Hi, I followed a different method: the distance between the point and the line is the distance between the point and the intersection between the line and the perpendicular plane to the line which passes through Q. 1) We know the vector of the direction of the line (2,-1,8) 2) We can calculate then the equation of the plane which is perpendicular to the line and passes through the external point Q (4,4,1). It is 2x-y+8z=12 [Eq. I] 3) We can transform the three parametric equations of the line into two symmetrical equations: x+2y=15 [Eq. II] and 8y+z=61[Eq. III] 4) Now we have three linear equations I, II and III and three unknowns x,y,z which are the coordinates of the point S where the plane calculated in paragraph 2) intersects the line. This point is S =(23/69, 506/69, 161/69). 5) The distance between the point Q and the point S is the module of the vector Q-S =(253/69,-230/69,92/69). The module is SQR(125,373)/69 ≈ 5.13
much better explanation than my math teacher, thank you very much! You have made my life a bit easier :)
This guy is the goat! Was havng trouble with this section but now its much more clearer
This is absolutely brilliant
These questions are tricky.
I thought they were direct.
Thanks for explaining the steps 👍
Hi, I followed a different method: the distance between the point and the line is the distance between the point and the intersection between the line and the perpendicular plane to the line which passes through Q.
1) We know the vector of the direction of the line (2,-1,8)
2) We can calculate then the equation of the plane which is perpendicular to the line and passes through the external point Q (4,4,1). It is 2x-y+8z=12 [Eq. I]
3) We can transform the three parametric equations of the line into two symmetrical equations: x+2y=15 [Eq. II] and 8y+z=61[Eq. III]
4) Now we have three linear equations I, II and III and three unknowns x,y,z which are the coordinates of the point S where the plane calculated in paragraph 2) intersects the line. This point is S =(23/69, 506/69, 161/69).
5) The distance between the point Q and the point S is the module of the vector Q-S =(253/69,-230/69,92/69). The module is SQR(125,373)/69 ≈ 5.13
Really a clear explanation. Really i like thiss
thank you boss!
thank you so much sir
V v thank u sir. Really it's helpful for me. I understood accurate.
my GOAT!
OMG this was so helpful! Thank you! :)
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you are life saver
Please recommend a textbook to study more. Thanks
Why can't you use dot product like in the plane and point video?
clearly cut explanation
i was given P and the line no Q
P(5,6,5) x=5+4t , y=6+4t , z=5
you will find q in your line