find the angle of plane 2x−y+2z=1 and x+3y−2z = 2


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Question Number 95509 by i jagooll last updated on 25/May/20

$find the angle of plane$ $2 x - y + 2 z = 1 and x + 3 y - 2 z = 2$
	Answered by bobhans last updated on 25/May/20

	$let β = angle the plane$ $\cos β = \frac{∣ 2.1 + (- 1) . 3 + 2 . (- 2) ∣}{\sqrt{9} \sqrt{14}} = \frac{5}{3 \sqrt{14}}$ $β = \cos^{- 1} (\frac{5}{3 \sqrt{14}})$ $the formula \cos β = \frac{∣ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} ∣}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
	Answered by mr W last updated on 25/May/20

	${\vec{n}}_{1} = (2, - 1, 2) = n o r m a l o f p l a n e 1$ ${\vec{n}}_{2} = (1, 3, - 2) = n o r m a l o f p l a n e 2$ ${\vec{n}}_{1} \cdot {\vec{n}}_{2} =∣ {\vec{n}}_{1} ∣∣ {\vec{n}}_{2} ∣ \cos θ$ $\cos θ = \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{∣ {\vec{n}}_{1} ∣∣ {\vec{n}}_{2} ∣} = \frac{2 \times 1 - 1 \times 3 - 2 \times 2}{\sqrt{2^{2} + 1^{2} + 2^{2}} \times \sqrt{1^{2} + 3^{2} + 2^{2}}} = \frac{5 \sqrt{14}}{42}$ $\Rightarrow θ = \cos^{- 1} \frac{5 \sqrt{14}}{42}$
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