find-vector-unit-perpendicular-to-vector-a-1-2-3-and-b-1-0-2-

Question Number 76524 by john santu last updated on 28/Dec/19

f i n d v e c t o r u n i t p e r p e n d i c u l a r

t o v e c t o r \bar{a} = (1, 2, 3) a n d \bar{b} = (- 1, 0, 2)

Answered by MJS last updated on 28/Dec/19

| \begin{matrix} 1 & - 1 & u_{x} \\ 2 & 0 & u_{y} \\ 3 & 2 & u_{z} \end{matrix} | = 4 u_{x} - 5 u_{y} + 2 u_{z} = (\begin{matrix} 4 \\ - 5 \\ 2 \end{matrix})

abs (\begin{matrix} 4 \\ - 5 \\ 2 \end{matrix}) = 3 \sqrt{5}

\frac{1}{3 \sqrt{5}} (\begin{matrix} 4 \\ - 5 \\ 2 \end{matrix}) = (\begin{matrix} \frac{4 \sqrt{5}}{15} \\ - \frac{\sqrt{5}}{3} \\ \frac{2 \sqrt{5}}{15} \end{matrix})

Commented by benjo 1/2 santuyy last updated on 28/Dec/19

w h y s i r n o t m a k e \pm ?

Commented by MJS last updated on 28/Dec/19

of course both + and - are ok, I thought one

of these is enough \dots

Commented by benjo 1/2 santuyy last updated on 28/Dec/19

o o o k s i r t h a n k s y o u

Answered by vishalbhardwaj last updated on 28/Dec/19

\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} and \vec{b} = - \hat{i} + 2 \hat{k}

\vec{a} \times \vec{b} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) \times (- \hat{i} + 2 \hat{k})

= - 2 (\hat{i} \times \hat{k}) - 2 (\hat{j} \times \hat{i}) + 4 (\hat{j} \times \hat{k})

- 3 (\hat{k} \times \hat{i})

= 2 \hat{j} + 2 \hat{k} + 4 \hat{i} - 3 \hat{j}

= 4 \hat{i} - \hat{j} + 2 \hat{k}

and ∣ \vec{a} \times \vec{b} ∣= \sqrt{{(4)}^{2} + {(- 1)}^{2} + {(2)}^{2}}

= \sqrt{16 + 1 + 4} = \sqrt{21}

let \vec{p} = \vec{a} \times \vec{b}, \hat{p} = \frac{\vec{p}}{∣ \vec{p} ∣} = \frac{4 \hat{i} - \hat{j} + 2 \hat{k}}{\sqrt{21}}

\Rightarrow \hat{p} = \frac{4}{\sqrt{21}} \hat{i} - \frac{1}{\sqrt{21}} \hat{j} + \frac{2}{\sqrt{21}} \hat{k}

Commented by benjo 1/2 santuyy last updated on 28/Dec/19

s o r r y s i r . b u t w r o n g a \times b

Commented by benjo 1/2 santuyy last updated on 28/Dec/19

i g o t a \times b = 4 i - 5 j + 2 k

Commented by JDamian last updated on 28/Dec/19

i \times 2 k \neq - 2 (i \times k)

i \times 2 k = - 2 j

s a n t u y y i s r i g h t

Commented by benjo 1/2 santuyy last updated on 28/Dec/19

t h a n k s y o u s i r

Answered by vishalbhardwaj last updated on 28/Dec/19

\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} and \vec{b} = - \hat{i} + 2 \hat{k}

\vec{a} \times \vec{b} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) \times (- \hat{i} + 2 \hat{k})

= - 2 (\hat{i} \times \hat{k}) - 2 (\hat{j} \times \hat{i}) + 4 (\hat{j} \times \hat{k})

- 3 (\hat{k} \times \hat{i})

= 2 \hat{j} + 2 \hat{k} + 4 \hat{i} - 3 \hat{j}

= 4 \hat{i} - \hat{j} + 2 \hat{k}

and ∣ \vec{a} \times \vec{b} ∣= \sqrt{{(4)}^{2} + {(- 1)}^{2} + {(2)}^{2}}

= \sqrt{16 + 1 + 4} = \sqrt{21}

let \vec{p} = \vec{a} \times \vec{b}, \hat{p} = \frac{\vec{p}}{∣ \vec{p} ∣} = \frac{4 \hat{i} - \hat{j} + 2 \hat{k}}{\sqrt{21}}

\Rightarrow \hat{p} = \frac{4}{\sqrt{21}} \hat{i} - \frac{1}{\sqrt{21}} \hat{j} + \frac{2}{\sqrt{21}} \hat{k}