Help with elementary linear algebra please(pretty easy stuff)

[mtlchris]

If the distance from the town of Bree to Weathertop is 10 miles on a 45 degree upward slope, what is the elevation gain (omit units)? I think its 10sin(45) but i have one attempt left.


What is the distance from the point (9, 7, -9) to the xz-plane? I Need the equation for the xz plane so I can use my formula





Find a vector perpendicular to the vectors 1
-3
4
and 1
-2
1
.




Is this the orthogonal projection of either vector?

[MaxWeiss]

Yes, you are right, it's 10*sin(45 degrees), which is 7.07. Alternatively, you can go about it by knowing that one angle is 45, and the triangle you are trying to make in order to figure out the answer must be a right triangle, with the hypotenuse being 10, and the remaining angle must be 45 also, since in triangles the degrees add to 180. Therefore, both legs of the triangle must be equal, so using the Pythagorean Theorem, you can deduce that 2*(x^2)=10^2, or that x^2=50.

I don't know the formula for the 3 dimensional plane, but what I would do is find the XY plan answers, and use those answers to form a plane of intersection for the third dimension, and then use the same formula, since now you again have a two dimensional plane, one being the XY answers and the other being the Z plane. I'm sure the equation is out there somewhere online, but for the time it takes, I can't imagine that just breaking up the problem into two parts would take much more time.

For example, on the [9,7,-9] distance to the origin, [0,0,0] (if that is the first question, which I don't know that it is--I can't tell from the post), I would start by finding the hypotenuse of the 9/7 triangle, which is 5.657. That answer is the new leg of the 5.657/-9 triangle, with a hypotenuse of 10.63, which is your answer.

For the last question, you just need a perpendicular line to the line created by those two points. To make life easy, you can simply use one of the points as the first point in your new line, and then you also already have one of the dimensions for your second point. However, it's late and I suck as visualizing things, and I can't find a 3d graphing thing online, so I'm going to give up and go to bed. The word orthogonal feels right, although I have no idea what it means, but it seemed to ring a bell somewhere when I read it in your post, which in my mind means it's the right starting point.

[bigpooch]

You should be familiar with the triangle with sides 1,1,
sqrt(2) that is isosceles with acute angle 45 degrees; the
answer then is 5sqrt(2) or equivalent to your answer.

The distance from (9,7,-9) to the xz-plane is just 7
(absolute value of the y-coordinate); the xz-plane is just
y=0.

The "cross product" of two vectors is always perpendicular
to each of the "input" vectors.

[jay_shark]

[ QUOTE ]


Find a vector perpendicular to the vectors 1
-3
4
and 1
-2
1
.




Is this the orthogonal projection of either vector?

[/ QUOTE ]

Just use the cross product :

i.(-3 -4*(-2)) -j.(1*1 -4*1) + k.(1*(-2) - -3*1)
= (5,3,1)

We can check this by using the dot product on the two direction vectors .

(5,3,1). (1,-3,4) = 0
(5,3,1). (1,-2,1) = 0

So the vector (5,3,1) is perpendicular to the two vectors .

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The projection of the vector a on b = a.b/|a|^2*b which is going to have the same slope as the vector b and so it cannot be perpendicular to b .