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JulijaS [17]
2 years ago
14

Given that u =< 2,12 >, and z =< -7,5 >

Mathematics
1 answer:
swat322 years ago
7 0

Using the dot product:

For any vector x, we have

||x|| = √(x • x)

This means that

||w|| = √(w • w)

… = √((u + z) • (u + z))

… = √((u • u) + (u • z) + (z • u) + (z • z))

… = √(||u||² + 2 (u • z) + ||z||²)

We have

u = ⟨2, 12⟩   ⇒   ||u|| = √(2² + 12²) = 2√37

z = ⟨-7, 5⟩   ⇒   ||z|| = √((-7)² + 5²) = √74

u • z = ⟨2, 12⟩ • ⟨-7, 5⟩ = -14 + 60 = 46

and so

||w|| = √((2√37)² + 2•46 + (√74)²)

… = √(4•37 + 2•46 + 74)

… = √314 ≈ 17.720

Alternatively, without mentioning the dot product,

w = u + z = ⟨2, 12⟩ + ⟨-7, 5⟩ = ⟨-5, 17⟩

and so

||w|| = √((-5)² + 17²) = √314 ≈ 17.720

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Explanation:

Giving the system of equations:

\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}

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Subtracting (2) from twice of (1), we have:

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\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\  \\ 3y+\frac{240}{13}=-3 \\  \\ 3y=-3-\frac{240}{13} \\  \\ 3y=-\frac{279}{13} \\  \\ y=-\frac{279}{39} \end{gathered}

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\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\  \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\  \\ x-\frac{231}{13}=-4 \\  \\ x=-4+\frac{231}{13} \\  \\ x=\frac{179}{13} \end{gathered}

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