Chapter 3 Lines and Planes 3.1 Lines in â 1. Find the equation of
Chapter 3 Lines and Planes 3.1 Lines in ℝ2 1. Find the equation of the line that passes through the point 𝑥 (1, 3) and is parallel to 𝐿: [𝑦] = [1] + 𝑡 [−4]. 0 1 Since the line is parallel to 𝐿, it must have the same −4 direction vector: [ ]. 1 𝑥 1 −4 So, the equation of the line is [𝑦] = [ ] + 𝑡 [ ]. 3 1 𝑥 = 1 − 4𝑡 In parametric/scalar equations: . 𝑦 =3+𝑡 −4 In standard form: the normal vector is orthogonal to [ ]. 1 1 1 1 𝑛⃗ = [ ] 1𝑥 + 4𝑦 = [ ] ∙ [ ] 𝑥 + 4𝑦 = 13 4 4 3 I am only stating the most common forms, you can try to write out the two-point form and point-normal form equations on your own.
2. Find the equation of the line that passes through the point 𝑥 (2, −3) and is orthogonal to 𝐿: [𝑦] = [−1] + 𝑡 [ 3 ]. −1 −1 3 1 The direction vector will be orthogonal to [ ] [ ]. −1 3 𝑥 2 1 So, the equation of the line is [𝑦] = [ ] + 𝑡 [ ]. −3 3 𝑥 =2+𝑡 In parametric/scalar equations: . 𝑦 = −3 + 3𝑡 3 2 In standard form: 3𝑥 − 𝑦 = [ ] ∙ [ ] 3𝑥 − 𝑦 = 9 −1 −3 3. Find the equation of the line that passes through the point (1,1) and is orthogonal to 𝐿: 2𝑥 − 𝑦 = 3. 2 The normal of 𝐿 is [ ]. Since the line we want is −1 2 orthogonal to 𝐿, [ ] is the direction vector of such a line. −1 𝑥 1 2 So, the equation of the line is [𝑦] = [ ] + 𝑡 [ ]. 1 −1 𝑥 = 1 + 2𝑡 In parametric/scalar equations: . 𝑦 =1−𝑡 1 1 In standard form: 1𝑥 + 2𝑦 = [ ] ∙ [ ] 𝑥 + 2𝑦 = 3 2 1
𝑥 −3 2 ( ) 4. Is the point 1, 3 on the line 𝐿: [𝑦] = [ ] + 𝑡 [ ]? −6 5 Let 1 be the 𝑥-component and 3 be the 𝑦-component, then we try to solve for 𝑡. 1 = −3 + 2𝑡 3 = −6 + 5𝑡 From equation 1: we see that 𝑡 = 2. From equation 2: we see that 𝑡 = 9/5. Since the 𝑡 values are different, the point is not on the line. (If the 𝑡 values were the same, then the point is on the line)
3.2 Lines and Planes in ℝ3 1. Find the point-parallel form equation of the line that passes through the points 𝐴(1, 1, 0) and 𝐵(4, −2, 1). 3 The direction vector ⃗⃗⃗⃗⃗ 𝐴𝐵 = [−3]. 1 So, the point-parallel form equation of the desired line is 𝑥 1 3 x⃗ = [𝑦] = [1] + 𝑡 [−3] 𝑧 0 1 2. Find a standard form equation of the plane in ℝ3 that passes through the point (1, −1, 2) and is perpendicular to 𝑥 3 −2 the line 𝐿: [𝑦 ] = [ 0 ] + 𝑡 [ 1 ] 𝑧 −1 −1 Since the line is perpendicular to 𝐿, its normal vector must −2 be a multiple of the direction vector of 𝐿 [ 1 ]. −1 Putting the equation into standard form 𝑎 3 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑, where 𝑛⃗ = [𝑏] and 𝑑 = 𝑛⃗ ∙ [0], 𝑐 1 −2 3 we have −2𝑥 + 1𝑦 − 1𝑧 = [ 1 ] ∙ [ 0 ] −1 −1 −2𝑥 + 𝑦 − 𝑧 = −6 + 0 + 1 −2𝑥 + 𝑦 − 𝑧 = −5 Make the first coefficient positive: we get 2𝑥 − 𝑦 + 𝑧 = 5.
