This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Welcome back — been a while! Let's pick up where you left off. Here are the solutions for Question 7: The position vectors are given as: a = 2i + j - 3k b = 4i - j + 2k c = i + 3j - k a) Find the vector AB × AC. Step 1: Find vector AB. AB = b - a = (4i - j + 2k) - (2i + j - 3k) AB = (4-2)i + (-1-1)j + (2-(-3))k AB = 2i - 2j + 5k Step 2: Find vector AC. AC = c - a = (i + 3j - k) - (2i + j - 3k) AC = (1-2)i + (3-1)j + (-1-(-3))k AC = -i + 2j + 2k Step 3: Calculate the cross product AB × AC. AB × AC = i & j & k \\ 2 & -2 & 5 \\ -1 & 2 & 2 = i((-2)(2) - (5)(2)) - j((2)(2) - (5)(-1)) + k((2)(2) - (-2)(-1)) = i(-4 - 10) - j(4 - (-5)) + k(4 - 2) = -14i - 9j + 2k The vector AB × AC is -14i - 9j + 2k. b) Hence, find the area of the triangle ABC. Step 1: Calculate the magnitude of AB × AC. |AB × AC| = sqrt((-14)^2 + (-9)^2 + (2)^2) = sqrt(196 + 81 + 4) = sqrt(281) Step 2: Calculate the area of triangle ABC. The area of triangle ABC is half the magnitude of the cross product. Area = (1)/(2) |AB × AC| = (1)/(2)sqrt(281) The area of the triangle ABC is (1)/(2)sqrt(281) units^2. c) Find the Cartesian equation of the plane ABC. Step 1: Use the normal vector from part a). The normal vector to the plane ABC is n = -14i - 9j + 2k. The equation of the plane is of the form n_x x + n_y y + n_z z = d. So, -14x - 9y + 2z = d. Step 2: Substitute the coordinates of point A (2, 1, -3) into the equation to find d. -14(