# I only knew how to get the direction vector of the point C but can't continue any further. The endpoints of the hypotenuse of a right triangle ABC are A(-10,10,9) and B(14,0,-4). The point C lies...

I only knew how to get the direction vector of the point C but can't continue any further.

The endpoints of the hypotenuse of a right triangle ABC are A(-10,10,9) and B(14,0,-4). The point C lies on the line that passes through the point A and is parallel to the vector 2i-2j-k. Determine the coordinates of the point C.

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The problem provides the information that vectors `2i - 2j - k` and `bar(AC)` are parallel, hence, you need to write the following equation, such that:

`(x_C - x_A)/2 = (y_C - y_A)/(-2) = (z_C - z_A)/(-1)`

Replacing the values of `x_A,y_A,z_A` , yields:

`(x_C + 10)/2 = (y_C - 10)/(-2) = (z_C - 9)/(-1)`

You also know that triangle ABC is rightg triangle, hence the vectors `bar (AC)` and `bar (BC)` are perpendicular, such that:

`(x_C - x_A)(x_C - x_B) + (y_C - y_A)(y_C - y_B) + (z_C - z_A)(z_C - z_B) = 0`

`(x_C + 10)(x_C - x_B) + (y_C - 10)(y_C - y_B) + (z_C - 9)(z_C - z_B) = 0`

Replace - `(x_C + 10)` for `(y_C - 10) ` and -(x_C + 10)/2 for (`z_C - 9)` , such that:

`(x_C + 10)(x_C - x_B) - (x_C + 10)(y_C - y_B) - (x_C + 10)/2*(z_C - z_B) = 0`

You need to factor out `(x_C + 10)` , such that:

`(x_C + 10)(x_C - 14 - y_C - (z_C+4)/2) = 0`

`x_C - 14 - y_C - (z_C+4)/2 = 0`

`x_C - y_C - (z_C)/2 = 16`

Use `(x_C + 10)/2 = (y_C - 10)/(-2) => x_C = -y_C`

Use `(y_C - 10)/(-2) = (z_C - 9)/(-1) => 10 - y_c = 18 - 2z_C`

`2z_C = 8 + y_C`

`z_C = (8 + y_C)/2`

Replace -`y_C ` for `x_C` and `(8 + y_C)/2` for `z_C` , such that:

`-2y_C - (8 + y_C)/4 = 16`

`-8y_C - 8 - y_C = 64`

`-9y_C = 72 => y_C= -8 => x_C = 8 => z_c = 0`

**Hence, evaluating the coordinates of vertex C, yields C(8,-8,0).**