Question

Find a unit vector perpendicular to each of the vectors `( -> a+ -> b)` and `( -> a- -> b)` , where ` -> a= hat i+ hat j+ hat k , -> b= hat i+2 hat j+3 hat k` .

Answer 1

We have
`(vec(a)+vec(b)) = (hat(i) + hat(j) + hat(k)) + (hat(i) + 2 hat(j) + 3 hat(k))=(2hat(i) + 3 hat(j) + 4 hat(k)) and`
`(vec(a) - vec(b)) = (hat(i) + hat(j) + hat(k))- (hat(i) + 2 hat(j) + 3 hat(k)) = (- hat(j) - 2 hat(k)).`
`:. (vec(a) + vec(b)) xx (vec(a) - vec(b))= |(hat(i),hat(j),hat(k)),(2,3,4),(0,-1,-2)|`
`=(-6+4)hat(i)-(-4-0)hat(j) + (-2-0)hat(k)`
`=(-2hat(i) + 4 hat(j)-2 hat(k)).`
`:. |(vec(a) + vec(b)) xx(vec(a)-vec(b))|= sqrt((-2)^(2)+4^(2)+(-2)^(2))=sqrt((24)) = 2sqrt(6).`
So, the vectors of magnitude 5 units and perpendicular to each the vectors `(vec(a)+vec(b)) and (vec(a)-vec(b))` are:
`pm(5{(vec(a) + vec(b)) xx(vec(a)- vec(b))})/(|(vec(a) + vec(b)) xx (vec(a) -vec(b))|)=pm (5(-2hat(i) +4hat(j) -2 hat(k)))/(2sqrt(6))=pm (5(-hat(i) + 2 hat(j) - hat(k)))/sqrt(6).`