3. For the following lines, state the point(s) of intersection (if any) and state whether the lines intersect, are identical, are skewed, or are parallel (but don’t intersect). 𝑥 = 3+𝑡 𝑥 = 7 − 2𝑟 a. 𝐿1 : 𝑦 = 2 − 2𝑡 𝐿2 : 𝑦 = −6 + 4𝑟 𝑧 = −7 + 𝑟 𝑧 = −5 + 𝑡 1 Quickly checking the direction vectors [−2] and 1 −2 [ 4 ], we find that they are not multiples of one 1 another, so the lines either intersect at a single point or are skewed. Now let’s equate the components 𝑥: 3 + 𝑡 = 7 − 2𝑟 𝑦: 2 − 2𝑡 = −6 + 4𝑟 𝑧: − 5 + 𝑡 = −7 + 𝑟 From the 𝑥 equation: 𝑡 = 4 − 2𝑟 Sub into the 𝑧 equation: −5 + (4 − 2𝑟) = −7 + 𝑟 𝑟=2 So, 𝑡 = 4 − 2𝑟 = 4 − 2(2) = 0 To check that these values are correct, we sub into the 𝑦 equation: 2 − 2(0) = −6 + 4(2) 2 = 2 it works! So the lines intersect at the point where 𝑡 = 0 or 𝑟 = 2 (3,2, −5)
𝑥 =1+𝑡 b. 𝐿1 : 𝑦 = 2 + 𝑡 𝑧 = 3 − 2𝑡
𝑥 =2+𝑟 𝐿2 : 𝑦 = 4 + 2𝑟 𝑧 =4−𝑟 1 1 Quickly checking the direction vectors [ 1 ] and [ 2 ], we −2 −1 find that they are not multiples of one another, so the lines either intersect at a single point or are skewed. Now let’s equate the components 𝑥: 1 + 𝑡 = 2 + 𝑟 𝑦: 1 + 𝑡 = 4 + 2𝑟 𝑧: 3 − 2𝑡 = 4 − 𝑟 From the 𝑥 equation: 𝑡 = 1 + 𝑟 Sub into the 𝑦 equation: 1 + (1 + 𝑟) = 2 + 𝑟 0=0 We don’t get anything useful, let’s sub it into the 𝑧 equation: 3 − 2(1 + 𝑟) = 4 − 𝑟 𝑟 = −3 So, 𝑡 = 1 + 𝑟 = 1 − 3 = −2 To check that these values are correct, we sub into the 𝑦 equation: 1 + (−2) = 4 + 2(−3) −1 = −2 it doesn’t work! So, the lines are skewed and don’t intersect.
𝑥 = −2 + 𝑡 c. 𝐿1 : 𝑦=𝑡 𝑧 = 12 − 2𝑡
𝑥 = 2 − 2𝑟 𝐿2 : 𝑦 = 4 − 2𝑟 𝑧 = 4 + 4𝑟 1 −2 Quickly checking the direction vectors [ 1 ] and [−2], we −2 4 find that they are multiples of one another, so the lines are either identical or parallel (but don’t intersect). Now let’s equate the components 𝑥: − 2 + 𝑡 = 2 − 2𝑟 𝑦: 𝑡 = 4 − 2𝑟 𝑧: 12 − 2𝑡 = 4 + 4𝑟 From the 𝑦 equation: 𝑡 = 4 − 2𝑟 Sub into the 𝑥 equation: −2 + (4 − 2𝑟) = 2 − 2𝑟 0=0 We don’t get anything too useful, let’s sub into 𝑧 equation: 12 − 2(4 − 2𝑟) = 4 + 4𝑟 0=0 Since we always get a valid solution (0 = 0 is always true), the lines have infinitely many points of intersection. Therefore, the lines are identical.
