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
Here's the solution to question 7: Given position vectors a = 2i + j - 3k, b = 3i - 2j + k, and c = i + 3j - k. a) Find the vector AB × AC. Step 1: Find the vector AB. AB = b - a = (3i - 2j + k) - (2i + j - 3k) AB = (3-2)i + (-2-1)j + (1-(-3))k AB = i - 3j + 4k Step 2: Find the 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 \\ 1 & -3 & 4 \\ -1 & 2 & 2 = i((-3)(2) - (4)(2)) - j((1)(2) - (4)(-1)) + k((1)(2) - (-3)(-1)) = i(-6 - 8) - j(2 - (-4)) + k(2 - 3) = i(-14) - j(6) + k(-1) = -14i - 6j - k The vector AB × AC is -14i - 6j - k. b) Hence, find the area of the triangle ABC. Step 1: The area of triangle ABC is half the magnitude of the cross product AB × AC. Area = (1)/(2) |AB × AC| Step 2: Calculate the magnitude of AB × AC. |AB × AC| = |-14i - 6j - k| = sqrt((-14)^2 + (-6)^2 + (-1)^2) = sqrt(196 + 36 + 1) = sqrt(233) Step 3: Calculate the area of the triangle. Area = (1)/(2) sqrt(233) The area of the triangle ABC is (1)/(2)sqrt(233) units^2. c) Find the Cartesian equation of the plane ABC. Step 1: The normal vector to the plane is n = AB × AC = -14i - 6j - k. The Cartesian equation of a plane is of the form Ax + By + Cz = D, where (A, B, C) are the components of the normal vector. So, the equation is -14x - 6y - z = D. Step 2: Use a point on the plane to find D. We can use point A with position vector a = 2i + j - 3k, so (x, y, z) = (2, 1, -3). -14(2) - 6(1) - (-3) = D -28 - 6 + 3 = D -31 = D Step 3: Write the Cartesian equation of the plane. -14x - 6y - z = -31 Multiplying by -1 to make the leading coefficient positive: 14x + 6y + z = 31 The Cartesian equation of the plane ABC is 14x + 6y + z = 31. d) Find the acute angle between the plane ABC and the plane x - y + z = 5. Step 1: Identify the normal vector of the plane ABC. From part c), n_1 = 14i + 6j + k. Step 2: Identify the normal vector of the second plane, x - y + z = 5. n_2 = i - j + k. Step 3: Calculate the magnitudes of the normal vectors. |n_1| = sqrt(14^2 + 6^2 + 1^2) = sqrt(196 + 36 + 1) = sqrt(233) |n_2| = sqrt(1^2 + (-1)^2 + 1^2) = sqrt(1 + 1 + 1) = sqrt(3) Step 4: Calculate the dot product of the normal vectors. n_1 · n_2 = (14)(1) + (6)(-1) + (1)(1) = 14 - 6 + 1 = 9 Step 5: Use the formula for the angle between two planes: = |n_1 · n_2||n_1| |n_2|. = (|9|)/(sqrt(233) 3) = (9)/(sqrt(699)) Step 6: Find the acute angle . = ^-1((9)/(sqrt(699))) The acute angle between the planes is ^-1((9)/(sqrt(699))).