Basic geometry - dot product/cart. lines

[boings]

Homework Statement

We consider two points, B and I and a line 'a'.
B(0,-4,-7) I(-2,-2,-5) and a: x = y+1 = (z-2)/2

Determine the summits of A and C of triangle ABC knowing that:

-Summit A belongs to the line 'a'
-I is the foot of the height from A (perpendicular to BC)
-The angle of A is equal to arccos(1/3)

Homework Equations

dot product

The Attempt at a Solution

I feel as if I'm really close, but keep getting the wrong answer. My process is as follows: The dot product of BI and AI is equal to zero. Knowing this, I can obtain an equation of the line AI and determine where it intersects with line 'a'. This is as far as I've gotten in my attempt to find A.

Thanks a lot for any help you can provide!

 

[I like Serena]

Homework Statement

We consider two points, B and I and a line 'a'.
B(0,-4,-7) I(-2,-2,-5) and a: x = y+1 = (z-2)/2

Determine the summits of A and C of triangle ABC knowing that:

-Summit A belongs to the line 'a'
-I is the foot of the height from A (perpendicular to BC)
-The angle of A is equal to arccos(1/3)

Homework Equations

dot product

The Attempt at a Solution

I feel as if I'm really close, but keep getting the wrong answer. My process is as follows: The dot product of BI and AI is equal to zero. Knowing this, I can obtain an equation of the line AI and determine where it intersects with line 'a'. This is as far as I've gotten in my attempt to find A.

Thanks a lot for any help you can provide!

Welcome to PF, boings!

Can you set up a parametric equation for the line 'a'?
That is, find a support vector and a direction vector?

If you fill that in for A in your dot product, you'll get an equation from which you can find A...

 

[boings]

Hi and thanks!

good point!

(if k=constant)
x= 0 + k
y= -1 + k
z= 2 + 2k

Although, when I plug that in it doesn't make much sense. I proceed like this:

AI (dot) BI = (-2, 2, 2)(dot)(-2, -2, -5)(0 + k, -1 + k, 2 + 2k)

I think that this is already flawed in some sense

 

[I like Serena]

Well, the vector AI is the difference of (-2, -2, -5) and (0 + k, -1 + k, 2 + 2k).
So that is (-2 - k, -1- k, -7 - 2k).
Then take the dot product and set it equal to zero...

 

[boings]

Alrighty, so I then get (4-2k, -2-2k, -14 - 4k)=0 which represents the line AI.

This is where I get a little tripped up. Should I solve for k and replace into original equation of the line? The answer should be: A(-3, -4, -4)

 

[I like Serena]

Alrighty, so I then get (4-2k, -2-2k, -14 - 4k)=0 which represents the line AI.

Hmm, (4-2k, -2-2k, -14 - 4k)=0 does not represent a line.
How come you think that?

This is where I get a little tripped up. Should I solve for k and replace into original equation of the line? The answer should be: A(-3, -4, -4)

The vector AI is represented by (-2 - k, -1- k, -7 - 2k) for some value of k.
The vector BI is (-2, 2, 2).
Their dot product has to be zero.

The dot product is (-2-k) x -2 + (-1-k) x 2 + (-7-2k) x 2 = 0.
Solve for k?

 

[boings]

Oh ok I get it! thank you so much.

I see how that's not a line now, but rather the equation that relies the orthogonality between AI and BI. It sure looks like a a parametric equation of a line though ^^

thanks again