Chapter 5 - Inner Product Spaces - 5.2 Orthogonal Sets of Vectors and Orthogonal Projections - Problems - Page 360: 25

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Given $P=(4,1,-1)\\ x(t)=(2t,-4-t,3t)$ We form vector $w=(4-0,1-(-4),-1-0)=(4,5,-1)$ The norm of $P(w,v)=\frac{}{||v||^2}v$ gives $P(w,v)=\frac{}{||(2,-1,3)||^2}(2,-1,3)=0$ The distance from $P$ to $L$ is $||w-P(w,v)||=||w-0||=||w||=\sqrt 4^2+5^2+(-1)^2=\sqrt 42$