Question 1207402
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I'll focus on problem 1 only.


You wrote that "line L passes through 4".
I'll assume the "4" should be "A". 
I've seen this strange error occur with OCR (optical character recognition) software. 
Admittedly it's not that strange considering "4"s and "A"s look somewhat alike.


This equation OA = -i+3j+5k tells us that to go from the origin to point A we will do the following:<ul><li>Move 1 unit in the negative x direction.</li><li>Move 3 units in the positive y direction.</li><li>Move 5 units in the positive z direction.</li></ul>The order of those steps does not matter.


This has us go from (0,0,0) to (-1,3,5) which is the location of point A.
This is the starting point or anchor point we'll use. 
Think of it like how the y intercept is a common start point in a 2D setting.


From this anchor point, we move in the same direction as vector OB. This is so we are parallel to line OB. Parallel lines in 3D space have equal direction vectors. 
Example: lines that point straight upward are parallel
It's similar to saying "lines in 2D are parallel only when they have equal slopes but different y intercepts".


The vector equation of line L would therefore be
<font color=red>r = -i+3j+5k + s(3i-j-4k)</font>


Note this template
r = startPoint + s*(directionVector)
which describes any line in 3-space.


In other words, we need to know where to start and which direction to aim at, to form a line.
It's analogous to the y intercept and slope in a 2D setting.


Side note: the component form of line L would be r = (-1+3s, 3-s, 5-4s) where "s" can be thought of as a timestamp.
See this <a href = "https://www.algebra.com/algebra/homework/playground/test.faq.question.1207403.html">similar problem</a> for more info.
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