Question

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

Answer 1

x = 2 , y = 1 , z = 2

Explanation:

`\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`

`\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`

Now ,we have to solve the system:

`\begin{cases}x=z&\\ -2x+2y+z=0&\end{cases}`

`\Longleftrightarrow \begin{cases}x=z&\\ -x+2y=0&\end{cases}`

`\Longleftrightarrow \begin{cases}x=z&\\ x=2y&\end{cases}`

then x = 2y = z

`\Longrightarrow \overrightarrow{c} \left( \begin{gathered}2y\\ y\\ 2y\end{gathered} \right)`

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

(x>0 ⇒ y>0)

`\Longrightarrow \overrightarrow{c} \left( \begin{gathered}2\\ 1\\ 2\end{gathered} \right)`