How do I find the angle between the planes 5(x+1) + 3(y+2) + 2z = 0 and x + 3(y-1) + 2(z+4) = 0?

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How do I find the angle between the planes 5(x+1) + 3(y+2) + 2z = 0 and x + 3(y-1) + 2(z+4) = 0?

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Answer 1 · 2016-09-04T19:47:13

$\approx {38.69}^{\circ}$ $~~38.69^@$

Explanation:

Let the given planes be $π_{1} and π_{2}$ $pi_1 and pi_2$ . Let ${\to n}_{1} and {\to n}_{2}$ $vecn_1 and vec n_2$

be their normals.

Recall that for a plane $: a x + b y + c z + d = 0$ $: ax+by+cz+d=0$ , its normal $\to n = (a, b, c)$ $vecn=(a,b,c)$ .

Clearly, ${\to n}_{1} = (5, 3, 2), and, n_{2} = (1, 3, 2)$ $vec n_1=(5,3,2), and, n_2=(1,3,2)$

the $∠ α$ $/_alpha$ btwn. $π_{1} and π_{2}$ $pi_1 and pi_2$ is, by defn.,

$α = a r c cos ⎧ ⎪ ⎨ ⎪ ⎩ \frac{∣ ∣ {\to n}_{1} \cdot {\to n}_{2} ∣ ∣}{∣ ∣ ∣ ∣ {\to n}_{1} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ {\to n}_{2} ∣ ∣ ∣ ∣} ⎫ ⎪ ⎬ ⎪ ⎭$ $alpha=arc cos{|vecn_1*vecn_2|/(||vecn_1||||vecn_2||)}$ .

$= a r c cos {\frac{| 5 \cdot 1 + 3 \cdot 3 + 2 \cdot 2 |}{(\sqrt{25 + 9 + 4} \sqrt{1 + 9 + 4})}}$ $=arc cos{|5*1+3*3+2*2|/((sqrt(25+9+4)sqrt(1+9+4))}}$

$= a r c cos (\frac{18}{\sqrt{2 \cdot 19 \cdot 2 \cdot 7}})$ $=arc cos(18/(sqrt(2*19*2*7)))$

$= a r c cos (\frac{9}{\sqrt{133}})$ $=arc cos(9/sqrt(133))$

$= a r c cos (\frac{9}{11.53})$ $=arc cos(9/11.53)$

$\approx a r c cos (0.7806)$ $~~arc cos (0.7806)$

$\approx {38.69}^{\circ}$ $~~38.69^@$

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How do I find the angle between the planes 5(x+1) + 3(y+2) + 2z = 0 and x + 3(y-1) + 2(z+4) = 0?

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