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Chapter 1: Problem 3

Determine the angle of intersection of the two planes \(5 x+3 y+\) \(2 z-4=0\) and \(3 x+4 y-7 z=0\).

Short Answer

Expert verified
The angle of intersection is approximately 75.80 degrees.

Step by step solution

01

- Identify normal vectors

For the plane equation in the form of \( Ax + By + Cz + D = 0 \), the normal vector is \( \textbf{n} = (A, B, C) \). For the first plane \( 5x + 3y + 2z - 4 = 0 \), the normal vector is \( \textbf{n}_1 = (5, 3, 2) \). For the second plane \( 3x + 4y - 7z = 0 \), the normal vector is \( \textbf{n}_2 = (3, 4, -7) \).
02

- Find the dot product of the normal vectors

The dot product of two vectors \( \textbf{a} = (a_1, a_2, a_3) \) and \( \textbf{b} = (b_1, b_2, b_3) \) is defined as \( \textbf{a} \bullet \textbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \). Therefore, for \( \textbf{n}_1 \bullet \textbf{n}_2 \), calculate \( 5 \times 3 + 3 \times 4 + 2 \times (-7) \). This results in \( 15 + 12 - 14 = 13 \).
03

- Calculate magnitudes of the normal vectors

The magnitude of a vector \( \textbf{a} = (a_1, a_2, a_3) \) is given by \( orm{\textbf{a}} = \sqrt{a_1^2 + a_2^2 + a_3^2} \). For \( \textbf{n}_1 = (5, 3, 2) \), the magnitude is \( orm{\textbf{n}_1} = \sqrt{5^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38} \). For \( \textbf{n}_2 = (3, 4, -7) \), the magnitude is \( orm{\textbf{n}_2} = \sqrt{3^2 + 4^2 + (-7)^2} = \sqrt{9 + 16 + 49} = \sqrt{74} \).
04

- Use the dot product formula to find the angle

The angle \( \theta \) between two vectors can be found using the dot product: \( \textbf{a} \bullet \textbf{b} = orm{\textbf{a}} orm{\textbf{b}} \, \theta \). Thus, \( \cos \, \theta = \frac{\textbf{a} \bullet \textbf{b}}{orm{\textbf{a}} orm{\textbf{b}}} \). Plugging in the values, we get \( \cos \, \theta = \frac{13}{\sqrt{38} \sqrt{74}} = \frac{13}{\sqrt{2812}} = \frac{13}{53} \approx 0.245 \). Hence, \( \theta = \cos^{-1}(0.245) \).
05

- Calculate the angle

Using a calculator, find the inverse cosine of 0.245 to get \( \theta \). Thus, \( \theta \approx 75.80 \) degrees.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

normal vectors
In the context of planes in 3D space, normal vectors are essential. A normal vector to a plane is a vector perpendicular to every line on that plane.

For the general plane equation of the form, \( Ax + By + Cz + D = 0 \)

The normal vector can be directly extracted. It is the vector made up of the coefficients of the variables \( A, B, \) and \( C \).

For instance, for the plane \( 5x + 3y + 2z - 4 = 0 \), the normal vector is \( \mathbf{n}_1 = (5, 3, 2) \).

Similarly, for the plane \( 3x + 4y - 7z = 0 \), the normal vector is \( \mathbf{n}_2 = (3, 4, -7) \).

These normal vectors are indispensable in finding the angle of intersection between the two planes.
dot product
The dot product, or scalar product, is a way to multiply two vectors that results in a scalar (a number, not a vector).

The formula for the dot product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is:
\( \mathbf{a} \bullet \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

In our problem, the normal vectors are \( \mathbf{n}_1 = (5, 3, 2) \) and \( \mathbf{n}_2 = (3, 4, -7) \).

Hence, their dot product is:

\( 5 \times 3 + 3 \times 4 + 2 \times -7 = 15 + 12 - 14 = 13 \).

This dot product helps in the calculation of the cosine of the angle between the two vectors later on.
vector magnitudes
The magnitude (or length) of a vector \( \mathbf{a} = (a_1, a_2, a_3) \) is calculated using the formula:

\( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

This is similar to finding the distance in 3D space. For the normal vector \( \mathbf{n}_1 = (5, 3, 2) \), the magnitude is:

\( \| \mathbf{n}_1 \| = \sqrt{5^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38} \)

Similarly, for \( \mathbf{n}_2 = (3, 4, -7) \), the magnitude is:

\( \| \mathbf{n}_2 \| = \sqrt{3^2 + 4^2 + (-7)^2} = \sqrt{9 + 16 + 49} = \sqrt{74} \)

The magnitudes are necessary to determine the cosine of the angle between the vectors.
cosine of angle
To determine the angle between two vectors from their dot product and magnitudes, we use the formula for the cosine of the angle \( \theta \):

\( \cos \theta = \frac{\mathbf{a} \bullet \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \)

Plugging in our values, we get:

\( \cos \theta = \frac{13}{\sqrt{38} \sqrt{74}} = \frac{13}{\sqrt{2812}} \)

This simplifies to:

\( \frac{13}{53} \approx 0.245 \)

This cosine value represents the cosine of the angle formed by our two normal vectors, and thus the planes.
inverse cosine
Finally, to find the actual angle \( \theta \), we take the inverse cosine (or arccosine) of the value obtained in the previous step.

