Here, plane, line and its image are parallel to each other. So, find any point on the normal to the plane from which the image line will be passed and then find equation of image line. Here, plane and line are parallel to each other. Equation of normal to the plane through the point (1, 3, 4) is $ \hspace40mm \frac{x-1}{2}=\frac{y-3}{-1} = \frac{z-4}{1} = k \, \, \, \, \, $ [say] Any point in this normal is (2k + 1, - k + 3,4 + k). Then, $\bigg(\frac{2k+1+1}{2},\frac{3-k+3}{2}, \frac{4 +k+4}{2}\bigg) $ lies on plane. $ \Rightarrow 2(k+1)-\bigg(\frac{6-k}{2}\bigg)+\bigg(\frac{8+k}{2}\bigg)+3 = 0 \, \, \Rightarrow \, \, \, k = -2$ Hence, point through which this image pass is $\, \, \, \, \, \, \, \, (2k + 1,3 - k, 4 + k)$ $i.e.\, [2(-2) + 1 ,3 + 2,4 - 2] = (-3 ,5 ,2 )$ Hence, equation of image line is $ \hspace40mm\frac{x+3}{3}=\frac{y-5}{1}= \frac{z-2}{-5}$
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Top Questions on introduction to three dimensional geometry
In mathematics, Geometry is one of the most important topics. The concepts of Geometry are defined with respect to the planes. So, Geometry is divided into three categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.
Direction Cosines and Direction Ratios of Line
Let's consider line ‘L’ is passing through the three-dimensional plane. Now, x,y, and z are the axes of the plane, and α,β, and γ are the three angles the line making with these axes. These are called the plane's direction angles. So, correspondingly, we can very well say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
