Answer 1

`veca = hat i - 7 hatj +7hatk`    `vecb = 3hati - 2hatj + 2hatk` .

perpandicular vector to both `veca  &  vecb   "is"    vecc` 

`hati = |[hati,hatj,hatk],[1,-7,7],[3,-2,2]|`

 = `hati(0) - hatj(2-21)+hatk(-2+21)`

= `0hati + 19hatj + 19hatk`

⇒ `vecc = 0hati + 19hatj + 19hatk`

`hatc = vecc/|vecc| = (0hati + 19hatj + 19hatk)/sqrt(0^2+19^2+19^2) = (19(hatj+hatk))/(19sqrt(2))`

= `(hatj + hatk)/sqrt(2)`

`vecc = 1/sqrt(2)(hatj+hatk)`

Answer 2

`veca = hat"i" - 7hat"j" + 7hat"k" and vecb = 3hat"i" - 2hat"j" + 2hat"k"`

let `vecn` be the vector perpendicular to `veca  "and"  vecb`

`vecn = veca xx vecb`

`vecn = |(hat"i", hat"j" ,hat"k") ,(1,-7,7),(3,-2,2)| = 19hat"j" + 19hat"k"`

Now, the unit vector perpendicular to `veca  "and"  vecb`

`hatn = (19hat"j" + 19hat"k")/sqrt(19^2 + 19^2) = 1/sqrt(2)(hat"j" + hat"k")`