4. State the 𝑥, 𝑦, and 𝑧-intercepts (if any) of the plane Π: 2𝑥 + 3𝑦 − 𝑧 = 10 𝑥-intercept: Let 𝑦 = 0 and 𝑧 = 0: 2𝑥 = 10 𝑥 = 5 So, the 𝑥-intercept is (5, 0, 0) 𝑦-intercept: Let 𝑥 = 0 and 𝑧 = 0: 3𝑦 = 10 𝑦 = 10/3 So, the 𝑦-intercept is (0, 10/3, 0) 𝑧-intercept: Let x= 0 and 𝑦 = 0: −𝑧 = 10 𝑧 = −10 So, the 𝑧-intercept is (0, 0, −10)
5. Which of the following lie on the line that passes through the points 𝐴(1, 2, 0) and 𝐵 (3, −4, 2)? a. (3, −4, 0) b. (4, −2,2) c. (3, −1, 1) d. (5, −10, 3) e. None of the above The line that passes through 𝐴 and 𝐵 have direction vector 3 1 2 1 parallel to ⃗⃗⃗⃗⃗ 𝐴𝐵 = [−4] − [2] = [−6] or [−3] (you don’t 2 0 2 1 have to reduce, I’m doing it so the numbers look “nicer”) So the line that passes through 𝐴 and 𝐵 is 𝑥 1 1 𝑥 = 1+𝑡 [𝑦] = [2] + 𝑡 [−3] or 𝑦 = 2 − 3𝑡. 𝑧 0 1 𝑧=𝑡 Now we can check each point to see if it fits: 3=1+𝑡 a. −4 = 2 − 3𝑡 equation 3 gives 𝑡 = 0, but that 0=𝑡 doesn’t work for equation 1 or 2. Not a point on the line. 4=1+𝑡 b. −2 = 2 − 3𝑡 equation 3 gives 𝑡 = 2, but that 2=𝑡 doesn’t work for equation 1. Not a point on the line.
3=1+𝑡 c. −1 = 2 − 3𝑡 equation 3 gives 𝑡 = 1, but that 1=𝑡 doesn’t work for equation 1. Not a point on the line. 5=1+𝑡 d. −10 = 2 − 3𝑡 equation 3 gives 𝑡 = 3, but that 3=𝑡 doesn’t work for equation 1 or 2. Not a point on the line. e. Therefore, none of the above points are on the line passing through 𝐴 and 𝐵
3.3 Extension to ℝ𝑛 1. Find a normal vector for the hyperplane 2𝑥1 + 𝑥2 − 𝑥3 + 𝑥4 − 3𝑥5 = 15 in ℝ5 . State any point on the hyperplane. 2 1 The normal is 𝑛⃗ = −1 . 1 [−3] To find a point on the hyperplane, set 4 of the 5 variables equal to a “nice” value (like 0 𝑜𝑟 1) and solve for the remaining one Let 𝑥1 = 𝑥2 = 𝑥3 = 𝑥4 = 0, then the hyperplane equation becomes −3𝑥5 = 15 𝑥5 = −5. So, a point on the hyperplane is (0, 0, 0, 0, −5).
End of Chapter 3 Exam-like Quiz (Lines and Planes) 1. Which of the following vectors is a normal vector of a line that is parallel to 2𝑥 + 𝑦 = −3? a. (2,1) b. (−2, 1) c. (2, −1) d. (−1, 2) e. (−1, −2) Answer a. The desired line is parallel to 2𝑥 + 𝑦 = −3. So, it should have the same normal vector as 2𝑥 + 𝑦 = −3 Therefore, the normal vector of the desired line is (2, 1). 2. Which of the following vectors is a normal vector of a line that is orthogonal/perpendicular to 2𝑥 + 𝑦 = −3? a. (2,1) b. (−2, 1) c. (2, −1) d. (−1, 2) e. (−1, −2) Answer d. The desired line is perpendicular to 2𝑥 + 𝑦 = −3, so their normal vectors should be orthogonal. Therefore, the normal vector of the desired line is orthogonal to (2, 1) we need a multiple of (−1, 2). The answer is d.
