Class 12 Maths Ex 10.4 Question 1.<\/strong> \nFind \\(|\\vec{a} \\times \\vec{b}|\\), if \\(\\vec{a}=\\hat{i}-7 \\hat{j}+7 \\hat{k}\\) and \\(\\vec{b}=3 \\hat{i}-2 \\hat{j}+2 \\hat{k}\\). \nSolution: \n <\/p>\nQuestion 2. \nFind a unit vector perpendicular to each of the vector \\(\\overrightarrow { a } +\\overrightarrow { b } \\quad and\\quad \\overrightarrow { a } -\\overrightarrow { b } \\), where \\(\\overrightarrow { a } =3\\hat { i } +2\\hat { j } +2\\hat { k } \\quad and\\quad \\overrightarrow { b } =\\hat { i } +2\\hat { j } -2\\hat { k } \\) \nSolution: \n <\/p>\n
Question 3. \nIf a unit vector \\(\\vec{a}\\) makes angles \\(\\frac { \u03c0 }{ 3 }\\) with \\(\\hat{i}\\), \\(\\frac { \u03c0 }{ 4 }\\) with \\(\\hat{j}\\) and an acute angle \u03b8 with k, then find \u03b8 and hence, the components of \\(\\vec{a}\\). \nSolution: \nThe direction cosines of \\(\\vec{a}\\) are \n \nSince \\(\\vec{a}\\) is a unit vector, its components are the direction cosines \n <\/p>\n
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Question 4. \nShow that \n\\((\\vec{a}-\\vec{b}) \\times(\\vec{a}+\\vec{b})=2(\\vec{a} \\times \\vec{b})\\) \nSolution: \n\\((\\vec{a}-\\vec{b}) \\times(\\vec{a}+\\vec{b})=\\vec{a} \\times \\vec{a}+\\vec{a} \\times \\vec{b}-\\vec{b} \\times \\vec{a}-\\vec{b} \\times \\vec{b}\\) \n= \\(\\overrightarrow{0}+\\vec{a} \\times \\vec{b}+\\vec{a} \\times \\vec{b}-\\overrightarrow{0}\\) \n= 2(\\(\\vec{a}\\) x \\(\\vec{b}\\))<\/p>\n
Question 5. \nFind \u03bb and \u03bc if \n\\(\\left( 2\\hat { i } +6\\hat { j } +27\\hat { k } \\right) \\times \\left( \\hat { i } +\\lambda \\hat { j } +\\mu \\hat { k } \\right)\\) = \\(\\vec{0}\\) \nSolution: \n\\((2 \\hat{i}+6 \\hat{j}+27 \\hat{k}) \\times(\\hat{i}+\\lambda \\hat{j}+\\mu \\hat{k})=\\overrightarrow{0}\\) \n\\(\\left|\\begin{array}{ccc} \n\\hat{i} & \\hat{j} & \\hat{k} \\\\ \n2 & 6 & 27 \\\\ \n1 & \\lambda & \\mu \n\\end{array}\\right|=\\overrightarrow{0}\\) \n\\(\\hat{i}(6 \\mu-27 \\lambda)-\\hat{j}(2 \\mu-27)+\\hat{k}(2 \\lambda-6)=\\overrightarrow{0}\\) \nEquating the corresponding components, we get \n6\u00b5 – 27\u03bb = 0 … (1) \n– 2\u00b5 + 27 = 0 … (2) \n2\u03bb – 6 = 0 … (3) \n(2) \u2192 \u00b5 = \\(\\frac { 27 }{ 2 }\\), (3) \u2192 \u03bb = 3 \nSubstituting the values of \u03bb and \u00b5 in (1), \nwe get 6(\\(\\frac { 27 }{ 2 }\\)) – 27(3) = 81 – 81 = 0 \n(1) satisfy \u03bb = 3 and \u00b5 = \\(\\frac { 27 }{ 2 }\\) \nHence \u03bb = 3, \u00b5 = \\(\\frac { 27 }{ 2 }\\)<\/p>\n
Question 6. \nGiven that \\(\\vec{a}\\).\\(\\vec{b}\\) = 0 and \\(\\vec{a}\\) x \\(\\vec{b}\\) = \\(\\vec{0}\\) What can you conclude about the vectors \\(\\vec{a}\\) and \\(\\vec{b}\\)? \nSolution: \n\\(\\vec{a}\\).\\(\\vec{b}\\) = 0 \u21d2 \\(\\vec{a}\\) = \\(\\vec{0}\\) or \\(\\vec{b}\\) = \\(\\vec{0}\\) or \\(\\vec{a}\\) \u22a5 \\(\\vec{b}\\) \n\\(\\vec{a}\\) x \\(\\vec{b}\\) = \\(\\vec{0}\\) \u21d2 \\(\\vec{a}\\) = \\(\\vec{0}\\) or \\(\\vec{b}\\) = \\(\\vec{0}\\) or \\(\\vec{a}\\) || \\(\\vec{b}\\) \nSince \\(\\vec{a}\\).\\(\\vec{b}\\) = 0 and \\(\\vec{a}\\) x \\(\\vec{b}\\) = 0, \nthen either \\(\\vec{a}\\) = \\(\\vec{0}\\) or \\(\\vec{b}\\).\\(\\vec{0}\\) \n\\(\\vec{a}\\) \u22a5 \\(\\vec{b}\\) and \\(\\vec{a}\\)||\\(\\vec{b}\\) are not possible at the same time.<\/p>\n
