The direction cosines of the line frac{x-1}{0 | vedclass

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(B) The given equation of the line is $\frac{x-1}{0}=\frac{y+1}{5}=\frac{z-3}{0}$.Comparing this with the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{

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$0, 1, 0$

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(B) The given equation of the line is $\frac{x-1}{0}=\frac{y+1}{5}=\frac{z-3}{0}$.Comparing this with the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$,we get the direction ratios as $(a, b, c) = (0, 5, 0)$.The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$.Here,$\sqrt{a^2+b^2+c^2} = \sqrt{0^2+5^2+0^2} = \sqrt{25} = 5$.Thus,$l = \frac{0}{5} = 0$,$m = \frac{5}{5} = 1$,and $n = \frac{0}{5} = 0$.Therefore,the direction cosines are $(0, 1, 0)$.

The direction cosines of the line $\frac{x-1}{0}=\frac{y+1}{5}=\frac{z-3}{0}$ are . . . . . . .

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