3. Which one of following vectors is parallel to the line 𝑥 3 1 [𝑦] = (1 − 𝑡) [ 1 ] + 𝑡 [ 1 ] 𝑧 −1 −2 −2 a. [ 1 ] −1 6 b. [0] 3 −3 c. [ 0 ] 1 1 d. [1] 0 e. None of the above Answer b. 𝑥 3 −2 Simplifying the given line, we get [𝑦] = [ 1 ] + 𝑡 [ 0 ]. A 𝑧 −1 −1 vector parallel to this line must be a multiple of the direction −2 vector [ 0 ]. −1 6 −2 Since [0] = −3 [ 0 ], answer is b. 3 −1
4. Which of the following lines pass through the point (−2,5) and is parallel to 3𝑥 − 𝑦 = 1? a. (−2, 5) + 𝑡(3, −1) b. (−2, 5) + 𝑡(1, 3) c. (3, −1) + 𝑡(−2, 5) d. (1, 3) + 𝑡(−2, 5) e. None of the above Answer b. Since the desired line is parallel to 3𝑥 − 𝑦 = 1, their direction vectors are multiples of one another. 3 The direction vector of 3𝑥 − 𝑦 = 1 is orthogonal to [ ] −1 1 [ ] is a good choice. 3 Therefore, the line parallel is (𝑥, 𝑦) = (−2,5) + 𝑡(1,3)
5. Find the normal to the plane that contains the lines 𝐿1 : (𝑥, 𝑦, 𝑧) = (1, 3,0) + 𝑡(2, 0, −1) and 𝐿2 : (𝑥, 𝑦, 𝑧) = (5, −7,1) + 𝑡(1, 1, 0) a. (−2, 0, 1) b. (−1, −1, 0) c. (2, 0, 0) d. (1, −1, 2) e. (1, −1, −1) Answer d. The normal vector to the plane must be orthogonal to both direction vectors of the lines. 2
1
1
𝑛⃗ = [ 0 ] × [1] = [−1] −1
0
2
2. Find the equation of the line that passes through the point 𝑥 (2, −3) and is orthogonal to 𝐿: [𝑦] = [−1] + 𝑡 [ 3 ]. −1 −1 3 1 The direction vector will be orthogonal to [ ] [ ]. −1 3 𝑥 2 1 So, the equation of the line is [𝑦] = [ ] + 𝑡 [ ]. −3 3 𝑥 =2+𝑡 In parametric/scalar equations: . 𝑦 = −3 + 3𝑡 3 2 In standard form: 3𝑥 − 𝑦 = [ ] ∙ [ ] 3𝑥 − 𝑦 = 9 −1 −3 3. Find the equation of the line that passes through the point (1,1) and is orthogonal to 𝐿: 2𝑥 − 𝑦 = 3. 2 The normal of 𝐿 is [ ]. Since the line we want is −1 2 orthogonal to 𝐿, [ ] is the direction vector of such a line. −1 𝑥 1 2 So, the equation of the line is [𝑦] = [ ] + 𝑡 [ ]. 1 −1 𝑥 = 1 + 2𝑡 In parametric/scalar equations: . 𝑦 =1−𝑡 1 1 In standard form: 1𝑥 + 2𝑦 = [ ] ∙ [ ] 𝑥 + 2𝑦 = 3 2 1
𝑥 −3 2 ( ) 4. Is the point 1, 3 on the line 𝐿: [𝑦] = [ ] + 𝑡 [ ]? −6 5 Let 1 be the 𝑥-component and 3 be the 𝑦-component, then we try to solve for 𝑡. 1 = −3 + 2𝑡 3 = −6 + 5𝑡 From equation 1: we see that 𝑡 = 2. From equation 2: we see that 𝑡 = 9/5. Since the 𝑡 values are different, the point is not on the line. (If the 𝑡 values were the same, then the point is on the line)
