Class 12 Maths Ncert Solutions Chapter 10.3 Question 1.<\/strong> \nFind the angle between two vectors \\(\\vec{a}\\) and \\(\\vec{b}\\) with magnitudes \\(\\sqrt{3}\\) and 2 respectively, and such that \\(\\vec{a}\\).\\(\\vec{b}\\) = \\(\\sqrt{6}\\) \nSolution: \n|\\(\\vec{a}\\)|= \\(\\sqrt{3}\\) \nand \\(\\vec{b}\\) = 2, \\(\\vec{a}\\).\\(\\vec{b}\\) = \\(\\sqrt{6}\\) \nLet \u03b8 be the angle between \\(\\vec{a}\\) and \\(\\vec{b}\\). Then \n\\(\\cos \\theta=\\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{a}||\\vec{b}|}=\\frac{\\sqrt{6}}{(\\sqrt{3})(2)}=\\frac{1}{\\sqrt{2}}\\) \n\u2234 \u03b8 = \\(\\cos ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)=\\frac{\\pi}{4}\\)<\/p>\nQuestion 2. \nFind the angle between the vectors \\(\\hat{i}-2 \\hat{j}+3 \\hat{k} \\text { and } 3 \\hat{i}-2 \\hat{j}+\\hat{k}\\) \nSolution: \nLet \\(\\vec{a}\\) = \\(\\hat{i}-2 \\hat{j}+3 \\hat{k}\\), \\(\\vec{b}\\) = \\(\\hat{3i}-2 \\hat{j}+ \\hat{k}\\) \n\\(\\vec{a}\\).\\(\\vec{b}\\) = \\((\\hat{i}-2 \\hat{j}+3 \\hat{k}) \\cdot(3 \\hat{i}-2 \\hat{j}+\\hat{k})\\) \n= 1(3) + (- 2)(- 2) + 3(1) = 3 + 4 + 3 = 10 \n <\/p>\n
Question 3. \nFind the projection of the vector \\(\\overrightarrow { i } -\\overrightarrow { j }\\), on the line represented by the vector \\(\\overrightarrow { i } +\\overrightarrow { j }\\). \nSolution: \n <\/p>\n
Question 4. \nFind the projection of the vector \\(\\hat{i}+3 \\hat{j}+7 \\hat{k}\\) on the vector \\(7 \\hat{i}-\\hat{j}+8 \\hat{k}\\) \nSolution: \n <\/p>\n
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Question 5. \nShow that each of the given three vectors is a unit vector \\(\\frac { 1 }{ 7 } \\left( 2\\hat { i } +3\\hat { j } +6\\hat { k } \\right) ,\\frac { 1 }{ 7 } \\left( 3\\hat { i } -6\\hat { j } +2\\hat { k } \\right) ,\\frac { 1 }{ 7 } \\left( 6\\hat { i } +2\\hat { j } -3\\hat { k } \\right)\\) \nAlso show that they are mutually perpendicular to each other. \nSolution: \nLet \\(\\vec{a}\\) = \\(\\frac{1}{7}(2 \\hat{i}+3 \\hat{j}+6 \\hat{k})\\), \\(\\vec{b}\\) = \\(\\frac{1}{7}(3 \\hat{i}-6 \\hat{j}+2 \\hat{k})\\) and \\(\\vec{c}\\) = \\(\\frac{1}{7}(6 \\hat{i}+2 \\hat{j}-3 \\hat{k})\\) \n \nHere \\(\\vec{a}\\).\\(\\vec{b}\\) = 0, \\(\\vec{b}\\).\\(\\vec{c}\\) = 0 and \\(\\vec{a}\\).\\(\\vec{c}\\) = 0 \n\u2234 The vectors \\(\\vec{a}\\).\\(\\vec{b}\\) and \\(\\vec{c}\\) are mutually per-pendicular vectors.<\/p>\n
Question 6. \n\\(Find\\left| \\overrightarrow { a } \\right| and\\left| \\overrightarrow { b } \\right| if\\left( \\overrightarrow { a } +\\overrightarrow { b } \\right) \\cdot \\left( \\overrightarrow { a } -\\overrightarrow { b } \\right) =8\\quad and\\left| \\overrightarrow { a } \\right| =8\\left| \\overrightarrow { b } \\right| \\) \nSolution: \n <\/p>\n
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Question 7. \nEvaluate the product : \n\\((3 \\vec{a}-5 \\vec{b}) \\cdot(2 \\vec{a}+7 \\vec{b})\\) \nSolution: \n\\(\\left( 3\\overrightarrow { a } -5\\overrightarrow { b } \\right) \\cdot \\left( 2\\overrightarrow { a } +7\\overrightarrow { b } \\right) \\) \n\\(=6\\overrightarrow { a } .\\overrightarrow { a } -10\\overrightarrow { b } \\overrightarrow { a } +21\\overrightarrow { a } .\\overrightarrow { b } -35\\overrightarrow { b } .\\overrightarrow { b } \\) \n\\(=6{ \\left| \\overrightarrow { a } \\right| }^{ 2 }-11\\overrightarrow { a } \\overrightarrow { b } -35{ \\left| \\overrightarrow { b } \\right| }^{ 2 }\\)<\/p>\n
Question 8. \nFind the magnitude of two vectors \\(\\vec{a}\\) and \\(\\vec{b}\\) having the same magnitude and such that the angle between them is 60\u00b0 and their scalar product is \\(\\frac { 1 }{ 2 }\\) \nSolution: \n <\/p>\n
Question 9. \nFind |\\(\\vec{a}\\)| , if for a unit vector \\(\\vec{a}\\), \\((\\vec{x}-\\vec{a}) \\cdot(\\vec{x}+\\vec{a})\\) = 12 \nSolution: \n <\/p>\n
