Find x so that the triangle with vertices A=(4, 4, −2), B=(2, −1, −5), and C=(x, 14, −6) has a right angle at A.
as with the other question just like this one, you want AB•AC = 0
So,
(2-4)(x-4) + (-1-4)(14-4) + (-5+2)(-6+2) = 0
x = -15
To find the value of x such that the triangle ABC has a right angle at A, we need to determine if the dot product between the vectors AB and AC is zero.
Here's how you can do it step by step:
Step 1: Calculate the vectors AB and AC
The vector AB can be obtained by subtracting the coordinates of point A from the coordinates of point B. Likewise, the vector AC is obtained by subtracting the coordinates of point A from the coordinates of point C.
AB = B - A = (2, -1, -5) - (4, 4, -2)
To calculate the coordinates, subtract corresponding values:
AB = (-2, -5, -3)
AC = C - A = (x, 14, -6) - (4, 4, -2)
AC = (x - 4, 10, -4)
Step 2: Calculate the dot product
The dot product between two vectors can be calculated by multiplying their corresponding components and then summing them up.
Dot product of AB and AC = AB · AC = (-2)(x-4) + (-5)(10) + (-3)(-4)
Simplifying, we get:
6 - 2x - 50 + 12
-2x - 32
Step 3: Set the dot product equal to zero and solve for x
Since we want the triangle to have a right angle at A, the dot product should be zero.
-2x - 32 = 0
To solve for x, we can add 32 to both sides:
-2x = 32
Finally, divide both sides by -2:
x = -16
Therefore, x = -16 is the value that makes the triangle ABC have a right angle at A.