ $c o s θ = ∣ \frac{{\vec{b}}_{1} \cdot {\vec{b}}_{2}}{∣ {\vec{b}}_{1} ∣∣ {\vec{b}}_{2} ∣} ∣=∣ \frac{(i - j + k) \cdot (2 i + j + 3 k)}{\sqrt{1 + 1 + 1} \sqrt{4 + 1 + 9}} ∣$ $\Rightarrow c o s θ =∣ \frac{2 - 1 + 3}{\sqrt{3} \sqrt{14}} ∣= \frac{4}{\sqrt{42}}$
	Commented by Rio Michael last updated on 07/Dec/19

	$t h a n k s s i r$
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