Porisma

Worked solution

Combine a known angle and a right-angle condition

The angle between a\mathbf a and b\mathbf b is π/3\pi/3 radians, and the angle between b\mathbf b and a−b\mathbf a-\mathbf b is a right angle. Show that ∣a∣=2∣b∣|\mathbf a|=2|\mathbf b|.

2003/NYJC/I/2

CJC · Vectors Basic · Example 11

Answer

∣a∣=2∣b∣|\mathbf a|=2|\mathbf b|.

Compare your reasoning

Use the given sixty-degree angle

a⋅b=∣a∣∣b∣cos⁡θ=∣a∣∣b∣cos⁡π3=12∣a∣∣b∣\begin{aligned}&\mathbf a\cdot\mathbf b\\ & =|\mathbf a||\mathbf b|\cos\theta\\ & =|\mathbf a||\mathbf b|\cos\frac\pi3\\ & =\frac12|\mathbf a||\mathbf b|\end{aligned}

The given angle requires nonzero a and b.

Use perpendicularity and expand

b⋅(a−b)=0b⋅a−b⋅b=0a⋅b−∣b∣2=0a⋅b=∣b∣2.\begin{aligned}\mathbf b\cdot(\mathbf a-\mathbf b)&=0\\\mathbf b\cdot\mathbf a-\mathbf b\cdot\mathbf b&=0\\\mathbf a\cdot\mathbf b-|\mathbf b|^2&=0\\\mathbf a\cdot\mathbf b&=|\mathbf b|^2.\end{aligned}

Commutativity and b dot b turn the second condition into a second expression for the same scalar product.

Equate and divide by the nonzero magnitude

12∣a∣∣b∣=∣b∣2⟹∣a∣=2∣b∣\frac12|\mathbf a||\mathbf b|=|\mathbf b|^2\quad\Longrightarrow\quad |\mathbf a|=2|\mathbf b|

The conclusion concerns lengths; it does not say the vectors are parallel.

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Solution notes

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