2025 JC 1 H2 Math Vectors 1 Summary Flashcards

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A comprehensive set of vocabulary flashcards covering the fundamental formulas and properties of vectors, including scalar and vector products, projections, and geometric distances for H2 Mathematics.

16 Terms

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Scalar Product (Dot Product)

a⋅b=∣a∣∣b∣cos⁡(θ)a \cdot b = |a||b|\cos(\theta)

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Scalar Product Rule for Self-Product

a⋅a=∣a∣2a \cdot a = |a|^2

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Condition for Perpendicularity (Scalar Product)

If aa and bb are perpendicular, then a⋅b=0a \cdot b = 0. Converesly, if a⋅b=0a \cdot b = 0, then either a=0a = 0, b=0b = 0, or aa and bb are perpendicular.

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Scalar Product of Standard Basis Vectors

i⋅i=j⋅j=k⋅k=1i \cdot i = j \cdot j = k \cdot k = 1 and i⋅j=0i \cdot j = 0, j⋅k=0j \cdot k = 0, k⋅i=0k \cdot i = 0

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Magnitude of Vector OP

∣OP∣=x2+y2+z2|OP| = \sqrt{x^2 + y^2 + z^2} for OP=xi+yj+zkOP = xi + yj + zk

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Unit Vector

A vector in the direction of uu defined as u^=u∣u∣\hat{u} = \frac{u}{|u|}, such that u=∣u∣u^u = |u|\hat{u}

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Length of Projection of a on b

∣a⋅b^∣=∣a⋅b∣∣b∣|a \cdot \hat{b}| = \frac{|a \cdot b|}{|b|}

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Ratio Theorem

Formula to find the position vector of point PP dividing segment ABAB in ratio λ:μ\lambda : \mu: OP=μOA+λOBλ+μOP = \frac{\mu OA + \lambda OB}{\lambda + \mu}

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Vector Product (Cross Product)

a×b=∣a∣∣b∣sin⁡(θ)n^a \times b = |a||b|\sin(\theta)\hat{n}, where n^\hat{n} is determined by the right hand rule.

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Vector Product Anti-commutativity

a×b=−(b×a)a \times b = -(b \times a)

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Vector Product Rule for Self-Product

a×a=0a \times a = 0

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Shortest Distance from point C to Line AB

Perpendicular distance = ∣AC×d^∣|AC \times \hat{d}| where d^\hat{d} is the unit direction vector of the line (d∣d∣\frac{d}{|d|}).

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Vector Product of Standard Basis Vectors

i×i=0i \times i = 0, j×j=0j \times j = 0, k×k=0k \times k = 0 and i×j=ki \times j = k, j×k=ij \times k = i, k×i=jk \times i = j

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Angle between 2 vectors (Cosine Formula)

θ=cos⁡−1(AB⋅AC∣AB∣∣AC∣)\theta = \cos^{-1}\left(\frac{AB \cdot AC}{|AB||AC|}\right)

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Area of Triangle ABC

12∣AB×AC∣=12∣BA×BC∣=12∣CA×CB∣\frac{1}{2} |AB \times AC| = \frac{1}{2} |BA \times BC| = \frac{1}{2} |CA \times CB|

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Area of Parallelogram ABPC

∣AB×AC∣=2×Area of triangle ABC|AB \times AC| = 2 \times \text{Area of triangle } ABC