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(B) Let the direction ratios be $a = 2, b = -3, c = 6$. The magnitude of the vector formed by these ratios is $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-

Vedclass · Accepted Answer

$\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$

Vedclass · Answer

(B) Let the direction ratios be $a = 2, b = -3, c = 6$. The magnitude of the vector formed by these ratios is $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$. 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 $l = \frac{2}{7}, m = \frac{-3}{7}, n = \frac{6}{7}$. Thus,the direction cosines are $\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$.

If the length of a vector is $21$ and its direction ratios are $2, -3, 6$,then its direction cosines are:

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