# Suppose vector u= <2,-1,0> , vector v= <-2,4,-5> and vector w= <2,1,2>Compute the following values: length of ( 2(vector u)-8(vector v)+6(vector w))

*print*Print*list*Cite

### 1 Answer

The given vectors are:

`vecu=lt2,-1,0gt ` `vecv=lt-2,4,-5gt` and `vecw=lt2,1,2gt`

To solve for length of `(2vecu - 8vecv + 6vecw)` , we need to perform scalar multiplication first.

`lambda` `veca` `= lambda <x,y,z>=<lambdax, lambday, lambdaz>`

where `lambda` is any real number.

So,

`2vecu` `=` `2lt2,-1,0gt` `=` `lt2*2 , 2(-1), 2*0gt` `=` `lt4,-2,0gt`

`8vecv` `=` `8lt-2,4,5gt` `=` `lt8(-2), 8*4, 8*5gt` `=` `lt-16, 32, 40gt`

`6vecw` `=` `6lt2,1,2gt` `=` `lt6*2,6*1,6*2gt` `=` `lt12,2,12gt`

Then, do subtraction and addition of vectors.

`vec a` `+-` `vec r` `=` `ltx_1, x_1, y_1gt` `+-` `ltx_2,y_2,z_2gt`

`vec a` `+-` `vecr` `=` `ltx_1+-x_2,y_1+-y_2,z_1+-z_2gt`

So,

`2vecu - 8vecv + 6vecw= lt4,-2,0gt - lt-16,32,40gt + lt12,2,12gt`

`2vecu - 8vecv + 6vecw=lt4-(-16)+12,-2-32+2,0-40+12gt`

`2vecu - 8vecv + 6vecw= lt32, -32,-28gt`

Then, let `vecr = <32,-32,-28>` .

Next, let's determine the length of `vecr` . Note that the length of a vector refers to its magnitude.

So,

`|vecr|=sqrt(x^2+y^2+z^2)=sqrt(32^2+(-32)^2+(-28)^2)`

`|vecr|=sqrt(2832)`

`|vecr|=53.22`

**Hence, the length of `(2vecu - 8vecv + 6vecw)` is 53.22 units.**