The projections of a vector on the three coor | vedclass

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(B) Let the vector be $\vec{v} = (a, b, c) = (6, -3, 2)$.The magnitude of the vector is given by $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$.$|\vec{v}| = \sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the vector be $\vec{v} = (a, b, c) = (6, -3, 2)$.The magnitude of the vector is given by $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$.$|\vec{v}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.The direction cosines $(l, m, n)$ of a vector are given by dividing the components by the magnitude of the vector:$l = \frac{a}{|\vec{v}|} = \frac{6}{7}$,$m = \frac{b}{|\vec{v}|} = \frac{-3}{7}$,$n = \frac{c}{|\vec{v}|} = \frac{2}{7}$.Thus,the direction cosines are $\left(\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}\right)$.

The projections of a vector on the three coordinate axes are $6, -3, 2$ respectively. The direction cosines of the vector are:

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