ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: boysenberry on September 11, 2009, 07:29:22 pm
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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.
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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
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;) ...what would you do next after the dot product?
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expand it and you should get two linear equations with two unknowns, for which you can solve.
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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