Vectors

[boysenberry]

I'm new at vectors. 


Vector xi - j + yk is perpendicular to vectors i + j + k and 2i + j - 3k. Find the values of x  and y.

(xi - j + yk).(i + j +k).(2i +j -3k) = 0
2x - 1 - 3y = 0

Is this right so far? If so, what do I do next? Please show complete working out.

[kamil9876]

The dot product is not done like that. Remember that a dot product turns two vectors into a real number. Hence 3 vectors producted would sort of be a vector since 2 vectors would give a real number which then when multiplied by the third vector would be a vector, though I don't even think this makes sense technically speaking.

Do it by focusing on the PAIRS of vectors.

ie: xi-j+yk and i+j+k would give (xi-j+yk).(i+j+k)=0

xi-j+yk and 2i + j -3k gives (xi-j+yk).(2i + j -3k)=0

[boysenberry]

  ...what would you do next after the dot product?

[kamil9876]

expand it and you should get two linear equations with two unknowns, for which you can solve.

[QuantumJG]

I'm new at vectors. 


Vector xi - j + yk is perpendicular to vectors i + j + k and 2i + j - 3k. Find the values of x  and y.

(xi - j + yk).(i + j +k).(2i +j -3k) = 0
2x - 1 - 3y = 0

Is this right so far? If so, what do I do next? Please show complete working out.

so we have (x,-1,y) perpendicular to (1,1,1) and (2, 1, -3)

so we can say that:

(x,-1,y) dot (1,1,1) = 0

(x,-1,y) dot (2,1,-3) = 0

so x - 1 + y = 0 ...1
   2x -1 - 3y = 0 ...2

(1) => x = 1 - y

.: 2(1 - y) - 1 - 3y = 0

.: 2 - 2y - 1 - 3y = 0 => -5y + 1 = 0 => y = 1/5 => x = 4/5