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Let vector A (1,0,-1), vector B (-2,2,1), vector C (x,y,z) (x>0) and | c | = 3 When vector C is perpendicular to both vector A and vector B, then

Let Vector A 101 Vector B 221 Vector C Xyz Xgt0 And C 3 When Vector C Is Perpendicular To Both Vector A And Vector B Then class=

Sagot :

Answer:

x = 2 , y = 1 , z = 2

Step-by-step explanation:

[tex]\overrightarrow{a} \perp \overrightarrow{c} \Longrightarrow \left( \begin{gathered}1\\ 0\\ -1\end{gathered} \right) \cdot \left( \begin{gathered}x\\ y\\ z\end{gathered} \right) =0 \Longleftrightarrow x - z=0 \Longleftrightarrow x = z[/tex]

[tex]\overrightarrow{b} \perp \overrightarrow{c} \Longrightarrow \left( \begin{gathered}-2\\ 2\\ 1\end{gathered} \right) \cdot \left( \begin{gathered}x\\ y\\ z\end{gathered} \right) =0 \Longleftrightarrow -2x +2y +z=0[/tex]

Now ,we have to solve the system:

[tex]\begin{cases}x=z&\\ -2x+2y+z=0&\end{cases}[/tex]

[tex]\Longleftrightarrow \begin{cases}x=z&\\ -x+2y=0&\end{cases}[/tex]

[tex]\Longleftrightarrow \begin{cases}x=z&\\ x=2y&\end{cases}[/tex]

then x = 2y = z

[tex]\Longrightarrow \overrightarrow{c} \left( \begin{gathered}2y\\ y\\ 2y\end{gathered} \right)[/tex]

|C| = 3 ⇒ (2y)² + y² + (2y)² = 9 ⇒ 9y² = 9 ⇒ y² = 1 ⇒ y = ±1 ⇒ y = 1

(x>0  ⇒ y>0)

[tex]\Longrightarrow \overrightarrow{c} \left( \begin{gathered}2\\ 1\\ 2\end{gathered} \right)[/tex]

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