what-is-vector-unit-orthogonal-to-1-2-2-and-parallel-to-yz-plane-

Question Number 81295 by jagoll last updated on 11/Feb/20

w h a t i s v e c t o r u n i t o r t h o g o n a l

t o (1, 2, - 2) a n d p a r a l l e l t o

y z - p l a n e ?

Commented by john santu last updated on 11/Feb/20

s a y t h i s v e c t o r \bar{b} = (x, y, z)

(i) x^{2} + y^{2} + z^{2} = 1

(i i) \bar{b} . (1, 2, - 2) = 0

x + 2 y - 2 z = 0

(i i i) \bar{b} p a r a l l e l t o y z - p l a n e

\Rightarrow \bar{b} ⊥ \hat{i} \Rightarrow x = 0

t h e n w e g e t y = z = \pm \frac{1}{2} \sqrt{2}

∴ \bar{b} = (0, \frac{1}{2} \sqrt{2}, \frac{1}{2} \sqrt{2})

o r \bar{b} = (0, - \frac{1}{2} \sqrt{2}, - \frac{1}{2} \sqrt{2})

Commented by jagoll last updated on 11/Feb/20

thx

Answered by MJS last updated on 11/Feb/20

parallel to y z - plane = orthogonal to x - axis

vector orthogonal to 2 given vectors

(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}), (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) can be calculated with a

determinant

u_{x} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), u_{y} = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), u_{z} = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})

| \begin{matrix} x_{1} & x_{2} & u_{x} \\ y_{1} & y_{2} & u_{y} \\ z_{1} & z_{2} & u_{z} \end{matrix} | = (y_{1} z_{2} - y_{2} z_{1}) u_{x} + (x_{2} z_{1} - x_{1} z_{2}) u_{y} + (x_{1} y_{2} - x_{2} y_{1}) u_{z} =

= (\begin{matrix} y_{1} z_{2} - y_{2} z_{1} \\ x_{2} z_{1} - x_{1} z_{2} \\ x_{1} y_{2} - x_{2} y_{1} \end{matrix})

in our case this gives

(\begin{matrix} 0 \\ - 2 \\ - 2 \end{matrix}) with length 2 \sqrt{2}

\Rightarrow answer is \pm (\begin{matrix} 0 \\ \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \end{matrix})

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