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\( \| \vec{x} \cdot \vec{y} \| \le \|\vec{x}\|\|\vec{y}\| \), aka Cauchy-Schwartz Inequality
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\( \| \vec{x} + \vec{y} \| \le \|\vec{x}\| + \|\vec{y}\| \), aka Triangular Inequality
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\( \vec{a} \cdot \vec{a} = \|\vec{a}\|^2 \), which can be easily verified.
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\( \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos{\theta} \), where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). Can be verified based on (3).
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\(\displaystyle \vec{a} \times \vec{b} = \|\vec{a}\| \|\vec{b}\| \sin{\theta} \, n \), where \(n\) is the unit vector perpendicular to the plane containing \(\vec{a}\) and \(\vec{b}\). Note the cross product between two vectors is applicable only in \(\mathbb{R}^3\). Based on (4), there is a nice proof at Khan Academy.
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\( \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} (\vec{a} \cdot \vec{c}) - \vec{c} (\vec{a} \cdot \vec{b}) \). Very optional with nice proof at Khan Academy.
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How to determine the linear equation, \(Ax + By + Cz = D\), of a plane in \(\mathbb{R}^3\) if given a point on the plane, and a normal vector, which is the vector perpendicular to the plane ?
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Conversely, how to determine the normal vector if given the linear equation of a plane in \(\mathbb{R}^3\) ? Turns out if the plan is \(Ax + By + Cz = D\), the normal vector \(\vec{n}\) is simply \(Ai + Bj + Ck\), where \(i, j, k\) are the unit vectors of the respective \(x, y, z\) coordinates.
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Given a plane, \(Ax + By + Cz = D\), and a point vector \(x_0i + y_0j + z_0k \) outside of the plane, how to determine the shortest distance \(d\) from the point vector to the plane ? Turns out: \(\displaystyle d = \frac{Ax_0 + By_0 + Cz_0 - D}{\sqrt{A^2 + B^2 + C^2}}\)
Footnotes:
- When is \( \| \vec{x} \cdot \vec{y} \| = \|\vec{x}\|\|\vec{y}\| \) ?
- When is \( \| \vec{x} + \vec{y} \| = \|\vec{x}\| + \|\vec{y}\| \) ?
- What is the angle \(\theta\) between two vectors when one of them is a zero vector ?
- The word "perpendicular" is meaningful only when the angle is defined.
- When the dot product of two vectors is zero, are they always perpendicular ? Or, are they always orthogonal ? Why ?
- Intuition of a dot product: how much are the two vectors moving in the same direction ?
- Intuition of a cross product: how much is the area (as a parallelogram) between the two vectors ?
- The dot product is a scalar, but the cross product is a vector.
- The cross product of two vectors is always orthogonal to each of the two original vectors. This can be used to find the normal vector of a plane.
- Video on finding the distance between planes as an interesting application.
# posted by rot13(Unafba Pune) @ 8:45 PM
