If the sum of the distances of the point P(3, | vedclass

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(B) Let the point be $P(3, 4, \alpha)$.The distance of $P$ from the $X$-axis is $d_X = \sqrt{4^2 + \alpha^2} = \sqrt{16 + \alpha^2}$.The distance of $

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$1$

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(B) Let the point be $P(3, 4, \alpha)$.The distance of $P$ from the $X$-axis is $d_X = \sqrt{4^2 + \alpha^2} = \sqrt{16 + \alpha^2}$.The distance of $P$ from the $Y$-axis is $d_Y = \sqrt{3^2 + \alpha^2} = \sqrt{9 + \alpha^2}$.The distance of $P$ from the $Z$-axis is $d_Z = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.Let $f(\alpha) = \sqrt{16 + \alpha^2} + \sqrt{9 + \alpha^2} + 5$.To minimize $f(\alpha)$,we find the derivative $f'(\alpha) = \frac{\alpha}{\sqrt{16 + \alpha^2}} + \frac{\alpha}{\sqrt{9 + \alpha^2}}$.Setting $f'(\alpha) = 0$,we get $\alpha \left( \frac{1}{\sqrt{16 + \alpha^2}} + \frac{1}{\sqrt{9 + \alpha^2}} \right) = 0$.Since the term in the bracket is always positive,the only solution is $\alpha = 0$.Thus,$\sec \alpha = \sec(0) = 1$.

If the sum of the distances of the point $P(3, 4, \alpha)$,where $\alpha \in R$,from the $X$-axis,$Y$-axis,and $Z$-axis is minimum,then $\sec \alpha =$

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