Please could you help

Oct 2012
4
0
UK
Hello Guys

I wonder if anyone can help me solving the attached question if possible by detailed steps.




Thanks in advance

Mike
 

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chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey mikejohn.

Can you show us what you have tried. As a starting hint, remember something is perpendicular if <a,b> = 0 and each line has a direction vector so in order to be perpendicular, both direction vectors must be orthogonal.

In terms of number 2, remember that an intersection is just when one thing equals another. What is the definition of the x-z plane in equation form? How can you make them equal one another and find a solution?

For number 3, you have the area of a triangle using normal high school math (something like 1/2absin(C) where a and b are the lengths and sin(C) is the angle formed between those lines) so |AB| = a and |AC| = b.
 
Oct 2012
4
0
UK
Hey mikejohn.

Can you show us what you have tried. As a starting hint, remember something is perpendicular if <a,b> = 0 and each line has a direction vector so in order to be perpendicular, both direction vectors must be orthogonal.

In terms of number 2, remember that an intersection is just when one thing equals another. What is the definition of the x-z plane in equation form? How can you make them equal one another and find a solution?

For number 3, you have the area of a triangle using normal high school math (something like 1/2absin(C) where a and b are the lengths and sin(C) is the angle formed between those lines) so |AB| = a and |AC| = b.


Thanks so much for giving me some hints.
But to be honest, the questions still vague for me, unfortunately because I did not understand the vector, lines, and planes topics from the professor and I do not have neither time right now to navigate through all the stuff that he gave us nor surfing internet to understand these topics


I would really really appreciate your time of further help with detailed steps if possible.
 
Last edited:

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
The equation for a plane in three dimensions is ax + by + cz + d = 0 and the equation for a line is At + B(1-t) = 0 or (x - x0)/a - (y-y0)/b - (z-z0)/c = 0 using this notation:

Pauls Online Notes : Calculus III - Equations of Lines

So now you have 1 = 2 in terms of equations and then you have to solve for this: if the line is not on the plane you will get one solution and if the line is on the plane then you will get infinitely many solutions unless the line is parallel to the plane (but not on it) which will give no solutions.