If a line has direction ratios 2, -1, -2,dete | vedclass

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(A) Let the direction ratios be $a = 2, b = -1, c = -2$. The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt

Vedclass · Accepted Answer

$\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}$

Vedclass · Answer

(A) Let the direction ratios be $a = 2, b = -1, c = -2$. The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$. Substituting the values,we get $l = \frac{2}{3}, m = -\frac{1}{3}, n = -\frac{2}{3}$. Thus,the direction cosines are $\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}$.

If a line has direction ratios $2, -1, -2$,determine its direction cosines.

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