How do you solve the equation 𝑤̅×<2,0,−2>=<4,0,4> where 𝑤̅=<𝑤1,𝑤2,𝑤3> is nonzero with a magnitude of 2?

Jul 27, 2018

$\overline{w} = \left(0 , - 2 , 0\right)$.

Explanation:

Given, $\overline{w} \times \left(2 , 0 , - 2\right) = \left(4 , 0 , 4\right) , \text{ where, } \overline{w} = \left({w}_{1} , {w}_{2} , {w}_{3}\right) .$

Now, $\overline{w} \times \left(2 , 0 , - 2\right) = \left(4 , 0 , 4\right)$.

$\therefore | \left(i , j , k\right) , \left({w}_{1} , {w}_{2} , {w}_{3}\right) , \left(2 , 0 , - 2\right) | = 4 i + 0 j + 4 k$.

;. -2w_2i-(-2w_1-2w_3)j+(-2w_2)k=4i+0j+4k.

$\therefore - 2 {w}_{2} = 4 , 2 {w}_{1} + 2 {w}_{3} = 0 \mathmr{and} - 2 {w}_{2} = 4$,

$i . e . , {w}_{2} - 2 , {w}_{1} = - {w}_{3} , \mathmr{and} , {w}_{2} = - 2. \ldots . \left({\ast}_{1}\right)$.

Finally, utilising $| | \overline{w} | | = 2 ,$ we have,

${w}_{1}^{2} + {w}_{2}^{2} + {w}_{3}^{2} = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\ast}_{2}\right)$.

From $\left({\ast}_{1}\right) \mathmr{and} \left({\ast}_{2}\right) , {\left(- {w}_{3}\right)}^{2} + {\left(- 2\right)}^{2} + {w}_{3}^{2} = 4 ,$

$\mathmr{and} , 2 {w}_{3}^{2} = 0 \Rightarrow {w}_{3} = 0 \Rightarrow {w}_{1} = 0. \ldots . . \left[\because \left({\ast}_{1}\right)\right]$.

Thus, $\overline{w} = \left({w}_{1} , {w}_{2} , {w}_{3}\right) = \left(0 , - 2 , 0\right)$, as desired!