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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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 $\sec \alpha =$

Similar Questions

The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as $(l_1, m_1, n_1)$,$(l_2, m_2, n_2)$ and $(l_3, m_3, n_3)$ are:

The direction cosines $l, m, n$ of two lines satisfy the equations $3l + m + 5n = 0$ and $6mn - 2nl + 5lm = 0$. If $\theta$ is the angle between these lines,then $|\cos \theta| = $

If the coordinates of a point $P$ with respect to the origin $O$ are $(3, 12, 4)$,then the direction cosines of $OP$ are ..........

The octant in which the point $(2, -4, -7)$ lies is

If a line makes an angle of $\frac{\pi}{3}$ with each of the $X$ and $Y$-axes,then the acute angle made by the line with the $Z$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module