Lines_ Attempt review _ eClass

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 1/15 Started on Wednesday, 17 January 2024, 10:08 AM State Finished Completed on Monday, 22 January 2024, 11:47 AM Time taken 5 days 1 hour Grade 56.00 out of 56.00 ( 100 %)

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 2/15 Question 1 Correct Mark 7.00 out of 7.00 Lines in R n For general explanations about lines in R 2 , watch Equations of Lines Part 1 between 00:00 and 13:00. The accompanying PDF file is here . Column vectors a b are used in this video instead of row vectors a b : this may seem strange, but columns vectors will be used more often than row vectors in this course for reasons that will become apparent later. Given a point P with position vector p and a non-zero vector d , both in R n , the set ℓ = { p + t d | t ∈ R} is a line in R n that passes through the point P and is parallel to the vector d . What that notation means is that the line ℓ consists of all the vectors of the form p + t d where t is any real number. More precisely, to be able to picture a line (in R 2 or R 3 ), we should consider the set above as the set of position vectors → OX of all the points X on the line ℓ . Here, → OX is the directed line segment (arrow) connecting the origin O to the point X . In particular, p = → OP and x = → OX . The vector d is called the direction vector for the line. The equation x = p + t d describing the position vector x of a point X on the line ℓ is called a vector equation for the line. To view the line as a function of t we write x ( t ) = p + t d . It can be useful to think of t as a variable representing time and x ( t ) as describing the position of a point that moves on the line ℓ . In particular, at time t = 0  , the point is at the position given by the vector p . Example: if p = [0, 1] and d = [1, 1] , then the line with vector equation x = p + t d is shown below. [ ] [ ]

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 4/15 Question 3 Correct Mark 7.00 out of 7.00 Finding a vector equation for a line through two points. For a video explaining one specific example of finding a vector equation using two points, see Equations of Lines Part 2 between 33:40 and 37:07. If two points \( P_1\) and\( P_2\), with position vectors \(\mathbf{p}_1\) and \(\mathbf{p}_2\), are on a line \(\ell\), then a vector equation for \ (\ell\) is \(\mathbf{x} = \mathbf{p}_1 + t\mathbf{d}\), where \(\mathbf{d} = \mathbf{p}_2 - \mathbf{p}_1\). That is, the line is described by \[ \mathbf{x}(t) = \mathbf{p}_1 + t(\mathbf{p}_2 - \mathbf{p}_1) ,\] where \(t \in \mathbb{R}\). The point \(\mathbf{p}_1\) is on the line since \(\mathbf{x}(t)=\mathbf{p}_1\) when \(t=\) 0  , and \(\mathbf{p}_2\) is on \(\ell\) since \(\mathbf{x}(t)=\mathbf{p}_2\) when \(t=\) 1  . Example: If \(\ell\) is the line in \(\mathbb{R}^3\) through the points \(\mathbf{p}_1=[1,2,1]\) and \(\mathbf{p}_2=[-2,1,3]\), by the formula above, a vector equation for \(\ell\) is \(\qquad \mathbf{x}(t) =\) [1,2,1]  \(+t\) [-3,-1,2]  . \(\qquad\) (Use square brackets, commas and no spaces for your vectors.) A third point on \(\ell\), this time with third coordinate \(7\), has position vector \(\mathbf{p}_3=[\) -8  , -1  ,\(7]\), which equals \( \mathbf{x}(t) \) when \( t=\) 3  . Correct Marks for this submission: 7.00/7.00.