3.2 Lines and Planes in ℝ3 1. Find the point-parallel form equation of the line that passes through the points 𝐴(1, 1, 0) and 𝐵(4, −2, 1). 3 The direction vector ⃗⃗⃗⃗⃗ 𝐴𝐵 = [−3]. 1 So, the point-parallel form equation of the desired line is 𝑥 1 3 x⃗ = [𝑦] = [1] + 𝑡 [−3] 𝑧 0 1 2. Find a standard form equation of the plane in ℝ3 that passes through the point (1, −1, 2) and is perpendicular to 𝑥 3 −2 the line 𝐿: [𝑦 ] = [ 0 ] + 𝑡 [ 1 ] 𝑧 −1 −1 Since the line is perpendicular to 𝐿, its normal vector must −2 be a multiple of the direction vector of 𝐿 [ 1 ]. −1 Putting the equation into standard form 𝑎 3 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑, where 𝑛⃗ = [𝑏] and 𝑑 = 𝑛⃗ ∙ [0], 𝑐 1 −2 3 we have −2𝑥 + 1𝑦 − 1𝑧 = [ 1 ] ∙ [ 0 ] −1 −1 −2𝑥 + 𝑦 − 𝑧 = −6 + 0 + 1 −2𝑥 + 𝑦 − 𝑧 = −5 Make the first coefficient positive: we get 2𝑥 − 𝑦 + 𝑧 = 5.
3. For the following lines, state the point(s) of intersection (if any) and state whether the lines intersect, are identical, are skewed, or are parallel (but don’t intersect). 𝑥 = 3+𝑡 𝑥 = 7 − 2𝑟 a. 𝐿1 : 𝑦 = 2 − 2𝑡 𝐿2 : 𝑦 = −6 + 4𝑟 𝑧 = −7 + 𝑟 𝑧 = −5 + 𝑡 1 Quickly checking the direction vectors [−2] and 1 −2 [ 4 ], we find that they are not multiples of one 1 another, so the lines either intersect at a single point or are skewed. Now let’s equate the components 𝑥: 3 + 𝑡 = 7 − 2𝑟 𝑦: 2 − 2𝑡 = −6 + 4𝑟 𝑧: − 5 + 𝑡 = −7 + 𝑟 From the 𝑥 equation: 𝑡 = 4 − 2𝑟 Sub into the 𝑧 equation: −5 + (4 − 2𝑟) = −7 + 𝑟 𝑟=2 So, 𝑡 = 4 − 2𝑟 = 4 − 2(2) = 0 To check that these values are correct, we sub into the 𝑦 equation: 2 − 2(0) = −6 + 4(2) 2 = 2 it works! So the lines intersect at the point where 𝑡 = 0 or 𝑟 = 2 (3,2, −5)
𝑥 =1+𝑡 b. 𝐿1 : 𝑦 = 2 + 𝑡 𝑧 = 3 − 2𝑡
𝑥 =2+𝑟 𝐿2 : 𝑦 = 4 + 2𝑟 𝑧 =4−𝑟 1 1 Quickly checking the direction vectors [ 1 ] and [ 2 ], we −2 −1 find that they are not multiples of one another, so the lines either intersect at a single point or are skewed. Now let’s equate the components 𝑥: 1 + 𝑡 = 2 + 𝑟 𝑦: 1 + 𝑡 = 4 + 2𝑟 𝑧: 3 − 2𝑡 = 4 − 𝑟 From the 𝑥 equation: 𝑡 = 1 + 𝑟 Sub into the 𝑦 equation: 1 + (1 + 𝑟) = 2 + 𝑟 0=0 We don’t get anything useful, let’s sub it into the 𝑧 equation: 3 − 2(1 + 𝑟) = 4 − 𝑟 𝑟 = −3 So, 𝑡 = 1 + 𝑟 = 1 − 3 = −2 To check that these values are correct, we sub into the 𝑦 equation: 1 + (−2) = 4 + 2(−3) −1 = −2 it doesn’t work! So, the lines are skewed and don’t intersect.
