Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :
Solution
Plane passing through (2, 1, 0), (4, 1, 1) and
(5, 0, 1) is
<br><br>$$\left| {\matrix{
{x - 2} & {y - 1} & {z - 0} \cr
{4 - 2} & {1 - 1} & {1 - 0} \cr
{5 - 2} & {0 - 1} & {1 - 0} \cr
} } \right|$$ = 0
<br><br>$\Rightarrow$ x + y – 2z = 3
<br><br>$\therefore$ Image of R(2, 1, 6) in this plane is
<br><br>$${{x - 2} \over 1} = {{y - 1} \over 1} = {{z - 6} \over { - 2}} = - 2{{\left( {2 + 1 - 12 - 3} \right)} \over {1 + 1 + 4}}$$
<br><br>$\therefore$ (x, y, z) = (6, 5, –2)
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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