The expression is:

\( \theta = \cos^{-1}(0.245) \)

Using a calculator, this yields:

\( \theta \approx 75.80 \) degrees.

This is the measure of the angle of intersection between the two planes.

The inverse cosine function is crucial here as it reverses the cosine function, allowing us to find the angle given the cosine value.

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Most popular questions from this chapter

Given the parametrized curve (helix) $$ \alpha(s)=\left(a \cos \frac{s}{c}, a \sin \frac{s}{c}, b \frac{s}{c}\right), \quad s \in R, $$ where \(c^{2}=a^{2}+b^{2}\), a. Show that the parameter \(s\) is the arc length. b. Determine the curvature and the torsion of \(\alpha\). c. Determine the osculating plane of \(\alpha\). d. Show that the lines containing \(n(s)\) and passing through \(\alpha(s)\) meet the \(z\) axis under a constant angle equal to \(\pi / 2\). e. Show that the tangent lines to \(\alpha\) make a constant angle with the \(z\) axis.

Given two planes \(a_{i} x+b_{i} y+c_{i} z+d_{i}=0, i=1,2\), prove that a necessary and sufficient condition for them to be parallel is $$ \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}, $$ where the convention is made that if a denominator is zero, the corresponding numerator is also zero (we say that two planes are parallel if they either coincide or do not intersect).

Let \(\alpha: I \rightarrow R^{3}\) be a parametrized regular curve (not necessarily by arc length) with \(k(t) \neq 0, \tau(t) \neq 0, t \in I\). The curve \(\alpha\) is called a Bertrand curve if there exists a curve \(\bar{\alpha}: I \rightarrow R^{3}\) such that the normal lines of \(\alpha\) and \(\bar{\alpha}\) at \(t \in I\) are equal. In this case, \(\bar{\alpha}\) is called a Bertrand mate of \(\alpha\), and we can write $$ \bar{\alpha}(t)=\alpha(t)+r n(t) $$ Prove that a. \(r\) is constant. b. \(\alpha\) is a Bertrand curve if and only if there exists a linear relation $$ A k(t)+B \tau(t)=1, \quad t \in I, $$ where \(A, B\) are nonzero constants and \(k\) and \(\tau\) are the curvature and torsion of \(\alpha\), respectively. c. If \(\alpha\) has more than one Bertrand mate, it has infinitely many Bertrand mates. This case occurs if and only if \(\alpha\) is a circular helix.

Let \(\alpha: I \rightarrow R^{3}\) be a regular parametrized curve (not necessarily by \(\operatorname{arc}\) length and let \(\beta: J \rightarrow R^{3}\) be a reparametrization of \(\alpha(I)\) by the arc length \(s=s(t)\), measured from \(t_{0} \in I\) (see Remark 2). Let \(t=t(s)\) be the inverse function of \(s\) and set \(d \alpha / d t=\alpha^{\prime}, d^{2} \alpha / d t^{2}=\alpha^{\prime \prime}\), etc. Prove that a. \(d t / d s=1 /\left|\alpha^{\prime}\right|, d^{2} t / d s^{2}=-\left(\alpha^{\prime} \cdot \alpha^{\prime \prime} /\left|\alpha^{\prime}\right|^{4}\right)\). b. The curvature of \(\alpha\) at \(t \in I\) is $$ k(t)=\frac{\left|\alpha^{\prime} \wedge \alpha^{\prime \prime}\right|}{\left|\alpha^{\prime}\right|^{3}} . $$ c. The torsion of \(\alpha\) at \(t \in I\) is $$ \tau(t)=-\frac{\left(\alpha^{\prime} \wedge \alpha^{\prime \prime}\right) \cdot \alpha^{\prime \prime \prime}}{\left|\alpha^{\prime} \wedge \alpha^{\prime \prime}\right|^{2}} $$ d. If \(\alpha: I \rightarrow R^{2}\) is a plane curve \(\alpha(t)=(x(t), y(t))\), the signed curvature (see Remark 1) of \(\alpha\) at \(t\) is $$ k(t)=\frac{x^{\prime} y^{\prime \prime}-x^{\prime \prime} y^{\prime}}{\left(\left(x^{\prime}\right)^{2}+\left(y^{\prime}\right)^{2}\right)^{3 / 2}} . $$

Find all unit vectors which are perpendicular to the vector \((2,2,1)\) and parallel to the plane determined by the points \((0,0,0),(1,-2,1)\), \((-1,1,1)\).
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