Determine whether the lines l1 and l2 are parallel, skew, or intersecting. If they intersect, find the point of intersection.<\/p>\n\n\n\n
l1: x=3-2t, y=7+4t, z=-3+8t<\/p>\n\n\n\n
l2: x=-1-u, y=18+3u, z=7+2u<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s analyze the given parametric equations for lines ( l_1 ) and ( l_2 ), and determine whether they are parallel, skew, or intersecting.<\/p>\n\n\n\n For line ( l_1 ): First, we extract the direction vectors of the lines. The direction vector of ( l_1 ) is the coefficient of the parameter ( t ) in the equations of ( l_1 ): Two lines are parallel if their direction vectors are proportional (i.e., one is a scalar multiple of the other). To check this, we compare the components of ( \\mathbf{d_1} ) and ( \\mathbf{d_2} ).<\/p>\n\n\n\n [ Since the components are not proportional, the lines are not parallel<\/strong>.<\/p>\n\n\n\n To determine if the lines intersect, we set the parametric equations of ( l_1 ) equal to the parametric equations of ( l_2 ):<\/p>\n\n\n\n [ Solving this system of equations, we first rearrange them: We now solve the system of equations:<\/p>\n\n\n\n From equation (1), solve for ( u ): Substitute ( u = 2t – 4 ) into equations (2) and (3):<\/p>\n\n\n\n Substitute into equation (2): Now substitute ( t = \\frac{1}{2} ) into ( u = 2t – 4 ): Now, we check the values of ( t = \\frac{1}{2} ) and ( u = -3 ) in the parametric equations of both lines.<\/p>\n\n\n\n For ( l_1 ), with ( t = \\frac{1}{2} ): For ( l_2 ), with ( u = -3 ): Since the coordinates of both lines match at ( (2, 9, 1) ), the lines intersect<\/strong> at this point.<\/p>\n\n\n\n The lines ( l_1 ) and ( l_2 ) are intersecting<\/strong> and the point of intersection is ( (2, 9, 1) ).<\/p>\n","protected":false},"excerpt":{"rendered":" Determine whether the lines l1 and l2 are parallel, skew, or intersecting. If they intersect, find the point of intersection. l1: x=3-2t, y=7+4t, z=-3+8t l2: x=-1-u, y=18+3u, z=7+2u The Correct Answer and Explanation is : Let’s analyze the given parametric equations for lines ( l_1 ) and ( l_2 ), and determine whether they are […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-190410","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190410","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=190410"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190410\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190410"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190410"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190410"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Parametric Equations:<\/h3>\n\n\n\n
[
\\begin{aligned}
x_1 &= 3 – 2t \\
y_1 &= 7 + 4t \\
z_1 &= -3 + 8t
\\end{aligned}
]
For line ( l_2 ):
[
\\begin{aligned}
x_2 &= -1 – u \\
y_2 &= 18 + 3u \\
z_2 &= 7 + 2u
\\end{aligned}
]<\/p>\n\n\n\nStep 1: Direction Vectors<\/h3>\n\n\n\n
[
\\mathbf{d_1} = (-2, 4, 8)
]
Similarly, for ( l_2 ), the direction vector is the coefficient of ( u ) in the equations of ( l_2 ):
[
\\mathbf{d_2} = (-1, 3, 2)
]<\/p>\n\n\n\nStep 2: Check if the Lines Are Parallel<\/h3>\n\n\n\n
\\frac{-2}{-1} = 2, \\quad \\frac{4}{3} \\neq 2, \\quad \\frac{8}{2} = 4
]<\/p>\n\n\n\nStep 3: Check if the Lines Intersect<\/h3>\n\n\n\n
\\begin{aligned}
3 – 2t &= -1 – u \\
7 + 4t &= 18 + 3u \\
-3 + 8t &= 7 + 2u
\\end{aligned}
]<\/p>\n\n\n\n
[
\\begin{aligned}
3 – 2t &= -1 – u \\quad \\Rightarrow \\quad 2t – u = 4 \\
7 + 4t &= 18 + 3u \\quad \\Rightarrow \\quad 4t – 3u = 11 \\
-3 + 8t &= 7 + 2u \\quad \\Rightarrow \\quad 8t – 2u = 10
\\end{aligned}
]<\/p>\n\n\n\n\n
[
u = 2t – 4
]<\/p>\n\n\n\n
[
4t – 3(2t – 4) = 11 \\quad \\Rightarrow \\quad 4t – 6t + 12 = 11 \\quad \\Rightarrow \\quad -2t = -1 \\quad \\Rightarrow \\quad t = \\frac{1}{2}
]<\/p>\n\n\n\n
[
u = 2\\left(\\frac{1}{2}\\right) – 4 = 1 – 4 = -3
]<\/p>\n\n\n\nStep 4: Check the Point of Intersection<\/h3>\n\n\n\n
[
\\begin{aligned}
x_1 &= 3 – 2\\left(\\frac{1}{2}\\right) = 3 – 1 = 2 \\
y_1 &= 7 + 4\\left(\\frac{1}{2}\\right) = 7 + 2 = 9 \\
z_1 &= -3 + 8\\left(\\frac{1}{2}\\right) = -3 + 4 = 1
\\end{aligned}
]<\/p>\n\n\n\n
[
\\begin{aligned}
x_2 &= -1 – (-3) = -1 + 3 = 2 \\
y_2 &= 18 + 3(-3) = 18 – 9 = 9 \\
z_2 &= 7 + 2(-3) = 7 – 6 = 1
\\end{aligned}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n