The direction cosines ell, m, n of the line f | vedclass

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(C) The given equation of the line is $\frac{x+2}{2} = \frac{2y-4}{3}$ and $z = -1$.We can rewrite the equation as $\frac{x+2}{2} = \frac{2(y-2)}{3} =

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$\ell= \pm \frac{4}{5}, m= \pm \frac{3}{5}, n=0$

Vedclass · Answer

(C) The given equation of the line is $\frac{x+2}{2} = \frac{2y-4}{3}$ and $z = -1$.We can rewrite the equation as $\frac{x+2}{2} = \frac{2(y-2)}{3} = \frac{y-2}{3/2}$ and $z = -1$.This represents a line in the $xy$-plane parallel to the $xy$-plane,where the $z$-coordinate is constant $(z = -1)$.The direction ratios of the line are $(a, b, c) = (2, \frac{3}{2}, 0)$.To find the direction cosines,we calculate the magnitude of the direction vector: $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (\frac{3}{2})^2 + 0^2} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{16+9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.The direction cosines $(\ell, m, n)$ are given by $(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}})$.Thus,$\ell = \pm \frac{2}{5/2} = \pm \frac{4}{5}$,$m = \pm \frac{3/2}{5/2} = \pm \frac{3}{5}$,and $n = \pm \frac{0}{5/2} = 0$.Therefore,the direction cosines are $\pm \frac{4}{5}, \pm \frac{3}{5}, 0$.

The direction cosines $\ell, m, n$ of the line $\frac{x+2}{2}=\frac{2y-4}{3}; z=-1$ are:

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