Questions and Answers #2 Vectors and Spaces

Questions and Answers Sheet 2 Vectors and Spaces Question #1 Use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work. 1 + 2𝑡 3 , 2 + 𝑡 − 3𝑡 2 , −𝑡 + 2𝑡 2 − 𝑡 3 1 + 2𝑡 Answer: Step 1 Write the polynomias as coordinate vectors with respect to the standard (ordered) basis for 𝑃𝟛 , namely 𝛽 = {1, 𝑡, 𝑡 2 , 𝑡 3 } 1 2 0 0 1 −1 [1 + 2t 3 ] = ( ),[2 + t − 3t 2 ] = ( ),[−t + 2t 2 − t 3 ] = ( ) 0 −3 2 2 0 −1 Step 2 Setup the matrix 𝐴 = ([1 + 2𝑡 3 ]𝛽 ),, and row reduce it. Since there is a pivot in each column, the coordinate vectors are linearly independant and so must our polynomials be by theorem 8 p. 237. 1 2 0 1 0 1 −1 0 A=( )~( 0 −3 2 0 2 0 −1 0 2 0 1 −1 ) 0 1 0 0 Result 1 + 2𝑡 3, 2 + 𝑡 − 3𝑡 2 , −𝑡 + 2𝑡 2 − 𝑡 3 are linearly independent. Question #2 List five vectors in Span {𝑣1 , 𝑣2 } . For each vector, show the weights on 𝑣1 and 𝑣2 used to generate the vector and list three entries of the vector. Do not make a sketch. 7  −5   v1 =  1  , v2 =  3   −6  0  Answer: Remember that span {𝑣1 , 𝑣2 } is the set of all linear combinations of the vectors 𝑣1 and 𝑣2 . That means that it is the collection of all vectors that can be written as 𝑎𝑣1 + 𝑏𝑣2 for some numbers a and b. The numbers a and b are called the weights of the vector relative to 𝑣1 , 𝑣2 So to get any vector in the span, we just pick any two numbers a and b and from the linear combination with those weights. Vector 1: Here is the vector in the span with weights 1 and 2. The three entries of the vector are given in red: 7  −5  7   −10  −3 1  1  + 2  3  =  1  +  6  =  7   −6  0   −6  0   −6 Vector 2: Weights: 0 and -2. The three entries of the vector are given in red: 7  −5 10    0  1  − 2  3  =  −6  −6  0   0  Vector 3: Weight 0 and 0. The three entries of the vector are given in red: 7  −5 0 0  1  + 0  3  = 0  −6  0  0 Vector 4: Weights 5 and 7. The three entries of the vector are given in red: 7  −5  0    5  1  + 7  3  =  26   −6  0   −30 Vector 4: Weights -3 and 1. The three entries of the vector are given in red:  7   −5  −26 −3  1  + 1  3  =  0   −6  0   18   −3 10  0  0   −26  7  ,  −6 , 0 ,  26  ,  0  . Result:            −6  0  0  −30  18  Question #3 Find the unit vector that has the same direction as the vector v. 