L1 and L2 are two lines whose vector equations are L1 : r¯=λ((cos⁡θ+3)i¯+(2sin⁡θ)j¯+(cos⁡θ−3)k¯) ; L2 : r¯=μ(ai¯+bj¯+ck¯)Where λ and μ are scalars and α is the acute angle between L1 and L2. If the angle α (is independent of θ ) is equal to πk then k = _____

Both the lines passes through the origin line L1 is parallel to the vector V¯1=(cos⁡θ+3)i¯+(2sin⁡θ)j¯+(cosθ−3)k¯And L2 is parallel to the vector V¯2=ai¯+bj¯+ck¯cos⁡α=a(cos⁡θ+3)+b(2sin⁡θ)+c(cos⁡θ−3a2+b2+c2(cos⁡θ+3)2+(2sin⁡θ)2(cos⁡θ−3)2 =(a+c)cos⁡θ+b2sin⁡θ+(a−c)3a2+b2+c2⋅8∴α is independent of θ⇒a+c=0.b=0cos⁡α=(a−c)32a⋅22=2a32a22=32;α=π6

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$L_{1} and L_{2}$ are two lines whose vector equations are $L_{1} : \bar{r} = λ ((\cos θ + \sqrt{3}) \bar{i} + (\sqrt{2} \sin θ) \bar{j} + (\cos θ - \sqrt{3}) \bar{k})$ ; $L_{2} : \bar{r} = μ (a \bar{i} + b \bar{j} + c \bar{k})$
Where $λ and μ$ are scalars and $α$ is the acute angle between $L_{1} and L_{2}$ . If the angle $α$ (is independent of $θ$ ) is equal to $\frac{π}{k}$ then k = _____

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answer is 6.

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Detailed Solution

Both the $lines passes through the origin line L_{1} is$ $parallel to the vector$

${\bar{V}}_{1} = (\cos θ + \sqrt{3}) \bar{i} + (\sqrt{2} \sin θ) \bar{j} + (\cos θ - \sqrt{3}) \bar{k}$

And $L_{2}$ is parallel to the vector ${\bar{V}}_{2} = a \bar{i} + b \bar{j} + c \bar{k}$

$\begin{array}{l} \cos α = \frac{a (\cos θ + \sqrt{3}) + b (\sqrt{2} \sin θ) + c (\cos θ - \sqrt{3}}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{(\cos θ + \sqrt{3})^{2} + (\sqrt{2 \sin θ})^{2} (\cos θ - \sqrt{3})^{2}}} \\ = \frac{(a + c) \cos θ + b \sqrt{2} \sin θ + (a - c) \sqrt{3}}{\sqrt{a^{2} + b^{2} + c^{2}} \cdot \sqrt{8}} \end{array}$

$∴ α is independent of θ$

$\begin{array}{l} \Rightarrow a + c = 0 . b = 0 \\ \cos α = \frac{(a - c) \sqrt{3}}{\sqrt{2} a \cdot 2 \sqrt{2}} = \frac{2 a \sqrt{3}}{\sqrt{2} a 2 \sqrt{2}} = \frac{\sqrt{3}}{2}; α = \frac{π}{6} \end{array}$