The distance of the point (1, –2, 3) from
the plane x – y + z = 5 measured parallel to
the line ${x \over 2} = {y \over 3} = {z \over { - 6}}$ is :
Solution
Equation of line parallel to ${x \over 2} = {y \over 3} = {z \over { - 6}}$ passes through $(1, - 2,3)$ is<br><br>${{x - 1} \over 2} = {{y + 2} \over 3} = {{z - 3} \over { - 6}} = r$<br><br>$x = 2r + 1$<br><br>$y = 3r - 2$, <br><br>$z = - 6r + 3$
<br><br>A point on whole line = (2r + 1, 3r – 2, – 6r + 3).
<br><br>This point lies on plane x – y + 2 = 5
<br><br>so, $2r + 1 - 3r + 2 - 6r + 3 = 5$<br><br>$\Rightarrow$ $r = {1 \over 7}$<br><br>$\therefore$ $x = {9 \over 7}$, $y = {{ - 11} \over 7}$, $z = {{15} \over 7}$<br><br>Distance is = $$\sqrt {{{\left( {{9 \over 7} - 1} \right)}^2} + {{\left( {2 - {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}} $$<br><br>$$ = \sqrt {{{\left( {{2 \over 7}} \right)}^2} + {{\left( {{3 \over 7}} \right)}^2} + {{\left( {{6 \over 7}} \right)}^2}} $$<br><br>$= {1 \over 7}\sqrt {4 + 9 + 36}$ = 1
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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