If a line has the direction ratios -18, 12, - | vedclass

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(A) Given direction ratios are $a = -18, b = 12, c = -4.$The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{(-18)^2 + (12)^2 + (-4)^2}.$$=

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$\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}$

Vedclass · Answer

(A) Given direction ratios are $a = -18, b = 12, c = -4.$The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{(-18)^2 + (12)^2 + (-4)^2}.$$= \sqrt{324 + 144 + 16} = \sqrt{484} = 22.$The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}.$$l = \frac{-18}{22} = \frac{-9}{11},$$m = \frac{12}{22} = \frac{6}{11},$$n = \frac{-4}{22} = \frac{-2}{11}.$Thus,the direction cosines are $\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}.$

If a line has the direction ratios $-18, 12, -4,$ then what are its direction cosines?

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