Question: "please solve this question as soon as possible"

Answered question

2022-01-04

please solve this question as soon as possible

alenahelenash

Expert2022-02-10Added 556 answers

a) Let the equation be of form,

$a (x - 3) + b (y - 0) + c (z - (- 1)) = 0$

Since it through $(- 2, - 2, 3)$ and $(7, 1, - 4)$

$a (- 2 - 3) + b (- 2) + c (3 + 1) = 0$

$\Rightarrow 5 a + 2 b - 3 c = 0$ --- (1)

$a (7 - 3) + b (1 - 0) + c (- 4 + 1) = 0$

$4 a + b - 3 c = 0$

$5 a + 2 b - 3 c = 0$

$\frac{a}{- 3 + 6} = \frac{b}{- 15 + 12} = \frac{c}{8 - 5}$

$\frac{a}{3} = \frac{b}{- 3} = \frac{c}{3}$

$∵ 3 (x - 3) - 3 (y) + 3 (2 + 1) = 0$

$\Rightarrow$ $x - y + z = 2$ Required plane

b) Since the given line is intersetting with the plane for $t = 1$ , hence distance $= 0$

c) Angle,

$θ = \cos^{- 1} (\frac{< 7, 2, 1 > \cdot < 1, - 1, 1 >}{\sqrt{7^{2} + 2^{2} + 1^{2}} \cdot \sqrt{1^{2} + 1^{2} + 1^{2}}})$

$= \cos^{- 1} (\frac{6}{\sqrt{54} \cdot \sqrt{3}})$

$= \cos^{- 1} (\frac{6}{9 \sqrt{2}})$

$= \cos^{- 1} (\frac{\sqrt{2}}{3})$

d) Since normal of plane will be direction ratio for line,

$x = - 6 + t, y = 2 - t, z = 8 + t$

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