If the direction ratios of a line are 1, -3, | vedclass

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(A) Given direction ratios are $a = 1, b = -3, c = 2$.The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9

Vedclass · Accepted Answer

$\frac{1}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{2}{\sqrt{14}}$

Vedclass · Answer

(A) Given direction ratios are $a = 1, b = -3, c = 2$.The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$.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 $\left( \frac{1}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right)$.

If the direction ratios of a line are $1, -3, 2$,what are its direction cosines?

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