𝑥 = −2 + 𝑡 c. 𝐿1 : 𝑦=𝑡 𝑧 = 12 − 2𝑡
𝑥 = 2 − 2𝑟 𝐿2 : 𝑦 = 4 − 2𝑟 𝑧 = 4 + 4𝑟 1 −2 Quickly checking the direction vectors [ 1 ] and [−2], we −2 4 find that they are multiples of one another, so the lines are either identical or parallel (but don’t intersect). Now let’s equate the components 𝑥: − 2 + 𝑡 = 2 − 2𝑟 𝑦: 𝑡 = 4 − 2𝑟 𝑧: 12 − 2𝑡 = 4 + 4𝑟 From the 𝑦 equation: 𝑡 = 4 − 2𝑟 Sub into the 𝑥 equation: −2 + (4 − 2𝑟) = 2 − 2𝑟 0=0 We don’t get anything too useful, let’s sub into 𝑧 equation: 12 − 2(4 − 2𝑟) = 4 + 4𝑟 0=0 Since we always get a valid solution (0 = 0 is always true), the lines have infinitely many points of intersection. Therefore, the lines are identical.
4. State the 𝑥, 𝑦, and 𝑧-intercepts (if any) of the plane Π: 2𝑥 + 3𝑦 − 𝑧 = 10 𝑥-intercept: Let 𝑦 = 0 and 𝑧 = 0: 2𝑥 = 10 𝑥 = 5 So, the 𝑥-intercept is (5, 0, 0) 𝑦-intercept: Let 𝑥 = 0 and 𝑧 = 0: 3𝑦 = 10 𝑦 = 10/3 So, the 𝑦-intercept is (0, 10/3, 0) 𝑧-intercept: Let x= 0 and 𝑦 = 0: −𝑧 = 10 𝑧 = −10 So, the 𝑧-intercept is (0, 0, −10)
5. Which of the following lie on the line that passes through the points 𝐴(1, 2, 0) and 𝐵 (3, −4, 2)? a. (3, −4, 0) b. (4, −2,2) c. (3, −1, 1) d. (5, −10, 3) e. None of the above The line that passes through 𝐴 and 𝐵 have direction vector 3 1 2 1 parallel to ⃗⃗⃗⃗⃗ 𝐴𝐵 = [−4] − [2] = [−6] or [−3] (you don’t 2 0 2 1 have to reduce, I’m doing it so the numbers look “nicer”) So the line that passes through 𝐴 and 𝐵 is 𝑥 1 1 𝑥 = 1+𝑡 [𝑦] = [2] + 𝑡 [−3] or 𝑦 = 2 − 3𝑡. 𝑧 0 1 𝑧=𝑡 Now we can check each point to see if it fits: 3=1+𝑡 a. −4 = 2 − 3𝑡 equation 3 gives 𝑡 = 0, but that 0=𝑡 doesn’t work for equation 1 or 2. Not a point on the line. 4=1+𝑡 b. −2 = 2 − 3𝑡 equation 3 gives 𝑡 = 2, but that 2=𝑡 doesn’t work for equation 1. Not a point on the line.