𝑣 = −𝑖 + 2𝑗 Answer: The goal of this is to find the unit vector that has the same direction as the vector |v|. To solve this problem we use the following formula: 𝑣 = √𝑎2 + 𝑏 2 ||𝑣|| 𝑣 = −𝑖 + 2𝑗 ||𝑣|| = √(−1)2 + 22 = √1 + 4 = √5 𝑣 ||𝑣|| = =− −𝑖+2𝑗 √5 √5 5 𝑣=− 𝑖+ √5 5 =− 2√5 5 𝑖+ 1 √5 𝑖+ 2 √5 𝑗 𝑗 2√5 5 Question #4 Find the scalar and vector projections of b onto a. 𝑎 = 𝑖 + 𝑗 + 𝑘, 𝑏 = 𝑖 − 𝑗 + 𝑘 Answer: The given vectors 𝑎⃗ and 𝑏⃗⃗ we can rewrite in the following form 𝑎⃗ =< 1,1,1 > 𝑏⃗⃗ =< 1, −1,1 > Since |𝑎| = √12 + 12 + 12 = √1 + 1 + 1 = √3 and 𝑎 ⋅ 𝑏 = 1 ⋅ 1 + 1(−1) + 1 ⋅ 1 = 1−1+1 =1 𝑡ℎ𝑒 𝑠𝑐alar and the vectors projection of b onto a, respectively give 1 𝑐𝑜𝑚𝑝𝑎 𝑏 = √3 √3 = 3 √3 <1,1,1> 𝑝𝑟𝑜𝑗𝑎 = 3 √3 √3 <1,1,1> 𝑝𝑟𝑜𝑗𝑎 𝑏 = 3 √3 1 1 1 =< , , > 3 3 3 𝑐𝑜𝑚𝑝𝑎 𝑏 = 1 1 1 √3 ; 𝑝𝑟𝑜𝑗𝑎 𝑏 =< , , > 3 3 3 3 Question #5 Force 𝐹⃗ = −10𝑗̂ N is exerted on a particle at 𝑟̂ = (5𝑖̂ + 5𝑗̂) m. What is the torque on the particle about the origin? Answer: Solve for the torque on the particle about the origin 𝐹⃗ = −10𝑗̂𝑁 𝑟⃗ = 5𝑖̂ + 5𝑗̂ Give these solutions 𝑖̂ × 𝑗̂ = 𝑘̂ 𝑗̂ × 𝑘̂ = 𝑖 𝑘̂ × 𝑖̂ 𝑟⃗ = 𝑟⃗ × 𝐹⃗ ˆj  iˆ kˆ    r  F = 5 −5 0  0 −10 0    𝑟⃗ × 𝐹⃗ = [(−5)(0) − (−10)(0)]𝑖̂ − [(5)(0) − (0)(0)]𝑗̂ + [(5)(−10) − (0)(−5)]𝑘̂ 𝑟̂ = (0𝑖̂ − 0𝑗̂ − 50𝑘̂ )𝑁 ⋅ 𝑚 Result 𝑟⃗ = (−50𝑘̂)𝑁 ⋅ 𝑚 Question #6 Find a basis for the space spanned by the given vectors, 𝑣1 … , 𝑣5 1   −2   6   5   0  0   1   −1  −3  3   , , , ,  0   −1  2   3   −1           1   1   −1  −4   1  Answer: This problem is equivalent to finding a basis for Col A, where A =  v1 v2 v3 v4 v5  Since the reduced echelon form A is 1 −2 6 5 0  1 0 1 −1 −3 3  0 ~ A= 0 −1 2 3 −1 0    1 1 −1 −4 1  0 0 1 0 0 0 −1 −2  0 −3 5  we see that the first, second and third columns of 1 0 2  0 0 0 A are its pivot columns. Thus a basis for the space spanned by the given vectors is 1   −2  6  0   1   −1 {  ,   ,  } 0   −1  2        1   1   −1 1   −2  6  0   1   −1 Result: Basis is {  ,   ,  } 0   −1  2        1   1   −1 Question #7 Find two vectors in opposite directions that are orthogonal to the vectot u. 𝑢 =< 3,5 > Answer: The vectors u and v are orthogonal if and only if 𝑢⋅𝑣 =0 We are given 𝑢 =< 3,5 > . We need to find 𝑣 =< 𝑣1 , 𝑣2 > such that: 𝑢⋅𝑣 =0 𝑟𝑖𝑔ℎ𝑡 ⇒ 3 ⋅ 𝑣1 + 5 ⋅ 𝑣2 = 0 5 𝑣1 = − 𝑣2 3 Now we can pick any value of 𝑣2 and well get 𝑣1 such that 𝑢 ⋅ 𝑣 = 0 For example, let 𝑣2 = 3 . Then, using (1), we have 𝑣1 = −5 For the vector in opposite direction we can take take 𝑤 = −𝑣 =< 5, −3 > Note that w also satisfies (1). Let's check that u and v are orthogonal. 