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 \(||v|| = 2\)

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