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.
ScanSolve AI Answer
a) Find the vector AB × AC. Step 1: Find the vectors AB and AC. The position vectors are given as: a = 2i + j - 3k b = 4i - j + 2k c = i + 3j - k Calculate AB: AB = b - a = (4i - j + 2k) - (2i + j - 3k) AB = (4-2)i + (-1-1)j + (2-(-3))k AB = 2i - 2j + 5k Calculate 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 2: Compute 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) = i(-14) - j(4 + 5) + k(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 the cross product. The area of triangle ABC is half the magnitude of the vector AB × AC. |AB × AC| = |-14i - 9j + 2k| = sqrt((-14)^2 + (-9)^2 + (2)^2) = sqrt(196 + 81 + 4) = sqrt(281) Step 2: Calculate the area of the triangle. Area = (1)/(2) |AB × AC| Area = (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: Identify the normal vector and a point on the plane. The vector AB × AC = -14i - 9j + 2k is a normal vector to the plane ABC. Let n = -14i - 9j + 2k. We can use point A (2, 1, -3) which lies on the plane. Step 2: Write the Cartesian equation of the plane. The equation of a plane with normal vector n = Ai + Bj + Ck passing through a point (x_0, y_0, z_0) is A(x-x_0) + B(y-y_0) + C(z-z_0) = 0. Substituting the values: