Vectors u = si + 2j - k and v = -3i + tj - 6k are perpendicular - Scottish Highers Maths - Question 1 - 2015

To determine when the two vectors are perpendicular, we calculate their scalar product and set it equal to zero. The scalar product of vectors u and v is given by:

ext{u} ullet ext{v} = (s)(-3) + (2)(t) + (-1)(-6) = -3s + 2t + 6

Setting this equal to zero gives:

$-3s + 2t + 6 = 0$

Rearranging yields:

$2t = 3s - 6$ t = rac{3s - 6}{2}