Analytic geometry #3 - Questions and Answers

Question 1 What is geometry? Answer Roughly speaking, I would say that geometry is the study of the properties of the objects defined in certain types of "spaces", for example, distance (a property) between points (objects) in the Euclidean plane (a space). As for what should go between the quotes, I would guess that they're asking about the etymology of the word itself. In that case, the "geo" part comes from an ancient Greek word meaning "Earth" and the "metry" part comes from an ancient Greek word meaning "measurement". So I'd think "Earth measurement" is what they're looking for. Question 2 How many lines of symmetry does the figure shown at the right have? A) 0 B) 1 C) 5 D) 10 Answer No. The shape has only 1 line of symmetry which is a vertical line passing through the upper and lower vertices. (Recall that a line of symmetry divides a figure into two figures that are mirror images.) Question 3 What are verticle angles? Answer Vertical angles are angles whose non-common sides form two pairs of opposite rays. Question 4 Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS. P(-2, 1, 0), Q(2, 3, 2), R(1, 4, -1), S(3, 6, 1) Answer 1 Any three vectors (a,b and c) we can find the volume of the parallelogram that the three vectors are shaping, using the triple product of these three vectors a · (b x c) Volume = |a · (b x c)| (1) a₁ a₂ a₃ [a · (b x c)] = | b₁ b₂ b₃ | c₁ c₂ c₃ The value Δ of the determinant can be computed, a₁ a₂ a₃ a₁ a₂ | b₁ b₂ b₃ | = a₁ | b₂ b₃ | - a₂ | b₁ b₃| + a₃ |b₁ b₂ | c₁ c₂ c₃ c₂ c₃ c₁ c₃ c₁ c₂ Δ = a₁((b₂c₃) - (b₃c₂)) - a₂((b₁c₃) - (b₃c₁)) + a₃((b₁c₂) - (b₂c₁)) Or the value of the determinant can be found using Sarrus's rule 3 x 3 determinant "This work for 3 x 3 determinant Only", where by duplicating the first two columns of the matrix after the third columns, a 3 x 5 determinant is formed as follows a₁ a₂ a₃ a₁ a₂ b₁ b₂ b₃ b₁ b₂ c₁ c₂ c₃ c₁ c₂ The value of the determinant is thus determined by Sarrus's rule to be the product of all the South-East diagonals minus the product of all the South-West diagonals, so that Δ = ((a₁b₂c₃) + (a₂b₃c₁) + (a₃b₁c₂)) - ((a₂b₁c₃) + (a₁b₃c₂) + (a₃b₂c₁)) In order to get the magnitude for the triple product, which would reflect the parallelogram's volume, we start by determining the value of the triple product, also known as "the value of the determinant." PQ = Q - P = (2+2, 3-1, 2-0) = (4, 2, 2) PS = S - P = (3+2, 6-1, 1-0) = (5, 5, 1) PR = R - P = (1+2, 4-1, -1-0) = (3, 3, -1) 5 5 1 PS · (PQ x PR) = | 4 2 2| 3 3 -1 Δ = |5| = 5 Answer 2 The scalar triple product is defined as follows: PQ · (PR x PS) where · denotes the dot product and × denotes the cross product. First, let's find the vectors PQ, PR, and PS. PQ = [2-(-2), 3-1, 2-0] = [4,2,2] PR = [1-(-2), 4-1, -1-0] = [3,3,-1] PS = [3-(-2), 6-1, 1-0] = [5,5,1] Next, let's calculate the cross product PR x PS PR x PS = | i j k | | 3 3 -1 | | 5 5 1 | = i(3-(-5)) -j(3-(-5)) + k(15-15) = 8i - 8j + 0k = [8,-8,0] Now, we can calculate the dot product of PQ and (PR x PS) PQ · (PR x PS) = [4,2,2] . [8,-8,0] = (48) + (2-8) + (2*0) = 32 - 16 + 0 = 16 Answer 3 To find the volume of a parallelepiped with adjacent edges PQ, PR, and PS, we can use the scalar triple product. The scalar triple product of three vectors is defined as follows: V = PQ · (PR x PS) where · denotes the dot product and × denotes the cross product. First, let's find the vectors PQ, PR, and PS. The vector PQ is given by: PQ = Q – P = (2, 3, 2) - (-2, 1, 0) = (4, 2, 2) Similarly, we can find the vectors PR and PS: PR = R – P = (1, 4, −1) – (-2, 1, 0) = (3,3,-1) PS = S – P = (3,6,1) - (-2, 1, 0) = (5,5, 1) Next, we can calculate the cross product of PR and PS: i j k PR x PS = | 3 3 -1 | | 5 5 1 | Expanding the determinant, we get: PR x PS = (3(1) - (-1)5)i - (3(1) - (-1)5)j + (3(5) - 5(3))k = 8i – 8j – 10k = (8, -8, -10) Now, we can calculate the dot product of PQ and (PR x PS): PQ · (PR x PS) = (4,2,2) · (8, -8, -10) = 4(8) + 2(-8) + 2(-10) = 32 - 16 - 20 = -4 Finally, we take the absolute value of the scalar triple product to get the volume of the parallelepiped: Volume = |V| = |-4| = 4 Therefore, the volume of the parallelepiped with adjacent edges PQ, PR, and PS is 4. Answer 4 |PQ · (PR x PS)| where PQ is the vector from point P to Q, and PR x PS is the cross product of vectors PR and PS. Let's calculate each component step by step. First, let's find the vector PQ: PQ = [2-(-2), 3-1, 2-0] = [4,2,2] Next, let's calculate the cross product PR x PS: PR x PS = | 1 - (-2) 3 - (-2) | = | 3 5 | | 4 - 1 6 - 1 | | 3 5 | | -1 - 0 1 - 0 | |-1 1 | PR x PS = [(3 * 1) - (-1 * 5)] = 8 = [ (3 * 1) - (-1 * 5)] = 8 = [(3 * 5) - (3 * 5)] = 0 = [8,8,0] Now, let's calculate the scalar triple product: PQ · (PR x PS) = [4 -12] [2 -12] = (4 * -12) + (2 * -12) + (2* 0) = -48 - 24 + 0 = -72 [2 0 ] |PQ · (PR x PS)| = |-72| = 72 Therefore, the volume of the parallelepiped is 72 cubic units. Question 5 What is the volume of this cone? Use π = 3.14 and round your answer to the nearest hundredth. Answer Given: Height of the cone (h) = 10 inches Diameter of cone (d) = 14 inches To find: Volume of the cone = ? Height of cone (h) = 10 inches Radius of cone = d/2 = 14/2 = 7 inches Now, volume of cone (v) v = (1/3) πr²h v = (1/3)(3.14)(7)²(10) ( π ≈ 3.14) v ≈ (1.046) * 49 * 10 v ≈ 512.54 in³