Given that vector p=i+2j-2k and q=2i-j+2k, find two vectors m and m satisfying the condition; m is perpendicular to p

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Step-by-step explanation:

There are an infinite number of possibilities. Any vector whose dot-product with p is zero will be perpendicular to p.

Let m = 0i +1j +ak. Then we require ...

  m·p = 0 = 0×1 +1×2 +a(-2)   ⇒   0 = 2 -2a   ⇒   a = 1

m = 0i +1j +1k

__

Let n = 2i +0j +bk

  n·p = 0 = 2×1 +0×2 +b(-2)   ⇒   2 -2b = 0   ⇒   b = 1

n = 2i +0j +1k

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!