# Lecture 2

### Topic: Vectors in 3D

Distance Between Points
$$P_1(x_1, y_1, z_1)$$ and $$P_2(x_2, y_2, z_2)$$

$$d(P_1, P_2) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2+(z_2-z_1)^2}$$

EX: Show that $$A(3,4,1)$$, $$B(4,4,6)$$, $$C(3,1,2)$$

$$d(AB) = \sqrt{1^2 + 0^2 + 5^2} = \sqrt{26}$$

The 2D Circle Equation becomes a Sphere Equation: $$(x-h)^2 + (y-k)^2 + (z-L)^2 = r^2$$

Midpoint Formula $$(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}$$ Center: $$(h, k, L)$$ Radius: $$r$$

Just as you would complete the square to find the center and radius of a circle, you’d do the same for spheres.

Position Vectors

$$\vec{v}$$ $$= <v_1, v_2, v_3>$$ or $$\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}$$

$$||\vec{v}|| = \sqrt{(v_1)^2 + (v_2)^2 + (v_3)^2}$$

Remember: If two vectors have the same scalar, they’re parallel.

EX: Find $$\vec{v}$$ such that $$||\vec{2}|| = 2$$ and $$\vec{v}$$ has the same direction as $$\vec{w} = \hat{i} - 2\hat{j} + 3\hat{k}$$

First, find the direction of $$\vec{w}$$: $$||w|| = \sqrt{1^2 + 2^2 + 3^3} = \sqrt{14}$$

Next, find the unit vector since the unit vector tells us the direction $$\vec{v}$$

$$\hat{u}$$ $$=$$ $$\frac{\vec{v}}{||\vec{v}||}$$

$$\hat{u} = \frac{\hat{i}-2\hat{j}+3\hat{k}}{\sqrt{14}}$$

Since, $$\vec{v} = \hat{u} \cdot ||\vec{v}||$$ and $$||v|| = 2$$

$$\vec{v} = \frac{\sqrt{14}}{7}(\hat{i}-2\hat{j}+3\hat{k})$$