𝑢 ⋅ 𝑣 =< 3,5 >⋅< −5,3 > = 3 ⋅ (−5) + 5 ⋅ 3 =0 We can easily see that u and w are also orthogonal. 𝑢 ⋅ 𝑤 = − 𝑢⏟ ⋅𝑣 =0 0 Result: 𝑣 =< −5,3 > and 𝑤 =< 5. −3 > Question #8 Find the gradient vector field of f. 𝑓(𝑥, 𝑦) = 𝑥𝑒 𝑥 𝑦 Answer: Recall that the gradient of the vector field is given as a vector its partial derivatives. That is: ∇𝑓 = ( 𝑑𝑓 𝑑𝑓 , 𝑑𝑥 𝑑𝑦 ) After calculating the partial derivatives we obtain the gradient: ∇𝑓 = (𝑥𝑒 𝑥𝑦 ⋅ 𝑦 + 𝑒 𝑥𝑦 , 𝑥 2 𝑒 𝑥𝑦 ) Result: ∇𝑓 = (𝑥𝑒 𝑥𝑦 ⋅ 𝑦 + 𝑒 𝑥𝑦 , 𝑥 2 𝑒 𝑥𝑦 ) Question #9 An object moving in the xy-plane is acted on by a conservative force described by the potential-energy function 𝑈(𝑥, 𝑦) = 𝛼(1/𝑥 2 + 1/𝑦 2 ) where 𝛼 is a positive constant. Derivative an expression for the force expressed terms of the unit vectors 𝑖⃗ and 𝑗⃗. Answer: First, we rewrite the potential energy produced by the force: 𝑈(𝑥, 𝑦) = 𝛼 ( 1 + 𝑥2 1 𝑦2 ) We know that, for motion in two dimentions, a conservative force is the negative partial derivative of its associated potential energy function with respect to each coordinate, multiplying by the corresponding unit vector: 𝑑𝑈 𝑑𝑈 𝐹⃗ = − ( 𝑖̂ + 𝑗̂) 𝑑𝑋 𝑑𝑦 So the force that acts on the object is: 𝑑 1 1 𝑑 1 𝐹⃗ = −𝛼 ( 2 + 2) 𝑖̂ − 𝛼 ( 2) 𝑗̂ = −𝛼 = 2𝛼 𝑑 𝑑𝑥 𝑥 1 ( ) 𝑖̂ − 𝛼 𝑑𝑥 𝑥 2 1 𝑥3 𝑦 𝑖̂ + 2𝛼 1 𝑦3 𝑑𝑦 𝑦 𝑑 1 ( ) 𝑗̂ 𝑑𝑦 𝑦 2 𝑗̂ 2𝛼 2𝛼 𝐹⃗ = 3 𝑖̂ + 3 𝑗̂ 𝑥 𝑦 Question #10 What is the angle between the given vector and the positive direction of the x-axis? 𝑖 + √3𝑗 Answer: Solution: Principles: The positive direction of x-axis can be represented using the following vector representation <1,0,0> Or In terms of unit vectors, we can express it as follows 1𝑖̂ + 0𝑗̂ + 0𝑘̂ Where, 𝑎̂ denotes the vector in the direction of the +ve x-axis The dot product between two vectors < 𝑥1 , 𝑦1 , 𝑧1 > and < 𝑥2 , 𝑦2 , 𝑧2 > is as follows < 𝑥1 , 𝑦1 , 𝑧1 >⋅< 𝑥2 , 𝑦2 , 𝑧2 ≥ 𝑥1 𝑦2 + 𝑧1 𝑧2 And the length of a vector , can be calculated using the following formula 2 | < 𝑎, 𝑏, 𝑐 > | = √𝑎 𝑏 + 𝑐 2 And the formula for the dot product between two vectors 𝑎⃗ and 𝑏⃗⃗ is as follows 𝑎⃗ ⋅ 𝑏⃗⃗ = |𝑎||𝑏| cos(𝜃) The dot produc of the two vectors using equation (1), is < 1,0 >⋅< 1, √3 ≥ 1 + 