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06/01/2022 1:09 pm
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A plane has equation x - y + z = 5 and a line has direction ratios as (2, 3, -6), then the distance of the point P(1, 3, 5) along the line from the given plane is
(a) 2 m
(b) 2√3 m
(c) 3m
(d) 3√2 m
1 Answer
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06/01/2022 1:15 pm
(a) 2 m
Explanation:
Equation of line \(\frac{x-1}{2}\)
= \(\frac{y-3}{3}\)
= \(\frac{z-5}{-6}\) = k
⇒ x = 2k + 1 , y = 3k + 3 , z = - 6k + 5
Equation of plane: x - y + z = 5
⇒ 2k + 1 - 3k - 3 - 6k + 5 = 5
⇒ -7k = 2
⇒ k = \(\frac{-2}{7}\)
(x, y, z) = (\(\frac{3}{7}, \frac{15}{7}, \frac{47}{7}\))
d = \(\sqrt{(\frac{4}{7})^2 + (\frac{6}{7})^2 + (\frac{-12}{7})^2}\)
= \(\frac{1}{7}\)\(\sqrt{16 + 26 + 144}\)
d = \(\frac{14}{7}\) = 2
This post was modified 3 years ago 2 times by admin