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 5/15 Question 4 Correct Mark 5.00 out of 5.00 A vector equation for a line is not unique If \(\mathbf{q}\) is the position vector of a point \( Q\) on the line \(\mathbf{x}(t)=\mathbf{p}+t\mathbf{d}\), then the line through \(Q\) with direction vector \(\mathbf{d}\) has vector equation \(\mathbf{x_1}(t)=\mathbf{q}+t\mathbf{d}\). The functions \(\mathbf{x} (t)=\mathbf{p}+t\mathbf{d}\) and \(\mathbf{x_1}(t)=\mathbf{q}+t\mathbf{d}\) produce the same line. The point \((-4,-4)\) is on the common line shown above. To get \([-4,-4]=\mathbf{x}(t)\) we let \(t=\) -4  , but to get \ ([-4,-4]=\mathbf{x_1}(t)\) we would have to let \(t=\) -2  . As functions of \(t\), \(\mathbf{x}(t)\) and \(\mathbf{x_1}(t)\) are different, but they trace out the same line \(\qquad \ell=\{[0,4]+t[1,2]\,:\,t\in\mathbb{R}\}=\{[-2,0]+t[1,2]\,:\,t\in\mathbb{R}\}\). Similarly, for any \(\lambda\neq 0\) the vector \(\mathbf{v}=\lambda\mathbf{d}\) is parallel to the direction vector in the line \(\mathbf{x} (t)=\mathbf{p}+t\mathbf{d}\), and hence the equation \(\mathbf{x_2}(t)=\mathbf{p}+t\mathbf{v}\) produces the same line.

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 7/15 Question 5 Correct Mark 5.00 out of 5.00 The Intersection of two lines The two lines in the image below intersect at (4,4)  . (Type your point with round brackets, commas and no spaces.) We see that this point is obtained on the blue line since \(\mathbf{x_1}(t)=[4,4]\) when \(t=\) 4  and that it is also on the purple line since \(\mathbf{x_2}(t)=[4,4]\) when \(t=\) 2  . The name of the variable \(t\) in the above two equations can be chosen to be anything, so the two \(t\)'s do not need to be equal for the line to intersect. It is more difficult to visualize the intersection of two lines in \(\mathbb{R}^3\), but if they do intersect, we can use linear algebra to find the point of intersection. For a video on this topic, see: Intersection of lines in R^3 . For the moment, you can watch it until 3:40. The example below can be worked out without the method described in the video after 3:48. The lines \(\mathbf{x_3}=[1,2,1]+s[1,0,-1]\) and \(\mathbf{x_4}=[2,2,0]+t[-1,0,1]\) intersect at the point \((-3,2,5)\) since \(\mathbf{x_3}(s)= [-3,2,5]\) when \(s=\) -4  and \(\mathbf{x_4}(t)=[-3,2,5]\) when \(t=\) 5  . Correct Marks for this submission: 5.00/5.00.

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 8/15 Question 6 Correct Mark 6.00 out of 6.00 Parametric equations for a line The part of the video Equations of Lines Part 1 between 13:00 and 20:10 is relevant here. Suppose \(\ell\) is the line in \(\mathbb{R}^2\) with vector equation \(\mathbf{x} = [-2,3] + t[4,-7]\). Let us find equations for the coordinates of a general point \( (x_1,x_2) \) on this line. The position vector of a point \( (x_1,x_2) \) on \(\ell\) takes the form \[ [x_1,x_2] = [-2,3] + t[4,-7] = [-2 + 4t,3 - 7t] \] with \(t \in \mathbb{R}\), so \[ x_1 = -2 + 4t ,\quad x_2 = 3 - 7t .\] These two equations for \(x_1\) and \(x_2\) are called parametric equations of the line. Parametric equations are not unique: others are possible for the same line \(\ell\). General forms for parametric equations in 2 and 3 dimensions In \(\mathbb{R}^2\), the parametric equations for a line \(\ell\) with vector equation \(\mathbf{x} = \mathbf{p} + t\mathbf{d}\), where \ (\mathbf{p} = [a_1,a_2]\) and \(\mathbf{d} = [b_1,b_2]\), are \[ x_1 = a_1 + tb_1 ,\quad x_2 = a_2 + tb_2 .\] In \(\mathbb{R}^3\), the parametric equations for a line \(\ell\) with vector equation \(\mathbf{x} = \mathbf{p} + t\mathbf{d}\), where \ (\mathbf{p} = [a_1,a_2,a_3]\) and \(\mathbf{d} = [b_1,b_2,b_3]\), are \[ x_1 = a_1 + tb_1 ,\quad x_2 = a_2 + tb_2 ,\quad x_3 = a_3 + tb_3 .\] Example Let \(\ell\) be the line in \(\mathbb{R}^2\) passing through the points with position vectors \(\mathbf{p}_1 = [-1,2]\) and \(\mathbf{p}_2 = [3,4]\). Then a vector equation for \(\ell\) is \(\mathbf{x} = [-1,2] + t[4,2]\), because \(\mathbf{p}_2 - \mathbf{p}_1 = [\) 4  \(,\) 2  \(]\). Therefore, \(\ell\) has parametric equations \[ x_1 = -1 + 4t ,\quad x_2 = 2 + rt ,\] where \(r = \) 2  . We will use these parametric equations to find where \(\ell\) crosses the \(x_1\)-axis. When \(x_2 = 0\), \(t = \) -1  . In this case, \ (x_1 = \) -5  . Therefore, \(\ell\) crosses the \(x_1\)-axis at the point \((\) -5  \(,0)\). Correct Marks for this submission: 6.00/6.00.

