The number of straight lines that are equally | vedclass

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(B) Let the direction angles of the line with the $x, y,$ and $z$ axes be $\alpha, \beta,$ and $\gamma$ respectively.Since the line is equally incline

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let the direction angles of the line with the $x, y,$ and $z$ axes be $\alpha, \beta,$ and $\gamma$ respectively.Since the line is equally inclined to the axes,we have $\alpha = \beta = \gamma$.The direction cosines of the line are $\cos \alpha, \cos \beta,$ and $\cos \gamma$.We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.Substituting $\alpha = \beta = \gamma$,we get $3 \cos^2 \alpha = 1$,which implies $\cos^2 \alpha = \frac{1}{3}$.Thus,$\cos \alpha = \pm \frac{1}{\sqrt{3}}$.Since $\cos \alpha = \cos \beta = \cos \gamma$,the possible direction cosines $(l, m, n)$ are $(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}})$.However,for a line to be equally inclined,the direction cosines must satisfy the condition that the line makes the same angle with each axis. The possible sets of direction cosines are:$1. (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$2. (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$3. (\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$4. (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}})$Thus,there are $4$ such lines.

The number of straight lines that are equally inclined to the three-dimensional coordinate axes is

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