Question

find the scalar product (F.d) of given vectors F = 2i - j + 3k , D = 3i + 2j - k​

Answer 1

Answer:

Scalar product F.D=5

Explanation:

F.D=(2i-j+3k) (3i+2j-k)

F.D=(2i-j+3k) (3i+2j-k) =2i(3i+2j-k)-j(3i+2j-k)+3k(3i+2j-k)

F.D=(2i-j+3k) (3i+2j-k) =2i(3i+2j-k)-j(3i+2j-k)+3k(3i+2j-k) =6*i.i+4*i.j-2*k.i-3*i.j+2j.j-j.k+9*k.i+6j.k-3k.k

F.D=(2i-j+3k) (3i+2j-k) =2i(3i+2j-k)-j(3i+2j-k)+3k(3i+2j-k) =6*i.i+4*i.j-2*k.i-3*i.j+2j.j-j.k+9*k.i+6j.k-3k.k =6*1+4*0+2*9-3*0+2*1-1*0+9*0+6*0-3*1

i.j=j.k=k.i=0,i.i=j.j=k.k=0

=6+2-3

F.D=5