If a line makes angles α, β, γ, δ with four diagonals of a cube, then cos2 α + cos2 β + cos2γ + cos2 δ is equal to

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KollegeApply · Accepted Answer

$\frac{4}{3}$ \text { Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure . Clearly, OP, AR, BS and CQ are the diagonals of the cube } . \text{ The direction ratios of OP, AR, BS and CQ are } a - 0, a - 0, a - 0, \text{ i . e } . a, a, a 0 - a, a - 0, a - 0, \text{ i . e } . - a, a, a a - 0, 0 - a, a - 0,\text{ i . e } . a, - a, a a - 0, a - 0, 0 - a\text{ i . e } . a, a, - a \text { Let the direction ratios of a line be proportional to l, m and n . Suppose this line makes angles} \alpha, \beta, \gamma \text { and } \delta \text{ with OP, AR, BS and CQ, respectively i . e } . \text{ Now} , \alpha \text{ is the angle between OP and the line whose direction ratios are proportional to l, m and n } . \cos \alpha = \frac{a . l + a . m + a . n}{$\sqrt$(a^2 + a^2 + a^2)$\sqrt$(l^2 + m^2 + n^2)} \Rightarrow \cos \alpha = \frac{l + m + n}{$\sqrt$(3)$\sqrt$(l^2 + m^2 + n^2)} \text{ Since } \beta \text{ is the angle between AR and the line with direction ratios proportional to l, m and n, we get } \cos \beta = \frac{- a . l + a . m + a . n}{$\sqrt$(a^2 + a^2 + a^2)$\sqrt$(l^2 + m^2 + n^2)} \Rightarrow \cos \beta = \frac{- l + m + n}{$\sqrt$(3)$\sqrt$(l^2 + m^2 + n^2)} \text{ Similarly }, \cos \gamma = \frac{a . l - a . m + a . n}{$\sqrt$(a^2 + a^2 + a^2)$\sqrt$(l^2 + m^2 + n^2)} \Rightarrow \cos \gamma = \frac{l - m + n}{$\sqrt$(3)$\sqrt$(l^2 + m^2 + n^2)} \cos \delta = \frac{a . l + a . m - a . n}{$\sqrt$(a^2 + a^2 + a^2)$\sqrt$(l^2 + m^2 + n^2)} \Rightarrow \cos \delta = \frac{l + m - n}{$\sqrt$(3)$\sqrt$(l^2 + m^2 + n^2)} \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{\left( l + m + n \right)^2}{3\left( l^2 + m^2 + n^2 \right)} + \frac{\left( - l + m + n \right)^2}{3\left( l^2 + m^2 + n^2 \right)} + \frac{\left( I - m + n \right)^2}{3\left( l^2 + m^2 + n^2 \right)} + \frac{\left( l + m - n \right)^2}{$\sqrt$(3)$\sqrt$(l^2 + m^2 + n^2)} = \frac{1}{3\left( l^2 + m^2 + n^2 \right)}\left\{ \left( l + m + n \right)^2 + \left( - l + m + n \right)^2 + \left( I - m + n \right)^2 + \left( l + m - n \right)^2 \right\} = \frac{1}{3\left( l^2 + m^2 + n^2 \right)}4\left( l^2 + m^2 + n^2 \right) = $\frac{4}{3}$ [Solution diagram: 45816_solution_1.png]

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