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Question 7. \nLet the vectors \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) be given as \\({ a }_{ 1 }\\hat { i } +{ a }_{ 2 }\\hat { j } +{ a }_{ 3 }\\hat { k } ,{ b }_{ 1 }\\hat { i } +{ b }_{ 2 }\\hat { j } +{ b }_{ 3 }\\hat { k } ,{ c }_{ 1 }\\hat { i } +{ c }_{ 2 }\\hat { j } +{ c }_{ 3 }\\hat { k }\\), then show that \\(\\vec{a} \\times(\\vec{b}+\\vec{c})=\\vec{a} \\times \\vec{b}+\\vec{a} \\times \\vec{c}\\). \nSolution: \n <\/p>\n
Question 8. \nIf either \\(\\vec{a}=\\overrightarrow{0} \\text { or } \\vec{b}=\\overrightarrow{0} \\text {, then } \\vec{a} \\times \\vec{b}=\\overrightarrow{0}\\). Is the converse true? Justify your answer with an example. \nSolution: \n\\(\\vec{a} \\times \\vec{b}=|\\vec{a}||\\vec{b}| \\sin \\theta \\hat{n}\\) \nIf \\(\\vec{a}=\\overrightarrow{0} \\text { or } \\vec{b}=\\overrightarrow{0} \\text {, then } \\vec{a} \\times \\vec{b}=\\overrightarrow{0}\\). \nThe converse need not be true. \nFor example, consider the non-zero parallel \n <\/p>\n
Question 9. \nFind the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5). \nSolution: \n <\/p>\n
Question 10. \nFind the area of the parallelogram whose adjacent sides are determined by the vectors \\(\\overrightarrow { a } =\\hat { i } -\\hat { j } +3\\hat { k } ,\\overrightarrow { b } =2\\hat { i } -7\\hat { j } +\\hat { k } \\) \nSolution: \n <\/p>\n
Question 11. \nLet the vectors \\(\\vec{a}\\) and \\(\\vec{b}\\) such that |\\(\\vec{a}\\)| = 3 and |\\(\\vec{b}\\)| = \\(\\frac{\\sqrt{2}}{3}\\), then \\(\\vec{a}\\) x \\(\\vec{b}\\) is a unit vector if the angle between \\(\\vec{a}\\) and \\(\\vec{a}\\) is \n(a) \\(\\frac { \\pi }{ 6 } \\) \n(b) \\(\\frac { \\pi }{ 4 } \\) \n(c) \\(\\frac { \\pi }{ 3 } \\) \n(d) \\(\\frac { \\pi }{ 2 } \\) \nSolution: \nLet \u03b8 be the angle between vectors \\(\\vec{a}\\) and \\(\\vec{b}\\). \nSince \\(\\vec{a}\\) x \\(\\vec{b}\\) is a unit vector, \n <\/p>\n
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Question 12. \nArea of a rectangles having vertices \n\\(\\left( -\\hat { i } +\\frac { 1 }{ 2 } \\hat { j } +4\\hat { k } \\right), \\left( \\hat { i } +\\frac { 1 }{ 2 } \\hat { j } +4\\hat { k } \\right)\\) \n\\(\\left( \\hat { i } -\\frac { 1 }{ 2 } \\hat { j } +4\\hat { k } \\right), \\left( -\\hat { i } -\\frac { 1 }{ 2 } \\hat { j } +4\\hat { k } \\right)\\) \n(a) \\(\\frac { 1 }{ 2 }\\) sq units \n(b) 1 sq.units \n(c) 2 sq.units \n(d) 4 sq.units \nSolution: \n <\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-10-ex-10-4\/ NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4 Class 12 Maths Ex 10.4 Question 1. Find , if and . Solution: Question 2. Find a unit …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","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":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n