what is vector unit orthogonal to (1,2,−2) and parallel to yz−plane?


Question and Answers Forum

All Questions Topic List
Vector Questions
Previous in All Question Next in All Question
Previous in Vector Next in Vector
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

	$parallel to y z - plane = orthogonal to x - axis$ $vector orthogonal to 2 given vectors$ $(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}), (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) can be calculated with a$ $determinant$ $u_{x} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), u_{y} = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), u_{z} = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$ $\| \begin{matrix} x_{1} & x_{2} & u_{x} \\ y_{1} & y_{2} & u_{y} \\ z_{1} & z_{2} & u_{z} \end{matrix} \| = (y_{1} z_{2} - y_{2} z_{1}) u_{x} + (x_{2} z_{1} - x_{1} z_{2}) u_{y} + (x_{1} y_{2} - x_{2} y_{1}) u_{z} =$ $= (\begin{matrix} y_{1} z_{2} - y_{2} z_{1} \\ x_{2} z_{1} - x_{1} z_{2} \\ x_{1} y_{2} - x_{2} y_{1} \end{matrix})$ $in our case this gives$ $(\begin{matrix} 0 \\ - 2 \\ - 2 \end{matrix}) with length 2 \sqrt{2}$ $\Rightarrow answer is \pm (\begin{matrix} 0 \\ \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \end{matrix})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum