Vectors and spaces #6 - Q&A

Question 1 Find a nonzero vector orthogonal to the plane through the points 𝑃 , 𝑄 , 𝑅 P,Q,R and the area of the triangle 𝑃 𝑄 𝑅 PQR. Given points: 𝑃 ( 1 , 0 , 1 ) , 𝑄 ( − 2 , 1 , 4 ) , 𝑅 ( 7 , 2 , 7 ) P(1,0,1),Q(−2,1,4),R(7,2,7) (a) Find a nonzero vector orthogonal to the plane through 𝑃 , 𝑄 , 𝑅 P,Q,R: Vectors 𝑃 𝑄 → PQ ​ and 𝑃 𝑅 → PR : 𝑃 𝑄 → = 𝑄 − 𝑃 = ( − 2 − 1 , 1 − 0 , 4 − 1 ) = ( − 3 , 1 , 3 ) , PQ ​ =Q−P=(−2−1,1−0,4−1)=(−3,1,3), 𝑃 𝑅 → = 𝑅 − 𝑃 = ( 7 − 1 , 2 − 0 , 7 − 1 ) = ( 6 , 2 , 6 ) . PR =R−P=(7−1,2−0,7−1)=(6,2,6). Find the cross product 𝑃 𝑄 → × 𝑃 𝑅 → PQ ​ × PR : 𝑃 𝑄 → × 𝑃 𝑅 → = ∣ 𝑖 𝑗 𝑘 − 3 1 3 6 2 6 ∣ . PQ ​ × PR = ​ i −3 6 ​ j 1 2 ​ k 3 6 ​ ​ . Expanding the determinant: 𝑃 𝑄 → × 𝑃 𝑅 → = 𝑖 ( 1 ⋅ 6 − 3 ⋅ 2 ) − 𝑗 ( − 3 ⋅ 6 − 3 ⋅ 6 ) + 𝑘 ( − 3 ⋅ 2 − 1 ⋅ 6 ) , PQ ​ × PR =i(1⋅6−3⋅2)−j(−3⋅6−3⋅6)+k(−3⋅2−1⋅6), = 𝑖 ( 6 − 6 ) − 𝑗 ( − 18 − 18 ) + 𝑘 ( − 6 − 6 ) , =i(6−6)−j(−18−18)+k(−6−6), = 𝑖 ( 0 ) − 𝑗 ( − 36 ) + 𝑘 ( − 12 ) , =i(0)−j(−36)+k(−12), = 0 𝑖 + 36 𝑗 − 12 𝑘 . =0i+36j−12k. Thus, a nonzero vector orthogonal to the plane is: 𝑛 = ⟨ 0 , 36 , − 12 ⟩ . n=⟨0,36,−12⟩. (b) Find the area of the triangle 𝑃 𝑄 𝑅 PQR: The area of the triangle is: Area = 1 2 ∥ 𝑃 𝑄 → × 𝑃 𝑅 → ∥ . Area= 2 1 ​ ​ PQ ​ × PR ​ . Compute the magnitude of 𝑃 𝑄 → × 𝑃 𝑅 → PQ ​ × PR : ∥ 𝑃 𝑄 → × 𝑃 𝑅 → ∥ = 0 2 + 3 6 2 + ( − 12 ) 2 = 0 + 1296 + 144 = 1440 . ​ PQ ​ × PR ​ = 0 2 +36 2 +(−12) 2 ​ = 0+1296+144 ​ = 1440 ​ . Area = 1 2 1440 = 1 2 ⋅ 12 10 = 6 10 . Area= 2 1 ​ 1440 ​ = 2 1 ​ ⋅12 10 ​ =6 10 ​ . Result: Nonzero vector orthogonal to the plane: 𝑛 = ⟨ 0 , 36 , − 12 ⟩ . n=⟨0,36,−12⟩. Area of the triangle: Area = 6 10 . Area=6 10 ​ . Question 2 Determine if the system of equations represented by the augmented matrix is consistent, and find its general solution: [ 1 − 2 6 0 0 0 ] . [ 1 0 ​ −2 0 ​ 6 0 ​ ]. Answer: The second row is all zeros, so there is no contradiction (e.g., 0 = 𝑐 0=c where 𝑐 ≠ 0 c  =0). The system is consistent. The first row represents the equation: 𝑥 1 − 2 𝑥 2 = 6. x 1 ​ −2x 2 ​ =6. Let 𝑥 2 = 𝑡 x 2 ​ =t (free variable). Then: 𝑥 1 = 6 + 2 𝑡 . x 1 ​ =6+2t. Write the solution in vector form: 𝑥 = [ 𝑥 1 𝑥 2 ] = [ 6 0 ] + 𝑡 [ 2 1 ] , 𝑡 ∈ 𝑅 . x=[ x 1 ​ x 2 ​ ​ ]=[ 6 0 ​ ]+t[ 2 1 ​ ],t∈R. Result: The system is consistent. General solution: 𝑥 = [ 6 0 ] + 𝑡 [ 2 1 ] , 𝑡 ∈ 𝑅 . x=[ 6 0 ​ ]+t[ 2 1 ​ ],t∈R. Question 3 Find the linear equation that models the height 𝑦 y (in inches) of a boy in terms of his age 𝑥 x (in years). Predict the average height of a 5-year-old boy. Given: ( 2 , 38 ) , ( 8 , 56 ) . (2,38),(8,56). Answer: Find the slope: 𝑚 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 56 − 38 8 − 2 = 18 6 = 3. m= x 2 ​ −x 1 ​ y 2 ​ −y 1 ​ ​ = 8−2 56−38 ​ = 6 18 ​ =3. The equation of the line is: 𝑦 = 𝑚 𝑥 + 𝑏 . y=mx+b. Substitute one point, e.g., ( 2 , 38 ) (2,38): 38 = 3 ( 2 ) + 𝑏    ⟹    38 = 6 + 𝑏    ⟹    𝑏 = 32. 38=3(2)+b⟹38=6+b⟹b=32. The linear equation is: 𝑦 = 3 𝑥 + 32. y=3x+32. Predict the height of a 5-year-old boy ( 𝑥 = 5 x=5): 𝑦 = 3 ( 5 ) + 32 = 15 + 32 = 47. y=3(5)+32=15+32=47. Result: The linear equation is: 𝑦 = 3 𝑥 + 32. y=3x+32. The predicted height of a 5-year-old boy is: 47   inches . 47inches. Question 4 Find the volume of the parallelepiped with vertices ( 1 , 3 , 0 ) (1,3,0), ( − 2 , 0 , 2 ) (−2,0,2), and ( − 1 , 3 , − 1 ) (−1,3,−1). Answer: The vectors defining the parallelepiped are: 𝑣 1 = ( − 2 − 1 , 0 − 3 , 2 − 0 ) = ( − 3 , − 3 , 2 ) , v 1 ​ =(−2−1,0−3,2−0)=(−3,−3,2), 𝑣 2 = ( − 1 − 1 , 3 − 3 , − 1 − 0 ) = ( − 2 , 0 , − 1 ) , v 2 ​ =(−1−1,3−3,−1−0)=(−2,0,−1), 𝑣 3 = 𝑣 1 × 𝑣 2 . v 3 ​ =v 1 ​ ×v 2 ​ . Construct the matrix: 𝐴 = [ 1 − 2 − 1 3 0 3 0 2 − 1 ] . A= ​ 1 3 0 ​ −2 0 2 ​ −1 3 −1 ​ ​ . The volume is: Volume = ∣ det ⁡ ( 𝐴 ) ∣ . Volume=∣det(A)∣. Calculate det ⁡ ( 𝐴 ) det(A): det ⁡ ( 𝐴 ) = 1 ∣ 0 3 2 − 1 ∣ − ( − 2 ) ∣ 3 3 0 − 1 ∣ + ( − 1 ) ∣ 3 0 0 2 ∣ . det(A)=1 ​ 0 2 ​ 3 −1 ​ ​ −(−2) ​ 3 0 ​ 3 −1 ​ ​ +(−1) ​ 3 0 ​ 0 2 ​ ​ . = 1 ( 0 − 6 ) − ( − 2 ) ( − 3 − 0 ) + ( − 1 ) ( 6 − 0 ) , =1(0−6)−(−2)(−3−0)+(−1)(6−0), = − 6 − 6 − 6 = − 18. =−6−6−6=−18. Volume = ∣ − 18 ∣ = 18. Volume=∣−18∣=18. Result: The volume of the parallelepiped is: 18. 18. Question 5 Find the equation of the plane through the origin and perpendicular to the vector ⟨ 1 , − 2 , 5 ⟩ ⟨1,−2,5⟩. Answer: The equation of a plane is: 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 = 0 , x+by+cz=0, where ⟨ 𝑎 , 𝑏 , 𝑐 ⟩ ⟨a,b,c⟩ is the normal vector. Substitute ⟨ 1 , − 2 , 5 ⟩ ⟨1,−2,5⟩: 1 𝑥 − 2 𝑦 + 5 𝑧 = 0. 1x−2y+5z=0. Result: The equation of the plane is: 𝑥 − 2 𝑦 + 5 𝑧 = 0. x−2y+5z=0. 2/2