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Angle between Two Lines

A unit vector...

Question

A unit vector $\to d$ is equally inclined at an angle $α$ with the vectors $\to a = cos θ^i + sin θ^j$ , $\to b = - sin θ^i + cos θ^j, \to c = \to k$ then $α =$

A

{cos}^{- 1} (\frac{1}{\sqrt{3}})

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B

{cos}^{- 1} (\frac{1}{3})

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C

\frac{π}{4}

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D

\frac{π}{2}

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Solution

The correct option is A ${cos}^{- 1} (\frac{1}{\sqrt{3}})$
$∣ ∣ \to a ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣ = 1$
$\Rightarrow \to d . \to a = \to d . \to b = \to d . \to c = cos α, \to c = \to k$
$\Rightarrow \to d . (\to a - \to k) = 0,, \to d . (\to b - \to k) = 0$
$∴ \to d$ is along $(\to a - \to k) \times (\to b - \to k)$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k cos θ & sin θ & - 1 - sin θ & cos θ & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= (cos θ - sin θ)^i + (cos θ + sin θ)^j +^k$
$∴ \to d = \frac{(cos θ - sin θ)^i + (cos θ + sin θ)^j +^k}{\sqrt{{(cos θ - sin θ)}^{2} + {(cos θ + sin θ)}^{2} + 1}}$
Consider ${(cos θ - sin θ)}^{2} + {(cos θ + sin θ)}^{2} + 1$
$= 2 {cos}^{2} θ + 2 {sin}^{2} θ + 1$
$= 2 ({cos}^{2} θ + {sin}^{2} θ) + 1$
$= 2 + 1 = 3$
$∴ \to d = \frac{(cos θ - sin θ)^i + (cos θ + sin θ)^j +^k}{\sqrt{3}}$
$cos α = \to d .^k$
$= \frac{1}{\sqrt{3}} [(cos θ - sin θ)^i + (cos θ + sin θ)^j +^k] .^k$
$= \frac{1}{\sqrt{3}}$ since $^k .^k = 1$ and $^i .^k =^j .^k = 0$
$cos α = \frac{1}{\sqrt{3}}$
$∴ α = {cos}^{- 1} (\frac{1}{\sqrt{3}})$

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Q. A unit vector

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is equally inclined at an angle

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