3=1+𝑡 c. −1 = 2 − 3𝑡 equation 3 gives 𝑡 = 1, but that 1=𝑡 doesn’t work for equation 1. Not a point on the line. 5=1+𝑡 d. −10 = 2 − 3𝑡 equation 3 gives 𝑡 = 3, but that 3=𝑡 doesn’t work for equation 1 or 2. Not a point on the line. e. Therefore, none of the above points are on the line passing through 𝐴 and 𝐵
3.3 Extension to ℝ𝑛 1. Find a normal vector for the hyperplane 2𝑥1 + 𝑥2 − 𝑥3 + 𝑥4 − 3𝑥5 = 15 in ℝ5 . State any point on the hyperplane. 2 1 The normal is 𝑛⃗ = −1 . 1 [−3] To find a point on the hyperplane, set 4 of the 5 variables equal to a “nice” value (like 0 𝑜𝑟 1) and solve for the remaining one Let 𝑥1 = 𝑥2 = 𝑥3 = 𝑥4 = 0, then the hyperplane equation becomes −3𝑥5 = 15 𝑥5 = −5. So, a point on the hyperplane is (0, 0, 0, 0, −5).
End of Chapter 3 Exam-like Quiz (Lines and Planes) 1. Which of the following vectors is a normal vector of a line that is parallel to 2𝑥 + 𝑦 = −3? a. (2,1) b. (−2, 1) c. (2, −1) d. (−1, 2) e. (−1, −2) Answer a. The desired line is parallel to 2𝑥 + 𝑦 = −3. So, it should have the same normal vector as 2𝑥 + 𝑦 = −3 Therefore, the normal vector of the desired line is (2, 1). 2. Which of the following vectors is a normal vector of a line that is orthogonal/perpendicular to 2𝑥 + 𝑦 = −3? a. (2,1) b. (−2, 1) c. (2, −1) d. (−1, 2) e. (−1, −2) Answer d. The desired line is perpendicular to 2𝑥 + 𝑦 = −3, so their normal vectors should be orthogonal. Therefore, the normal vector of the desired line is orthogonal to (2, 1) we need a multiple of (−1, 2). The answer is d.
3. Which one of following vectors is parallel to the line 𝑥 3 1 [𝑦] = (1 − 𝑡) [ 1 ] + 𝑡 [ 1 ] 𝑧 −1 −2 −2 a. [ 1 ] −1 6 b. [0] 3 −3 c. [ 0 ] 1 1 d. [1] 0 e. None of the above Answer b. 𝑥 3 −2 Simplifying the given line, we get [𝑦] = [ 1 ] + 𝑡 [ 0 ]. A 𝑧 −1 −1 vector parallel to this line must be a multiple of the direction −2 vector [ 0 ]. −1 6 −2 Since [0] = −3 [ 0 ], answer is b. 3 −1
4. Which of the following lines pass through the point (−2,5) and is parallel to 3𝑥 − 𝑦 = 1? a. (−2, 5) + 𝑡(3, −1) b. (−2, 5) + 𝑡(1, 3) c. (3, −1) + 𝑡(−2, 5) d. (1, 3) + 𝑡(−2, 5) e. None of the above Answer b. Since the desired line is parallel to 3𝑥 − 𝑦 = 1, their direction vectors are multiples of one another. 3 The direction vector of 3𝑥 − 𝑦 = 1 is orthogonal to [ ] −1 1 [ ] is a good choice. 3 Therefore, the line parallel is (𝑥, 𝑦) = (−2,5) + 𝑡(1,3)
5. Find the normal to the plane that contains the lines 𝐿1 : (𝑥, 𝑦, 𝑧) = (1, 3,0) + 𝑡(2, 0, −1) and 𝐿2 : (𝑥, 𝑦, 𝑧) = (5, −7,1) + 𝑡(1, 1, 0) a. (−2, 0, 1) b. (−1, −1, 0) c. (2, 0, 0) d. (1, −1, 2) e. (1, −1, −1) Answer d. The normal vector to the plane must be orthogonal to both direction vectors of the lines. 2
1
1
𝑛⃗ = [ 0 ] × [1] = [−1] −1
0
2