0 = Add the length of the vector <1,0> using equation (2), is | < 1,0 > | = √12 + 02 And the length of the vector < 1, √3 > , is | 1, 3 |= 12 + ( 3) 2 = 4 = 2 Thus, using the product formula (3), the cosine of the angle between both vectors is ⃗⃗ 𝑎⃗⃗⋅𝑏 1 cos(𝜃) = |𝑎||𝑏| = 1×2 = 0.5 Hence we can us the cousine inverse to find the angle \theta both vectors, where cos −1 (cos(𝜃)) = 𝜃 Thus, 𝜃 = cos −1 (0.5) = 60∘ Therefore, the angle between both vectors is 60∘ Question #11 Find the unit tangent and unit normal vectors T(t) and N(t). 𝑟(𝑡) =< 𝑡, 3 cos 𝑡 , 3 sin 𝑡 > Answer: Let us start with the given equation of the curve, that is, 𝑟(𝑡) =< 𝑡, 3 cos 𝑡 , 3 sin 𝑡 > Now, calculate the first derivative of the above equation of the curve component-wise, we get, |𝑟 ′ (𝑡)| = √12 + (−3 sin 𝑡)2 + (3 cos 𝑡)2 = √10 Calculate the unit tangent vector. It simplifies the later calculations if we leave the vector in this form, with 1 the coefficient on the outside of the vector, rather than distributing it within each component of the √𝑥 vector. 𝑟 ′(𝑡) 𝜏(𝑡) = |𝑟 ′(𝑡)| = 1 √10 ⋅< 1, −3 sin 𝑡 , 3 cos 𝑡 > Next, let us calculate the first derivative of this unit tangent vector. 1 𝜏 ′ (𝑡) = < 0, −3 cos 𝑡 , −3 sin 𝑡 > √10 Simplify by moving the common coefficient of 3 outside of the vector. 3 𝜏 ′ (𝑡) = < 0, − cos 𝑡 , − sin 𝑡 > √10 Calculate the magnitude of 𝜏 and simplify |𝜏 ′ (𝑡)| = √( = = = 3 √10 3 √10 3 3 √10 2 ) ⋅ ((− cos 𝑡)2 + (− sin 𝑡)2 ) √sin2 𝑡 + cos 2 𝑡 (1) √10 Calculate the unit normal vector and simplify. 𝜏′ (𝑡) 𝑁(𝑡) = |𝜏′ (𝑡)| = 3 <0,− cos 𝑡,− sin 𝑡> √10 3 √10 =< 0, − cos 𝑡 , − sin 𝑡 > Question #12 Find the directional derivative of the function at the given point in the direction of the vector v. 𝑓(𝑥, 𝑦) = 𝑥/𝑥 2 + 𝑦 2 , (1,2), 𝑢 =< 3,5 > Answer: Knowning that the directional derivative is 𝐷𝑢⃗⃗ 𝑓(𝑥, 𝑦) = ∇𝑓(𝑥, 𝑦) ⋅ 𝑢 ⃗⃗ Where ∇𝑓(𝑥, 𝑦) =< 𝑓_𝑥(𝑥, 𝑦), 𝑓_𝑦(𝑥, 𝑦)> and 𝑢 ⃗⃗ is a unit vector. 𝑥 𝑓(𝑥, 𝑦) = 2 2 , (1,2), 𝑣⃗ =< 3,5 > 𝑥 +𝑦 Using the quotient rule to find the partial derivative with respect to x 𝑓𝑥 = (1)(𝑥 2 +𝑦 2 )−(𝑥)(2𝑥) (𝑥 2 +𝑦 2 )2 𝑦 2 −𝑥 2 𝑓𝑥 = (𝑥 2 +𝑦 2 )2 Rewriting f(x,y) and using the chain rule to acquire the partial derivative with respect to y. 