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Linear Algebra Solutions 91

# Linear Algebra Solutions 91 - or x−2 y−1 z−4 −1...

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Unformatted text preview: or x−2 y−1 z−4 −1 −2 −2 2 −2 −3 = 2x − 7y + 6z − 21 = 0. 16. Non–parallel lines L and M in three dimensional space are given by equations P = A + sX, Q = B + tY. E (i) Suppose P Q is orthogonal to both X and Y . Now E E P Q= Q − P = (B + tY ) − (A + sX ) =AB +tY − sX. Hence E (AB +tY + sX ) · X = 0 E (AB +tY + sX ) · Y = 0. More explicitly E t(Y · X ) − s(X · X ) = − AB ·X E t(Y · Y ) − s(X · Y ) = − AB ·Y. However the coeﬃcient determinant of this system of linear equations in t and s is equal to Y · X −X · X Y · Y −X · Y = −(X · Y )2 + (X · X )(Y · Y ) = ||X × Y ||2 = 0, as X = 0, Y = 0 and X and Y are not proportional (L and M are not parallel). (ii) P and Q can be viewed as the projections of C and D onto the line P Q, where C and D are arbitrary points on the lines L and M, respectively. Hence by equation (8.14) of Theorem 8.5.3, we have P Q ≤ CD. Finally we derive a useful formula for P Q. Again by Theorem 8.5.3 E E E | AB · P Q | ˆ = | AB ·n|, PQ = PQ 94 ...
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