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(A) Let the direction cosines of the vector $\vec{r}$ be $l, m, n$. The projections of the vector on the coordinate axes are given by $l|\vec{r}|, m|\

Vedclass · Accepted Answer

$m = -\frac{3}{7}, n = \frac{2}{7}$

Vedclass · Answer

(A) Let the direction cosines of the vector $\vec{r}$ be $l, m, n$. The projections of the vector on the coordinate axes are given by $l|\vec{r}|, m|\vec{r}|, n|\vec{r}|$.Therefore,$l|\vec{r}| = 6, m|\vec{r}| = -3, n|\vec{r}| = 2$  $(i)$.Squaring and adding these equations:$(l|\vec{r}|)^2 + (m|\vec{r}|)^2 + (n|\vec{r}|)^2 = 6^2 + (-3)^2 + 2^2$$|\vec{r}|^2(l^2 + m^2 + n^2) = 36 + 9 + 4$Since $l^2 + m^2 + n^2 = 1$,we have $|\vec{r}|^2 = 49$,which implies $|\vec{r}| = 7$.Substituting $|\vec{r}| = 7$ into equation $(i)$:$l = \frac{6}{7}, m = -\frac{3}{7}, n = \frac{2}{7}$.

Find the direction cosines and the length of a vector whose projections on the coordinate axes are $6, -3, 2$.

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