If l, m, n are the direction cosines of a lin | vedclass

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(B) We know that the direction cosines are given by $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. Also, the sum of squares of direction

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$\alpha=\beta=\gamma$

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(B) We know that the direction cosines are given by $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. Also, the sum of squares of direction cosines is $l^2 + m^2 + n^2 = 1$. We want to maximize the expression $S = lm + mn + nl$. Using the identity $(l + m + n)^2 = l^2 + m^2 + n^2 + 2(lm + mn + nl)$, we have $2(lm + mn + nl) = (l + m + n)^2 - (l^2 + m^2 + n^2) = (l + m + n)^2 - 1$. To maximize $lm + mn + nl$, we must maximize $(l + m + n)^2$. By the Cauchy-Schwarz inequality, $(l + m + n)^2 \leq (1^2 + 1^2 + 1^2)(l^2 + m^2 + n^2) = 3(1) = 3$. The equality holds when $l = m = n$. Since $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$, this implies $\cos \alpha = \cos \beta = \cos \gamma$, which means $\alpha = \beta = \gamma$ (given the angles are between $0$ and $\pi$). Thus, the maximum value is attained when $\alpha = \beta = \gamma$.

If $l, m, n$ are the direction cosines of a line which makes angles $\alpha, \beta$ and $\gamma$ with the coordinate axes $X, Y, Z$, respectively, then $l m+m n+n l$ takes the maximum value when

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