𝑓(𝑥, 𝑦) = 𝑥(𝑥 2 + 𝑦 2 )−1 𝑓𝑦 = (−1)(𝑥)(𝑥 2 + 𝑦 2 )−2 (2𝑦) −2𝑥𝑦 𝑓𝑦 = (𝑥 2 +𝑦 2 )2 Plugging in the given point into the partial derivatives gives us ∇𝑓(1,2) =< (2)2 −(1)2 𝑓𝑥 = ((1)2 +(2)2 )2 = 3 , −4 25 25 > 3 25 −2(1)(2) −4 𝑓𝑦 = = ((1)2 + (2)2 ) 25 Need a unit vector, so have to divide the components of the given vector by its length |𝑣⃗| = √32 + 52 = √34 𝑣⃗⃗ So 𝑢 ⃗⃗ = |𝑣⃗⃗| =< 3 , 5 √34 √34 > Dotting the gradient vector with our newly found unit vector −11 𝐷𝑢⃗⃗ 𝑓(1,2) = Result: 25√34 −11 25√34 Question #13 Find the domain of the vector function. 𝑟(𝑡) = 𝑡 − 2/𝑡 + 2𝑖 + sin 𝑡 𝑗 + ln(9 − 𝑡 2 ) 𝑘 Answer: 𝑡−2 𝑖 + sin 𝑡 𝑗 + ln(9 − 𝑡 2 ) 𝑘 𝑡+2 The domain has to exclude anything that will make the function undefined. 𝑡−2 For , the denominator can't be 0 so we exclude t=-2 𝑟(𝑡) = 𝑡+2 For \sin it is defined for all real numbers so there is nothing to exclude for this part. For ln(9 − 𝑡 2 ) , we can only take the logarithm of positive numbers so 9 − 𝑡 2 must be >0 0 = 9 − 𝑡2 = (3 − 𝑡)(3 + 𝑡) 𝑇 = −3,3 Since 𝑦 = 9 − 𝑡 2 is a downward opening parabola we know that it is only >0 in between but not including the zeros, -3 Answer: Recall that: Rate of change of at P(a,b,c) in the direction of u is 𝐷𝑢 f Where 𝐷𝑢 𝑓(𝑎, 𝑏, 𝑐) = ∇𝑓(𝑎, 𝑏, 𝑐) ⋅ 𝑢 Where u is unit vector From part (b), we know that ∇𝑓(0,1, −1) =< −1,2,0 > 3 4 12 Given that 𝑢 =< , , >, Note that it is already a unit vector 13 13 13 𝐷𝑢 𝑓(0,1, −1) = ∇𝑓(0,1, −1) ⋅ 𝑢 𝐷𝑢 𝑓(0,1, −1) =< −1,2,0 >⋅< 𝐷𝑢 𝑓(0,1, −1) = − 3 13 + 8 13 Result: 𝐷𝑢 𝑓(0,1, −1) = Question #16 3 +0= 5 13 , 4 , 12 13 13 13 5 13 > Find the best approximation to z by vectors of the form 𝑐1 𝑣1 + 𝑐2 𝑣2 3 2 1  −7   −1 1     z= ,v = v =   2  1  −3 2  0        3 1  −1 Answer: Find the best approximation to z by vectors of the form 𝑐1 𝑣1 + 𝑐2 𝑣2 3 2 1  −7   −1 1 z =   , v1 =   v2 =   2  −3 0       3 1  −1 The formula of orthigonal projection of z onto span 𝑣1 , 𝑣2 and this projection is the best approximiation to z by vectors 𝑐1 𝑣1 + 𝑐2 𝑣2 because 𝑧, 𝑣1 = 10, 𝑣1 ⋅ 𝑣1 = 15, 𝑧 ⋅ 𝑣2 = −7, 𝑣2 ⋅ 𝑣2 = 3 2 1     10 −1 −7  1  zˆ =   + 15  −3 3  0      1  −1 The required best approximation to z by vectors 𝑐1 𝑣1 + 𝑐2 𝑣2  −1  −3  ˆz =    −2    3 Result: 2  1   −1  −1     10   −7  1   −3 zˆ = + = 15  −3 3  0   −2       1  −1  3  Question #17 ⃗⃗ has magnitude 16.0 m. The scalar product 𝐴⃗ ⋅ 𝐵 ⃗⃗ is Vector 𝐴⃗ has magnitude 12.0 m, and vector 𝐵 2 112.0𝑚 . What is the magnitude of