Question 1086193
The cross product of the two vectors is perpendicular to both vectors,
{{{A=(matrix(3,3,
i,j,k,
1,-2,3,
2,5,0))}}}
{{{A=(-2(0)-3(5))i+(3(2)-0(1))j+(1(5)-2(-2))k}}}
{{{A=-15i+6j+9k}}}
Now find the magnitude of A,
{{{abs(A)=sqrt((-15)^2+6^2+9^2)=sqrt(225+36+81)=sqrt(342)=3sqrt(38)}}}
To make the unit vector, divide A by its magnitude,
{{{u[A]= (-5/(sqrt(38)))i+(2/(sqrt(38)))j+(3/(sqrt(38)))k}}}


{{{u[A]=(sqrt(38)/38)(-5i+2j+3k)}}}