{"id":24673,"date":"2020-12-19T07:05:26","date_gmt":"2020-12-19T07:05:26","guid":{"rendered":"https:\/\/samacheer-kalvi.com\/?p=24673"},"modified":"2021-12-06T15:56:30","modified_gmt":"2021-12-06T10:26:30","slug":"samacheer-kalvi-12th-maths-guide-chapter-6-ex-6-8","status":"publish","type":"post","link":"https:\/\/samacheer-kalvi.com\/samacheer-kalvi-12th-maths-guide-chapter-6-ex-6-8\/","title":{"rendered":"Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.8"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Maths Guide<\/a> Pdf Chapter 6 Applications of Vector Algebra Ex 6.8 Textbook Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8<\/h2>\n

Question 1.
\nShow that the straight lines
\n\\(\\overline { r }\\) = (5\\(\\hat { i }\\) + 7\\(\\hat { j }\\) – 3\\(\\hat { k }\\)) + s(4\\(\\hat { i }\\) + 4\\(\\hat { j }\\) – 5\\(\\hat { k }\\)) and
\n\\(\\overline { r }\\) = (8\\(\\hat { i }\\) + 4\\(\\hat { j }\\) + 5\\(\\hat { k }\\)) + t(7\\(\\hat { i }\\) + \\(\\hat { j }\\) + 3\\(\\hat { k }\\)) are coplanar. Find the vector equation of the, plane in which they lie.
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nShow that the lines \\(\\frac { x-2 }{ 1 }\\) = \\(\\frac { y-3 }{ 1 }\\) = \\(\\frac { z -4}{ 3 }\\) and \\(\\frac { x-1 }{ -3 }\\) = \\(\\frac { y-4 }{ 2 }\\) = \\(\\frac { z-5 }{ 1 }\\) are coplanar. Also, find the plane containing these lines.
\nSolution:
\n\"Samacheer
\nCartesian equation
\nx + 2y – 2z = 4
\nx + 2y – 2z – 4 = 0<\/p>\n

Question 3.
\nIf the straight lines \\(\\frac { x-1 }{ 1 }\\) = \\(\\frac { y-2 }{ 2 }\\) = \\(\\frac { z-3}{ m^2 }\\) and \\(\\frac { x-3 }{ 1 }\\) = \\(\\frac { y-2 }{ m^2 }\\) = \\(\\frac { z-1 }{ 2 }\\) are coplanar, find the distinct real values of m
\nSolution:
\n\"Samacheer
\n2(4 – m4<\/sup>) – 2(m\u00b2 – 2) = 0
\n8 – 2m4<\/sup> – 2m\u00b2 + 4 = 0
\n12 – 2m4<\/sup> – 2m\u00b2 = 0
\n(\u00f7 -2) -6 + m4<\/sup> + m\u00b2 = 0
\nm4<\/sup> + m\u00b2 – 6 = 0
\n(m\u00b2 – 2)(m\u00b2 + 3) = 0
\nm\u00b2 – 2 = 2; m\u00b2 = -3 (not possible)
\nm\u00b2 = 2
\nm = \u00b1\u221a2<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nIf the straight lines \\(\\frac { x-1 }{ 2 }\\) = \\(\\frac { y+1 }{ \u03bb }\\) = \\(\\frac { z }{ 2 }\\) and \\(\\frac { x+1 }{ 5 }\\) = \\(\\frac { y+1 }{ 2 }\\) = \\(\\frac { z }{ \u03bb }\\) are coplanar, find \u03bb and equations of the planes containing these two lines.
\nSolution:
\nIf the two lines are coplanar
\n\"Samacheer
\nWhen \u03bb = 2
\n(x1<\/sub>, y1<\/sub>, z1<\/sub>) = (1, -1, 0)
\n(b1<\/sub>, b2<\/sub>, b3<\/sub>) = (2, 2, 2)
\n(d1<\/sub>, d2<\/sub>, d3<\/sub>) = (5, 2, 2)
\n\\(\\left|\\begin{array}{ccc}
\nx-x_{1} & y-y_{1} & z-z_{1} \\\\
\nb_{1} & b_{2} & b_{3} \\\\
\nd_{1} & d_{2} & d_{3}
\n\\end{array}\\right|\\) = 0
\n\u21d2 \\(\\left|\\begin{array}{ccc}
\nx-1 & y+1 & z-0 \\\\
\n2 & 2 & 2 \\\\
\n5 & 2 & 2
\n\\end{array}\\right|\\) = 0
\n\u21d2 (x – 1)(0) – (y + 1)(-6) + z(6) = 0
\n\u21d26(y + 1) – 6z = 0
\n\u21d2 6y + 6 – 6z = 0
\n\u21d2 y – z + 1 = 0
\nWhen \u03bb = 2
\n(b1<\/sub>, b2<\/sub>, b3<\/sub>) = (2, -2, 2)
\n(d1<\/sub>, d2<\/sub>, d3<\/sub>) = (5, 2, -2)
\n\u21d2 \\(\\left|\\begin{array}{ccc}
\nx-1 & y+1 & z-0 \\\\
\n2 & -2 & 2 \\\\
\n5 & 2 & -2
\n\\end{array}\\right|\\) = 0
\n\u21d2 (x – 1)(0) – (y + 1)(-14) + z(4 + 10) = 0
\n\u21d2 14(y + 1) + 14z = 0
\n\u21d2 14y + 14 + 14z = 0
\n\u21d2 y + z + 1 = 0<\/p>\n

\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.8 Textbook Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8 Question 1. Show that the straight lines = (5 + 7 – 3) + s(4 + 4 – …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/24673"}],"collection":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/comments?post=24673"}],"version-history":[{"count":1,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/24673\/revisions"}],"predecessor-version":[{"id":47970,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/24673\/revisions\/47970"}],"wp:attachment":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/media?parent=24673"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/categories?post=24673"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/tags?post=24673"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}