{"id":32454,"date":"2021-08-24T10:30:42","date_gmt":"2021-08-24T05:00:42","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32454"},"modified":"2021-08-24T10:30:42","modified_gmt":"2021-08-24T05:00:42","slug":"ncert-solutions-for-class-12-maths-chapter-10-ex-10-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-10-ex-10-3\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 10 Vector Algebra Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\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>\n

Question 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\"NCERT<\/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\"NCERT<\/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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT
\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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT<\/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\"NCERT<\/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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT
\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\"NCERT<\/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\"NCERT
\n\u2234 \\(\\overrightarrow{AB}\\)\\(\\overrightarrow{AC}\\) are parallel and A is a common point. Therefore A, B, C are collinear.<\/p>\n

\"NCERT<\/p>\n

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\"NCERT
\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. NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3 Question 1. Find the angle between two vectors and with magnitudes and 2 respectively, and such that . = …<\/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":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-10-ex-10-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3 Question 1. 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