Question

Find all values of x so that the triangle with vertices A=(x,2,−3),B=(−6,1,−6), and C=(−4,−1,−4) has area 5. Enter your answer as a commaseparated list of values. Use the square root symbol 'V" where needed to give an exact value for your answer.

Answer 1

To find the values of x that make the area of the triangle 5 units², calculate the cross product of vectors AB and AC, find the magnitude, set it equal to twice the area, and then solve the resulting quadratic equation for x, considering both positive and negative solutions.

Step-by-step explanation:

To find all values of x that make the triangle with vertices A=(x,2,−3),B=(−6,1,−6), and C=(−4,−1,−4) have an area of 5, we can use the formula for the area of a triangle:

A = 1/2 * √(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²

We substitute the coordinates of the vertices and the given area into the formula:

5 = 1/2 * √(x-(-6))² + (2-1)² + (-3-(-6))²

Simplifying the equation:

5 = 1/2 * √(x+6)² + 1 + 9

Next, we isolate the square root by multiplying both sides of the equation by 2:

10 = √(x+6)² + 10

Squaring both sides of the equation:

100 = (x+6)² + 10

Expanding and rearranging the equation:

0 = x² + 12x + 64

Applying the quadratic formula to solve for x:

x = (-12 ± √(12)² - 4(1)(64))/(2(1))

x = (-12 ± √144 - 256)/2

x = (-12 ± √-112)/2

Since the square root of a negative number is not a real number, there are no values of x that make the triangle have an area of 5.