There is given a vector $2 \vec i + \vec j - 3 \vec k$ and now I want to find the equation of a line that is perpendicular to the given vector and passing through a known point $(1,1,1)$. How can I solve this?
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$\begingroup$ Instead of Cross product why can't we use the Dot product.Because the Dot product of any two perpendicular vectors are equal to 0. $\endgroup$– ThusithaCommented May 13, 2012 at 11:54
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3$\begingroup$ 10k views? How??? $\endgroup$– evil999manCommented Apr 29, 2014 at 18:41
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$\begingroup$ 15k views and 3 upvotes :( $\endgroup$– Anubian NoobCommented Apr 29, 2015 at 17:04
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1$\begingroup$ @AlexM. Not a duplicate. Perpendicular to a plane is relatively straightforward, perpendicular to a vector in 3D is a tricky question indeed. $\endgroup$– user147263Commented Feb 1, 2016 at 20:24
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2 Answers
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So, you are given the vector $(2,1,-3)$. Let $(2k,k,-3k)$ be the orthogonal projection of $(1,1,1)$ on $(2,1,-3)$. Then, $(2k-1,k-1,-3k-1)$ and $(2,1,-3)$ are orthogonal, giving: $4k-2+k-1+9k+3=0$ i.e. $k=0$. So, $(0,0,0)$, the origin is the projection. Hence, the line contains the points $(0,0,0)$ and $(1,1,1)$, so its equation is $x=y=z$, if my calculations are correct!
answered May 13, 2012 at 11:55
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ai + b j + ck is the linear combination notation for a point in 3-space..It is meaningless to speak of a line being "perpendicular" to a point. ...What you want is the line perpendicular to the line L through (2 , 1 , -3) and the origin. The direction vector of L is (2,1,-3)..a vector orthogonal to that is (3,O 2)..so the formula of the line you want is of the form (1,1,1) + k(3,O,2)..k a real number.....which you may rewrite in cartesian notation
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1$\begingroup$ Perhaps using MathJAX to format your mathematical expressions would help, such as \$ai+bj+ck\$ to produce $ai+bj+ck$. $\endgroup$– abiessuCommented Apr 29, 2014 at 19:15
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$\begingroup$ The answer is not quite correct...because in 3-space, the thing perpendicular to a line L is a plane. So any line in that plane is perpendicular to the givenline. $\endgroup$– kozenkoCommented Apr 29, 2014 at 19:45
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$\begingroup$ That would be a good comment to add to the answer that isn't correct. Please note that my comment only applies to how your answer is presented, not to its content. Note also that the question specifically asked for a line perpendicular to the given vector which passes through a given point, so it appears that the answer responds correctly to the question. $\endgroup$– abiessuCommented Apr 29, 2014 at 19:49