Saved Bookmarks
| 1. |
Find the symetrical form, the equation of the line \( x+y+z+1=0,4 x+y-2 z=0 \) and. find the direction of cosines. |
|
Answer» Let direction cosines of line are l, m and n. \(\therefore\) l + m + n = 0 4l + m - 2n = 0 (\(\because\) Line is present in both planes, therefore line is perpendicular to normals of both planes) ⇒ \(\frac{l}{-2-1}=\frac{m}{4+2}=\frac{n}{1-4}\) = k (let) ⇒ l = -3k, m = 6k, n = -3k But l, m, n are direction cosines. l = \(\frac{-3k}{\sqrt{(-3k)^2+(6k)^2+(-3k)^2}}\) \(=\frac{-3k}{\sqrt{9k^2+36k^2+9k^2}}\) \(=\frac{-3k}{\sqrt{54k^2}}=\frac{-3}{3\sqrt6}=\frac{-1}{\sqrt 6}\) m = \(\frac{6k}{\sqrt{(-3k)^2+(6k)^2+(-3k)^2}}\) = \(\frac{6k}{3\sqrt6}\) = \(\frac{2}{\sqrt6}\) n = \(\frac{-3k}{\sqrt{(-3k)^2+(6k)^2+(-3k)^2}}\) = \(\frac{-3k}{3\sqrt6}=\frac{-1}{\sqrt6}\) \(\therefore\) Direction cosines of line are \((\mp\frac1{\sqrt6},\pm\frac{2}{\sqrt6},\mp\frac1{\sqrt6}).\) |
|