Direction cosines of the line frac{x+2}{2} = | vedclass

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Question

(A) The given equation of the line is $\frac{x+2}{2} = \frac{2y-5}{3}, z = -1$. First,rewrite the line in standard symmetric form: $\frac{x+2}{2} = \f

Vedclass · Accepted Answer

$\frac{4}{5}, \frac{3}{5}, 0$

Vedclass · Answer

(A) The given equation of the line is $\frac{x+2}{2} = \frac{2y-5}{3}, z = -1$. First,rewrite the line in standard symmetric form: $\frac{x+2}{2} = \frac{y - 5/2}{3/2} = \frac{z+1}{0}$. The direction ratios of the line are $(a, b, c) = (2, 3/2, 0)$. To simplify,multiply by $2$: $(4, 3, 0)$. The magnitude of the direction vector is $\sqrt{4^2 + 3^2 + 0^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. The direction cosines $(l, m, n)$ are given by $\left( \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \right)$. Thus,the direction cosines are $\left( \frac{4}{5}, \frac{3}{5}, 0 \right)$.

Direction cosines of the line $\frac{x+2}{2} = \frac{2y-5}{3}, z = -1$ are $.......$

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