First slide

Straight lines in 3D

Question

Given lines are
$L_{1} : \vec{r} = λ ((\cos θ + \sqrt{3}) \hat{i} + (\sqrt{2} \sin θ) \hat{j} + (\cos θ - \sqrt{3}) \hat{k})$
$L_{2} : \vec{r} = μ (a \hat{i} + b \hat{j} + c \hat{k})$ where $λ$ and $μ$ pare scalars and a is the acute angle between L₁and L₂.If the angle $α$ is independent of $θ$ then the value of $α$ is

Moderate

A

$\frac{π}{6}$
B

$\frac{π}{4}$
C

$\frac{π}{3}$
D

$\frac{π}{2}$

Solution

Both the lines pass through the origin. Line L₁, is parallel to the vector $\vec{V_{1}}$
${\vec{V}}_{1} = (\cos θ + \sqrt{3}) \hat{i} + (\sqrt{2} \sin θ) \hat{j} + (\cos θ - \sqrt{3}) \hat{k}$
Line L₁, is parallel to the vector $\vec{V_{2}}$
$\begin{array}{l} {\vec{V}}_{2} = a \hat{i} + b \hat{j} + c \hat{k} \\ ∴ \cos α = \frac{{\vec{V}}_{1} \cdot {\vec{V}}_{2}}{|{\vec{V}}_{1}| |{\vec{V}}_{2}|} \\ = \frac{a (\cos θ + \sqrt{3}) + (b \sqrt{2}) \sin θ + c (\cos θ - \sqrt{3})}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{(\cos θ + \sqrt{3})^{2} + 2 \sin^{2} θ + (\cos θ - \sqrt{3})^{2}}} \\ = \frac{(a + c) \cos θ + b \sqrt{2} \sin θ + (a - c) \sqrt{3}}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{2 + 6}} \end{array}$
For cos a to be independent of $θ$ we get
$\begin{array}{l} a + c = 0 and b = 0 \\ ∴ \cos α = \frac{2 a \sqrt{3}}{a \sqrt{2} 2 \sqrt{2}} = \frac{\sqrt{3}}{2} \\ or α = \frac{π}{6} \end{array}$

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