Question 10. \nIf \\(\\overrightarrow { a } =2\\hat { i } +2\\hat { j } +3\\hat { k } ,\\overrightarrow { b } =-\\hat { i } +2\\hat { j } +\\hat { k } and \\overrightarrow { c } =3\\hat { i } +\\hat { j } \\) such that \\(\\overrightarrow { a } +\\lambda \\overrightarrow { b } \\bot \\overrightarrow { c } \\) , then find the value of \u03bb. \nSolution: \n <\/p>\n
Question 11. \nShow that \\(|\\vec{a}| \\vec{b}+|\\vec{b}| \\vec{a} \\), is per-pendicular to \\(|\\vec{a}| \\vec{b}-|\\vec{b}| \\vec{a} \\) for any two non-zero vectors \\(\\vec{a} \\text { and } \\vec{b}\\) \nSolution: \n <\/p>\n
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Question 12. \nIf \\(\\overrightarrow { a } \\cdot \\overrightarrow { a } =0\\quad and\\quad \\overrightarrow { a } \\cdot \\overrightarrow { b } =0\\), then what can be concluded about the vector \\(\\overrightarrow { b } \\) ? \nSolution: \n\\(\\vec{a}\\).\\(\\vec{a}\\) = 0 \u21d2 \\(\\vec{a}\\) is a zero vector. \nsince \\(\\vec{a}\\) = \\(\\vec{0}\\), \\(\\vec{a}\\).\\(\\vec{b}\\) = 0 for any vector \\(\\vec{b}\\) \nVector \\(\\vec{b}\\) be any vector.<\/p>\n
Question 13. \nIf \\(\\overrightarrow { a } ,\\overrightarrow { b } ,\\overrightarrow { c } \\) are the unit vector such that \\(\\overrightarrow { a } +\\overrightarrow { b } +\\overrightarrow { c } =0\\) , then find the value of \\(\\overrightarrow { a } .\\overrightarrow { b } +\\overrightarrow { b } .\\overrightarrow { c } +\\overrightarrow { c } .\\overrightarrow { a } \\) \nSolution: \n <\/p>\n
Question 14. \nIf either vector \\(\\vec{a}=\\overrightarrow{0} \\text { or } \\vec{b}=\\overrightarrow{0}\\), then \\(\\vec{a} \\cdot \\vec{b}\\). But the converse need not be true. Justify your answer with an example. \nSolution: \n \nThus two non-zero vectors \\(\\vec{a}\\) and \\(\\vec{b}\\) may have \\(\\vec{a}\\).\\(\\vec{a}\\) zero.<\/p>\n
Question 15. \nIf the vertices A,B,C of a triangle ABC are (1, 2, 3) (-1, 0, 0), (0, 1, 2) respectively, then find \u2220ABC. \nSolution: \n <\/p>\n
Question 16. \nShow that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are collinear. \nSolution: \nA (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are the points. \n \n\u2234 \\(\\overrightarrow{AB}\\)\\(\\overrightarrow{AC}\\) are parallel and A is a common point. Therefore A, B, C are collinear.<\/p>\n
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Question 17. \nShow that the vectors \\(2 \\hat{i}-\\hat{j}+\\hat{k}, \\hat{i}-3 \\hat{j}-5 \\hat{k} \\text { and } \\quad 3 \\hat{i}-4 \\hat{j}-4 \\hat{k}\\) from the vertices of a right angled triangle. \nSolution: \nLet A, B and C be the vertices of the triangle with the vectors \\(2 \\hat{i}-\\hat{j}+\\hat{k}, \\hat{i}-3 \\hat{j}-5 \\hat{k} \\text { and } \\quad 3 \\hat{i}-4 \\hat{j}-4 \\hat{k}\\) \n\u2234 \\(\\overrightarrow{AB}\\) = p.v. of B – p.v. of A \n \n\u2234 A, B, C are the vertices of \u2206 ABC \n\\(\\overrightarrow{BC}\\).\\(\\overrightarrow{CA}\\) = (2)(-1) + (-1)(3) + (1)(5) \n= – 2 – 3 + 5 = 0 \n\u2234 \\(\\overrightarrow{BC}\\)\u22a5\\(\\overrightarrow{CA}\\) \nHence triangle ABC is a right triangle<\/p>\n
Question 18. \nIf \\(\\overrightarrow { a } \\) is a non-zero vector of magnitude \u2018a\u2019 and \u03bb is a non- zero scalar, then \u03bb \\(\\overrightarrow { a } \\) is unit vector if \n(a) \u03bb = 1 \n(b) \u03bb = – 1 \n(c) a = |\u03bb| \n(d) a = \\(\\frac { 1 }{ \\left| \\lambda \\right| } \\) \nSolution: \n\\(\\left| \\overrightarrow { a } \\right| =a\\) \nGiven : \\(\\lambda \\overrightarrow { a } \\) is a unit vectors. \n\\(|\\lambda \\vec{a}|=1 \\Rightarrow|\\lambda||\\vec{a}|=1 \\Rightarrow|\\lambda| a=1 \\Rightarrow a=\\frac{1}{|\\lambda|}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-10-ex-10-3\/ NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3 Class 12 Maths Ncert Solutions Chapter 10.3 Question 1. Find the angle between two vectors and with magnitudes …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3<\/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 10.3 - 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