Related Formula
A · B = | A| | B| φ$$\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \phi$$
Core Logic
Let the given baseline vectors be p₁, p₂, p₃$\vec{p}_1, \vec{p}_2, \vec{p}_3$. Note that | p₁| = | p₂| = | p₃| = 1$|\vec{p}_1| = |\vec{p}_2| = |\vec{p}_3| = 1$.
- Angle with p₁$\vec{p}_1$ is (π)/(2)$\frac{\pi}{2}$:
u · p₁ = 0 x√(2) + z√(2) = 0 x + z = 0 (i)$$\hat{u} \cdot \vec{p}_1 = 0 \implies \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \implies x + z = 0 \quad \dots (i)$$
- Angle with p₂$\vec{p}_2$ is (π)/(3)$\frac{\pi}{3}$:
u · p₂ = (π)/(3) y√(2) + z√(2) = (1)/(2) y + z = 1√(2) (ii)$$\hat{u} \cdot \vec{p}_2 = \cos\frac{\pi}{3} \implies \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \implies y + z = \frac{1}{\sqrt{2}} \quad \dots (ii)$$
- Angle with p₃$\vec{p}_3$ is (2π)/(3)$\frac{2\pi}{3}$:
u · p₃ = (2π)/(3) x√(2) + y√(2) = -(1)/(2) x + y = - 1√(2) (iii)$$\hat{u} \cdot \vec{p}_3 = \cos\frac{2\pi}{3} \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \implies x + y = -\frac{1}{\sqrt{2}} \quad \dots (iii)$$
Step 1: Finding Vector Coordinates
Subtracting (ii) from (iii):
(x + y) - (y + z) = - 1√(2) - 1√(2) x - z = -√(2) (iv)$$(x + y) - (y + z) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \implies x - z = -\sqrt{2} \quad \dots (iv)$$
Solving (i) and (iv):
2x = -√(2) x = - 1√(2)$$2x = -\sqrt{2} \implies x = -\frac{1}{\sqrt{2}}$$
z = 1√(2)$$z = \frac{1}{\sqrt{2}}$$
From (ii): y = 1√(2) - 1√(2) = 0$y = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$.
So, u = - 1√(2) i + 0 j + 1√(2) k$\hat{u} = -\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.
Step 2: Evaluating the Norm Difference
Given v = 1√(2) i + 1√(2) j + 1√(2) k$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$:
u - v = (- 1√(2) - 1√(2)) i + (0 - 1√(2)) j + ( 1√(2) - 1√(2)) k$$\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right)\hat{j} + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{k}$$
u - v = -√(2) i - 1√(2) j + 0 k$$\hat{u} - \vec{v} = -\sqrt{2}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}$$
Evaluating the squared magnitude:
| u - v|² = (-√(2))² + (- 1√(2))² = 2 + (1)/(2) = (5)/(2)$$|\hat{u} - \vec{v}|^2 = (-\sqrt{2})^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 2 + \frac{1}{2} = \frac{5}{2}$$
Pattern Recognition
Setting up dot products systematically transforms descriptive geometric angles into solvable sets of linear equations.
Chapter Mix
Class 12 Mathematics: Vector Algebra
