If the direction cosines of a line L are (pq, | vedclass

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(D) The direction cosines of a line are denoted by $(l, m, n)$. Given $(l, m, n) = (pq, q, q)$.We know that for any line,$l^2 + m^2 + n^2 = 1$.Substit

Vedclass · Accepted Answer

$16 : 9$

Vedclass · Answer

(D) The direction cosines of a line are denoted by $(l, m, n)$. Given $(l, m, n) = (pq, q, q)$.We know that for any line,$l^2 + m^2 + n^2 = 1$.Substituting the given values: $(pq)^2 + q^2 + q^2 = 1$,which simplifies to $p^2q^2 + 2q^2 = 1$,or $q^2(p^2 + 2) = 1$.Also,the direction cosine $l$ is given by $\cos(\alpha)$,where $\alpha$ is the angle with the $X$-axis.Given $\alpha = \frac{\pi}{3}$,so $l = \cos(\frac{\pi}{3}) = \frac{1}{2}$.Since $l = pq$,we have $pq = \frac{1}{2}$,which implies $p^2q^2 = \frac{1}{4}$.Substitute $p^2q^2 = \frac{1}{4}$ into the equation $p^2q^2 + 2q^2 = 1$:$\frac{1}{4} + 2q^2 = 1$$2q^2 = 1 - \frac{1}{4} = \frac{3}{4}$$q^2 = \frac{3}{8}$.Now,find $p^2$ using $p^2q^2 = \frac{1}{4}$:$p^2(\frac{3}{8}) = \frac{1}{4}$$p^2 = \frac{1}{4} 	imes \frac{8}{3} = \frac{2}{3}$.Finally,the ratio $p^2 : q^2 = \frac{2}{3} : \frac{3}{8} = \frac{2}{3} 	imes \frac{8}{3} = \frac{16}{9}$.

If the direction cosines of a line $L$ are $(pq, q, q)$ and the angle between the line $L$ and the positive direction of the $X$-axis is $\frac{\pi}{3}$,then $p^2 : q^2 =$

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