Question

find the cosine of the angle between the planes 3x + 4y + 4z = 4 and -2x - 3y - 3z = 0. cosθ = input box submit answer next item

Explanation:

Step1: Recall the formula for the cosine of the angle between two planes.

The formula for the cosine of the angle \(\theta\) between two planes \(A_1x + B_1y + C_1z = D_1\) and \(A_2x + B_2y + C_2z = D_2\) is given by:
\[ \cos\theta=\frac{\vert A_1A_2 + B_1B_2 + C_1C_2\vert}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}} \]
For the planes \(3x + 4y + 4z = 4\) and \(-2x - 3y - 3z = 0\), we have \(A_1 = 3\), \(B_1 = 4\), \(C_1 = 4\), \(A_2=-2\), \(B_2 = -3\), \(C_2=-3\).

Step2: Calculate the numerator.

First, calculate \(A_1A_2 + B_1B_2 + C_1C_2\):
\[ \begin{align*} A_1A_2 + B_1B_2 + C_1C_2&=(3)(-2)+(4)(-3)+(4)(-3)\\ &=-6-12 - 12\\ &=-30 \end{align*} \]
Take the absolute value: \(\vert - 30\vert=30\)

Step3: Calculate the denominators.

Calculate \(\sqrt{A_1^2 + B_1^2 + C_1^2}\):
\[ \begin{align*} \sqrt{A_1^2 + B_1^2 + C_1^2}&=\sqrt{3^2+4^2 + 4^2}\\ &=\sqrt{9 + 16+16}\\ &=\sqrt{41} \end{align*} \]
Calculate \(\sqrt{A_2^2 + B_2^2 + C_2^2}\):
\[ \begin{align*} \sqrt{A_2^2 + B_2^2 + C_2^2}&=\sqrt{(-2)^2+(-3)^2+(-3)^2}\\ &=\sqrt{4 + 9+9}\\ &=\sqrt{22} \end{align*} \]

Step4: Calculate \(\cos\theta\).

Now, substitute the values into the formula:
\[ \cos\theta=\frac{30}{\sqrt{41}\sqrt{22}}=\frac{30}{\sqrt{41\times22}}=\frac{30}{\sqrt{902}}\approx\frac{30}{30.03}\approx0.999 \]
But we can also rationalize or keep it as \(\frac{30}{\sqrt{902}}\) or simplify \(\sqrt{902}\) is \(\sqrt{41\times22}\), but let's check the calculation again. Wait, maybe I made a mistake in signs? Wait, the formula is \(\vert A_1A_2 + B_1B_2 + C_1C_2\vert\), so:

Wait, \(A_1 = 3\), \(A_2=-2\), \(B_1 = 4\), \(B_2=-3\), \(C_1 = 4\), \(C_2=-3\)

So \(A_1A_2=3\times(-2)=-6\), \(B_1B_2 = 4\times(-3)=-12\), \(C_1C_2=4\times(-3)=-12\)

Sum: \(-6-12 - 12=-30\), absolute value is \(30\)

Denominator 1: \(\sqrt{3^2 + 4^2+4^2}=\sqrt{9 + 16 + 16}=\sqrt{41}\)

Denominator 2: \(\sqrt{(-2)^2+(-3)^2+(-3)^2}=\sqrt{4 + 9+9}=\sqrt{22}\)

So \(\cos\theta=\frac{30}{\sqrt{41}\times\sqrt{22}}=\frac{30}{\sqrt{902}}\approx\frac{30}{30.033}\approx0.999\) or we can write it as \(\frac{30}{\sqrt{902}}\) and rationalize:

\(\frac{30\sqrt{902}}{902}=\frac{15\sqrt{902}}{451}\approx\frac{15\times30.033}{451}\approx\frac{450.495}{451}\approx0.999\)

Wait, but maybe there is a miscalculation. Wait, let's check the planes again. The first plane is \(3x + 4y + 4z = 4\), normal vector \(\vec{n_1}=(3,4,4)\). The second plane is \(-2x - 3y - 3z = 0\), normal vector \(\vec{n_2}=(-2,-3,-3)\). The formula for the angle between two planes is the angle between their normal vectors, so \(\cos\theta=\frac{\vert\vec{n_1}\cdot\vec{n_2}\vert}{\vert\vec{n_1}\vert\vert\vec{n_2}\vert}\)

\(\vec{n_1}\cdot\vec{n_2}=3\times(-2)+4\times(-3)+4\times(-3)=-6 - 12-12=-30\), absolute value is \(30\)

\(\vert\vec{n_1}\vert=\sqrt{3^2 + 4^2+4^2}=\sqrt{9 + 16 + 16}=\sqrt{41}\)

\(\vert\vec{n_2}\vert=\sqrt{(-2)^2+(-3)^2+(-3)^2}=\sqrt{4 + 9+9}=\sqrt{22}\)

So \(\cos\theta=\frac{30}{\sqrt{41}\sqrt{22}}=\frac{30}{\sqrt{902}}\approx0.999\), but let's compute \(\sqrt{41}\approx6.403\), \(\sqrt{22}\approx4.690\), so \(\sqrt{41}\times\sqrt{22}\approx6.403\times4.690\approx30.03\), so \(30\div30.03\approx0.999\), which is approximately \(1\), but that seems odd. Wait, maybe I made a mistake in the normal vectors? Wait, the plane equation is \(Ax + By + Cz = D\), so the normal vector is \((A,B,C)\), that's correct.

Wait, let's check the arithmetic again. \(3\times(-2)=-6\), \(4\times(-3)=-12\), \(4\times(-3)=-12\). Sum: \(-6-12 = -18\), \(-18-12=-30\). Absolute value is \(30\). \(\vert\vec{n_1}\vert=\sqrt{9 + 16 + 16}=\sqrt{41}\approx6.403\). \(\vert\vec{n_2}\vert=\sqrt{4 + 9 + 9}=\sqrt{22}\approx4.690\). Then \(6.403\times4.690\approx30.03\), so \(30\div30.03\approx0.999\), which is very close to \(1\), meaning the angle between the planes is very close to \(0\) degrees, i.e., the planes are almost parallel.

Answer:

\(\frac{30}{\sqrt{902}}\) (or approximately \(0.999\))