the vector product between these two vectors? Answer: ⃗⃗ = 112𝑚2 𝐴⃗ ⋅ 𝐵 ⃗⃗ , we will use the scalar product equation. To find the angle between 𝐴⃗ and 𝐵 ⃗ ⃗⃗ 𝐴 ⋅ 𝐵 = 𝐴𝐵𝑐𝑜𝑠 𝜙 112=(12)(16)\cos𝜙 112 cos 𝜙 = (12)(16) 112 𝜙 = cos −1 ((12)(16)) = 54.3∘ To calculate the magnitude of the vector product, use the following equation: C = AB sin  𝐶 = (12)(16) sin 5 4.3∘ = 156𝑚2 Result: 𝐶 = 156𝑚2 Question #18 Find 𝑎⃗ + 𝑏⃗⃗, 2𝑎⃗ + 3𝑏⃗⃗, |𝑎⃗| and |𝑎⃗ − 𝑏⃗⃗|. 𝑎⃗ =< 5, −12 >, 𝑏⃗⃗ =< −3, −6 > Answer: Using the properties of the vectors, we obtain 𝑎⃗ + 𝑏⃗⃗ =< 5 + (−3), −12 + (−6) > =< 2, −18 > 2𝑎⃗ + 3𝑏⃗⃗ =< 2 ⋅ 5, 2(−12) > +< 3(−3), 3(−6) > =< 10, −24 > +< −9, −18 > < 10 + (−9), −24 + (−18) > =< 1, −42 > The length of the vector 𝑎⃗ is |𝑎⃗| = √52 + (−12)2 = √25 + 144 = √169 =13 Determine the vector as the difference between the vectors 𝑎⃗ and 𝑏⃗⃗ , and then determine length of it 𝑎⃗ − 𝑏⃗⃗ =< 5 − (−3), −12 − (−6) > =<8,-6> Hence, |𝑎⃗ − 𝑏⃗⃗| = |< 8, −6 >| = √82 + (−6)2 = √64 + 36 = √100 = 10 Question #19 Determine whether or not F is a conservative vector field. If it is, find a function f such that F=delf. 𝐹(𝑥, 𝑦) = (3𝑥 2 − 2𝑦 2 )𝑖 + (4𝑥𝑦 + 3)𝑗 Answer: The vector field F=Pi+Qj is convervative if and only if 𝑑𝑃 𝑑𝑦 = 𝑑𝑄 𝑑𝑥 The given vector field is 𝐹(𝑥, 𝑦) = (3𝑥 2 − 2𝑦 2 )𝑖 + (4𝑥𝑦 + 3)𝑗 𝑑𝑃 Here 𝑃 = 3𝑥 2 − 2𝑦 2 , Therefore And 𝑄 = 4𝑥𝑦 + 3 , Therefore 𝑑𝑃 𝑑𝑦 ≠ 𝑑𝑄 𝑑𝑥 𝑑𝑦 = −4𝑦 = 4𝑦 𝑑𝑄 𝑑𝑥 Therefore F is NOT conservative. Question #20 Describe the zero vector (the additive identity) of the vector space. 𝑅4 Answer: Every vector in 𝑅𝟜 can be represented as 𝑣 = (𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 ) We have to find the additive identity vector - the vector that has following property: 𝑣+𝑥 = 𝑥+𝑣 =𝑣 In coordinate notation: (𝑣1 , 𝑣2 , 𝑣3 , 𝑣3 , 𝑣4 ) We have to find the additive identity vector - the vector that has following property: 𝑣+𝑥 = 𝑥+𝑣 =𝑣 In coordinate notation: (𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 ) + (𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 ) = (𝑣1 + 𝑥1 , 𝑣2 + 𝑥2 , 𝑣3 + 𝑥3 , 𝑣4 + 𝑥4 ) = (𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 ) 𝑣1 + 𝑥1 = 𝑣1 𝑣2 + 𝑥2 = 𝑣2 𝑣3 + 𝑥3 = 𝑣3 𝑣4 + 𝑥4 = 𝑣4 From properties of addition in 𝑅, we know that 𝑥1 = 𝑥2 = 𝑥3 = 𝑥4 = 0 So, additive identity vector is 𝑥 = (0,0,0,0). Result: (0,0,0,0)