Projections and Planes

Chapter 4 • Section 4-2

"Shadows and Surfaces... where geometry meets algebra! 📐"

⚫ The Dot Product

For $\vec{u} = (u_1, \dots, u_n)$ and $\vec{v} = (v_1, \dots, v_n)$:

$$ \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + \dots + u_nv_n $$

$\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ (Commutative)
$\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$ (Distributive)
$(k\vec{u}) \cdot \vec{v} = k(\vec{u} \cdot \vec{v})$ (Associative)
$\vec{u} \cdot \vec{u} = \norm{\vec{u}}^2 \geq 0$

The dot product relates to the angle $\theta$ between vectors:

$$ \vec{u} \cdot \vec{v} = \norm{\vec{u}} \norm{\vec{v}} \cos \theta $$

If $\vec{u} \cdot \vec{v} = 0$, the vectors are orthogonal (perpendicular).

Are $\vec{u}=(1, -2, 3)$ and $\vec{v}=(4, 5, 2)$ orthogonal?

$$ \vec{u} \cdot \vec{v} = 1(4) + (-2)(5) + 3(2) = 4 - 10 + 6 = 0 $$

Yes, they are orthogonal!

Vector $\vec{u}$

Vector $\vec{v}$

The projection of $\vec{u}$ onto $\vec{d}$ is the "shadow" of $\vec{u}$ on the line along $\vec{d}$.

$$ \proj_{\vec{d}} \vec{u} = \frac{\vec{u} \cdot \vec{d}}{\norm{\vec{d}}^2} \vec{d} $$

Find the shortest distance from point $P(3, 2, -1)$ to the line $L$ passing through $P_0(2, 1, 3)$ with direction $\vec{d}=(3, -1, -2)$.

Vector $\vec{v} = \overrightarrow{P_0P} = (3-2, 2-1, -1-3) = (1, 1, -4)$.
Projection $\vec{w} = \proj_{\vec{d}} \vec{v} = \frac{\vec{v} \cdot \vec{d}}{\norm{\vec{d}}^2} \vec{d}$.
$\vec{v} \cdot \vec{d} = 1(3) + 1(-1) + (-4)(-2) = 10$.
$\norm{\vec{d}}^2 = 9 + 1 + 4 = 14$.
$\vec{w} = \frac{10}{14}(3, -1, -2) = \frac{5}{7}(3, -1, -2)$.
Distance is $\norm{\vec{v} - \vec{w}}$. (Calculation omitted for brevity, but it's the length of the orthogonal component).

A plane with normal vector $\vec{n} = (a, b, c)$ passing through point $P_0(x_0, y_0, z_0)$ has the equation:

$$ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 $$

or $ax + by + cz = d$

Find the distance from $P(2, 3, 0)$ to the plane $5x + y + z = -1$.

Formula: $D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$

$$ D = \frac{|5(2) + 1(3) + 1(0) - (-1)|}{\sqrt{25 + 1 + 1}} = \frac{|10 + 3 + 1|}{\sqrt{27}} = \frac{14}{3\sqrt{3}} $$

Find the equation of the plane passing through $(1, 0, 0)$ with normal vector $\vec{n} = (2, 3, -1)$.