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astarmathsandphysics:
Yes use augmented matrices. Row reduce and the inverse is on the right hand side. The multiple the vector on right hand side of the original system of equations by the inverse you have found and the answer is the vector  you have found

astarmathsandphysics:

--- Quote from: astarmathsandphysics on February 28, 2009, 08:25:26 am ---Yes use augmented matrices. Row reduce and the inverse is on the right hand side. The multiple the vector on right hand side of the original system of equations by the inverse you have found and the answer is the vector  you have found

--- End quote ---
suppose the equation of a line is r(t) =2i+4j-k+t(5i-2j+7k) the arbitrary point vector is (2+5t,4-2t,-1+7t)

avrila:
P=(4,1,-1) Q=(3,3,5) R=(1,0,2c) S=(1,1,2)

Find C so that QR and PR are orthogonal.
I know that I have to use dot product. QR.PR=0
But I got a weird value of C. Is my method correct? How do you do this.

astarmathsandphysics:

--- Quote from: avrila on March 01, 2009, 01:05:33 am ---P=(4,1,-1) Q=(3,3,5) R=(1,0,2c) S=(1,1,2)

Find C so that QR and PR are orthogonal.
I know that I have to use dot product. QR.PR=0
But I got a weird value of C. Is my method correct? How do you do this.

--- End quote ---
QR=(-2,-3,2c-5) and PR=(-3,-1,2c+1)
QR.PR=6+3+(2c-5)(2c+1)=9+4c^2-8c-5=4c^2-8c+4=0 solution is c=1

avrila:
Ya.. I know where my mistake is. I misplace one of the sign and ruin the question.. Thanks.

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