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 10/15 Question 8 Correct Mark 6.00 out of 6.00 Using parametric equations to find intersections: An example similar to the one below is presented in the video Intersection of lines in \(\mathbb{R}^3\) starting at 3:48. The lines \(\mathbf{x}(s)=[1,2,3]+s[2,-1,1]\) and \(\mathbf{x}(t)=[1,-1,4]+t[3,0,1]\) have parametric equations \[ \begin{array}{lclclcl} x_1&=&1+2s &\qquad& x_1=1+3t \\ x_2&=&2-s &\qquad& x_2=-1 \\ x_3&=&3+s &\qquad& x_3=4+t. \end{array}\] These lines intersect if and only if we can find \(s\) and \(t\) that produce the same coordinates \(x_i\) for \( i=1,2,3\). This happens exactly when \[ \begin{array}{lcl} 1+2s &=& 1+3t \\ 2-s &=& -1 \\ 3+s &=& 4+t \end{array}\] which is equivalent to \[ \begin{array}{lcl} 2s-3t &=& 0 \\ -s &=& -3 \\ s-t &=& 1 \end{array}\] which happens \( \Longleftrightarrow \) \(s=\) 3  and \(t=\) 2  . The point of intersection has position vector \([1,2,3]+\) 3  \([2,-1,1]=\) [7,-1,6]  \(=[1,-1,4]+\) 2  \([3,0,1]\). (Type your vector answer with square brackets, commas and no spaces.) Repeat the above process for the lines \(\mathbf{x}(s)=[1,2,3]+s[2,-1,1]\) and \(\mathbf{x}(r)=[1,1,2]+r[1,2,1]\). We see that these lines  intersect. (If you need help seeing why this is the case then please join an office hour to ask!) do not Correct Marks for this submission: 6.00/6.00.

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 11/15 Question 9 Correct Mark 6.00 out of 6.00 How to find the general equation of a line in \(\mathbb{R}^2\) with vector equation \(\mathbf{x}=\mathbf{p}+t\mathbf{d}\): 1. Find a non-zero vector \(\mathbf{n}=[a,b]\) that is orthogonal to \(\mathbf{d}\) (there are many to choose from). 2. Expand the normal form of the equation with position vector \(\mathbf{p}\) and normal vector \(\mathbf{n}\). This is explained in the video Equations of Lines Part 2 between 16:08 and 23:35. Example The line \(\mathbf{x}=[1,3]+t[4,-2]\) has normal vector \(\mathbf{n}=[2,\) 4  \(]\), so a general equation for the line is \(\qquad 2x+\) 4  \(y=\) 14  . Note: to find one vector orthogonal to \( [c,d] \), flip \( c \) and \( d\), and change one sign: you can check that both \( [-d,c] \) and \( [d,-c] \) are orthogonal to \( [c,d] \) because their dot products with \( [c,d] \) equal 0. The general equation of a line is unique up to non-zero scalar multiple. Example The vector \(\mathbf{n}=[1,\) 2  \(]\) is also a normal vector for the line described by the equation \(\mathbf{x}=[1,3]+t[4,-2]\), so a general equation for the line is \(\qquad x+\) 2  \(y=\) 7  , which is just \(1/2\) times the previous general equation for the same line. Correct Marks for this submission: 6.00/6.00.

