First slide

Direction ratios and direction cosines

Question

A line makes angles $60 °, 60 °, 45 °$ and $θ$ with the four diagonals of a cube. Then $\sin^{2} θ =$

Moderate

A

$\frac{1}{2}$
B

$\frac{2}{3}$
C

$\frac{1}{4}$
D

$\frac{3}{4}$

Solution

If a line makes angles   $α, β, γ, δ$ with the four diagonals of a cube then $\cos^{2} α + \cos^{2} β + \cos^{2} γ + \cos^{2} δ = \frac{4}{3}$
Here given   $α = 60 °, β = 60 °, γ = 45 °$ and $δ = θ$
Hence,
$\begin{matrix} \cos^{2} 60 ° + \cos^{2} 60 ° + \cos^{2} 45 ° + \cos^{2} δ = \frac{4}{3} \\ \frac{1}{4} + \frac{1}{4} + \frac{1}{2} + \cos^{2} θ = \frac{4}{3} \\ \cos^{2} θ = \frac{1}{3} \end{matrix}$
Therefore, $\sin^{2} θ = 1 - \frac{1}{3} = \frac{2}{3}$

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