Question

find a nonzero vector x perpendicular to the vector v= [-2,-8,-7,2] u= [6,7,-2,8] x= [answer,answer,answer,answer]

Answer 1

Answer:

18ij-28kl+46lj

Step-by-step explanation:

A vector is a quantity which has both magnitude and direction.

For vectors u and v, vector perpendicular to both u and v is equal to u×v

Given:

$v= [-2,-8,-7,2]\\u= [6,7,-2,8]$

Solution:

$u\times v=\left | \begin{matrix}i&j&k&l\\-2&-8&-7&2\\6&7&-2&8 \end{matrix} \right |\\=i\left | \begin{matrix}j&k&l\\-8&-7&2\\7&-2&8 \end{matrix} \right |-j\left | \begin{matrix}i&k&l\\-2&-7&2\\6&7&8 \end{matrix} \right |+k\left | \begin{matrix}i&j&l\\-2&-8&2\\6&7&8 \end{matrix} \right |-l\left | \begin{matrix}i&j&k\\-2&-8&-7\\6&7&-2 \end{matrix} \right |$

Solving: $i\left | \begin{matrix}j&k&l\\-8&-7&2\\7&-2&8 \end{matrix} \right |$

$i\left | \begin{matrix}j&k&l\\-8&-7&2\\7&-2&8 \end{matrix} \right |=i\left [ j(-56+4)-k(-64-14)+l(16+49) \right ]=i(-52j+78k+65l)$

Solving: $-j\left | \begin{matrix}i&k&l\\-2&-7&2\\6&7&8 \end{matrix} \right |$

$-j\left | \begin{matrix}i&k&l\\-2&-7&2\\6&7&8 \end{matrix} \right |=-j\left [ i(-56-14)-k(-16-12)+l(-14+42) \right ]=j(70i-28k-28l)$

Solving: $k\left | \begin{matrix}i&j&l\\-2&-8&2\\6&7&8 \end{matrix} \right |$

$k\left | \begin{matrix}i&j&l\\-2&-8&2\\6&7&8 \end{matrix} \right |=k\left [ i(-64-14)-j(-16-12)+l(-14+48) \right ]=k(-78i+28j+34l)$

Solving: $-l\left | \begin{matrix}i&j&k\\-2&-8&-7\\6&7&-2 \end{matrix} \right |$

$-l\left | \begin{matrix}i&j&k\\-2&-8&-7\\6&7&-2 \end{matrix} \right |=-l\left [ i(16+49)-j(4+42)+k(-14+48) \right ]=-l(65i-46j+34k)$

Therefore,

u×v = $i(-52j+78k+65l)+j(70i-28k-28l)+k(-78i+28j+34l)-l(65i-46j+34k)\\=-52ij+78ik+65il+70ij-28jk-28kl-78ik+28jk+34kl-65li+46lj-34kl\\=18ij-28kl+46lj$

So, $18ij-28kl+46lj$ is perpendicular to both u and v