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 13/15 Question 11 Correct Mark 5.00 out of 5.00 In this question we will learn how to use projections to locate the point on a line \(\ell\) that is closest to a given point \( P \) that is not on \ (\ell\). We can use this to also find the distance from \(\mathbf{p}\) to \(\ell\). We will start using column vectors. These may seem more strange to you than row vectors, but we will use mostly columns vectors in this course because they are more compatible with systems of linear equations. One can easily pass from a row vector to a column vector and vice-versa: just write the components of the vector vertically instead of horizontally, and vice-versa. Let us assume that the position vector of \( P \) is \(\mathbf{p}={\left[\begin{array}{c} 4 \\ 3 \\ 1 \end{array}\right]}\), and let \(\ell\) be the line through the point with position vector \(\mathbf{q}={\left[\begin{array}{c} -3 \\ 5 \\ -1 \end{array}\right]}\) that is parallel to the vector \ (\mathbf{v}={\left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right]}\), so \[\ell=\{\mathbf{q}+t\mathbf{v}\,|\,t\in\mathbb{R}\}.\] Join the line to the point: We find a vector \(\mathbf{w}\) that connects the line to the point \(P\). The vector \(\mathbf{w}=\)\(\mathbf{p}-\mathbf{q}=\) 7 -2 2 Your last answer was interpreted as follows: \[ \left[\begin{array}{c} 7 \\ -2 \\ 2 \end{array}\right] \] Correct answer, well done. Marks for this submission: 1.00/1.00. has this property. Project the join onto the line: We project our \(\mathbf{w}\) onto the vector \(\mathbf{v}\), which is a direction vector of the line. This gives us the vector \ (\mathbf{w}_{\mathbf{v}}=\) \(\text{proj}_{\mathbf{v}} {\mathbf{w}}=\) \(\left( \dfrac{\mathbf{w}\cdot\mathbf{v}} {\mathbf{v}\cdot\mathbf{v}} \right) \mathbf{v} =\) 6 -3 3 Your last answer was interpreted as follows: \[ \left[\begin{array}{c} 6 \\ -3 \\ 3 \end{array}\right] \]

3/29/24, 1:37 PM Lines: Attempt review | eClass https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=14583624&cmid=7601843 14/15 Correct answer, well done. Marks for this submission: 1.00/1.00. The closest point: The point \( S\) with position vector given by \(\mathbf{s}=\)\(\mathbf{q}+\mathbf{w}_{\mathbf{v}}=\) 3 2 2 is the point on \(\ell\) that is closest to the point \( P \). Your last answer was interpreted as follows: \[ \left[\begin{array}{c} 3 \\ 2 \\ 2 \end{array}\right] \] Correct answer, well done. Marks for this submission: 1.00/1.00. The distance: The distance between the point \( P \) and the line \(\ell\) is the distance between the vectors \(\mathbf{s}\) and \(\mathbf{p}\), which is \(\qquad \text{dist}(\mathbf{s},\mathbf{p})=\|\mathbf{s}-\mathbf{p}\|=\) sqrt(3) . Your last answer was interpreted as follows: \[ \sqrt{3} \] Correct answer, well done. Marks for this submission: 1.00/1.00. (To enter something line \(\sqrt{5}\), type sqrt(5).) This same distance can be calculated as the length of the vector \(\mathbf{w}_{\mathbf{v}^{\perp}}=\)\(\mathbf{w}-\mathbf{w}_{\mathbf{v}}=\) 1 1 -1