Show that the two vectors A= 2i+3j-k and B=3i-j+3k are perpendicular

We know that two vectors are perpendicular when their dot product is equal to 0. As you can see from the SQA Higher Maths formula sheet, the equation for the dot product is; A.B=a1b1+a2b2+a3b3.To answer this question we just substitute in the values of both vectors. Since A = (2,3,-1) we have a1 = 2, a2 = 3 and a3 = -1 (if the student doesn't understand how to convert i,j,k vectors into column or row form I would explain this here) and for B =(3,-1,3) we have b1=3,b2=-1,b3=3. Thus putting this into the equation we get A.B= (2x3)+(3x-1)+(-1x3), by multiplying this out we get A.B= 6 -3 -3 = 0. Thus since A.B=0, the two vectors are perpendicular.