2<\/sub>).
\nSolution:
\n<\/p>\nQuestion 3.
\nA girl walks 4 km towards west, then she walks 3 km in a direction 30\u00b0 east of north and stops. Determine the girl\u2019s displacement from her initial point of departure.
\nSolution:
\n
\nLet the girl starts from O and stops at P. Let the coordinates of P be (x, y).
\n\\(\\vec{r}\\) = \\(\\overrightarrow{OP}\\)
\n\u2234 \\(\\vec{r}\\) = – x\\(\\hat{i}\\) + y \\(\\hat{j}\\) … (1)
\nIn \u2206 QMP, sin60\u00b0 = \\(\\frac { y }{ 3 }\\)
\n<\/p>\n
Question 4.
\nIf \\(\\vec{a}\\) = \\(\\vec{b}\\) + \\(\\vec{c}\\), then is it true that |\\(\\vec{a}\\)| = |\\(\\vec{b}\\)| + |\\(\\vec{c}\\)|? Justify your answer.
\nSolution:
\n<\/p>\n
Question 5.
\nFind the value of x for which x\\((\\hat{i}+\\hat{j}+\\hat{k})\\) is a unit vector.
\nSolution:
\n<\/p>\n
<\/p>\n
Question 6.
\nFind a vector of magnitude 5 units, and parallel to the resultant of the vectors \\(\\vec{a}=2 \\hat{i}+3 \\hat{j}-\\hat{k} \\text { and } \\vec{b}=\\hat{i}-2 \\hat{j}+\\hat{k}\\).
\nSolution:
\n\\(\\vec{a}+\\vec{b}\\) be the resultant of \\(\\vec{a}\\) and \\(\\vec{b}\\).
\n<\/p>\n
Question 7.
\nIf \\(\\vec{a}\\) = \\(\\hat{i}+\\hat{j}+\\hat{k}\\), \\(\\vec{b}\\) = \\(2 \\hat{i}-\\hat{j}+3 \\hat{k}\\) and \\(\\vec{c}\\) = \\(\\hat{i}-2 \\hat{j}+\\hat{k}\\), find a unit vector parallel to the vector \\(2 \\vec{a}-\\vec{b}+3 \\vec{c}\\).
\nSolution:
\n<\/p>\n
Question 8.
\nShow that the points A (1, – 2, – 8), B (5, 0, – 2)and C (11, 3, 7) are collinear and find the ratio in which B divides AC.
\nSolution:
\n
\n\\(\\overrightarrow{AB}\\) is a scalar multiple of \\(\\overrightarrow{BC}\\)
\n\u2234 \\(\\overrightarrow{AB}\\) || \\(\\overrightarrow{BC}\\) and B is the common point.
\nHence A, B, C are collinear.
\nAlso \\(\\frac{|\\overrightarrow{\\mathrm{AB}}|}{|\\overrightarrow{\\mathrm{BC}}|}=\\frac{2}{3}\\)
\n\u2234 The point B divides AC in the ratio 2 : 3.<\/p>\n
Another method:
\nLet \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\) be the position vectors of A, B and C.
\n\u2234 \\(\\vec{a}\\) = \\(\\hat{i}-2 \\hat{j}-8 \\hat{k}\\), \\(\\vec{b}\\) = \\(5 \\hat{i}+0 \\hat{j}-2 \\hat{k}\\), \\(\\vec{c}\\) = \\(11 \\hat{i}+3 \\hat{j}+7 \\hat{k}\\)
\nLet B divide AC in the ratio \u03bb : 1
\nBy section formula, the position vector of
\n
\nEquating components of \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\), we get
\n
\n\u2234 \u03bb = \\(\\frac{2}{3}\\) satisfies (1) and (3)
\nHence A, B and C are collinear and B divides AC in the ratio \\(\\frac{2}{3}\\) : 1 That is, in the ratio 2 : 3.<\/p>\n
<\/p>\n
Question 9.
\nFind the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2\\(\\vec{a}\\) + \\(\\vec{b}\\)) and (\\(\\vec{a}\\)-3\\(\\vec{b}\\)) externally in the ratio 1 : 2. Also, show that P is the midpoint of the line, segment RQ.
\nSolution:
\n\\(\\overrightarrow{\\mathrm{OP}}=2 \\vec{a}+\\vec{b}, \\overrightarrow{\\mathrm{OQ}}=\\vec{a}-3 \\vec{b}\\),
\nR divides PQ in the ratio 1 : 2 externally.
\n
\n\u2234 P is the midpoint of the line segment RQ.<\/p>\n
Question 10.
\nThe two adjacent sides of a parallelogram are \\(2 \\hat{i}-4 \\hat{j}+5 \\hat{k} \\text { and } \\hat{i}-2 \\hat{j}-3 \\hat{k}\\). Find the unit vector parallel to its diagonal. Also, find its area.
\nSolution:
\nLet \\(\\vec{a}=2 \\hat{i}-4 \\hat{j}+5 \\hat{k} \\text { and } \\vec{b}=\\hat{i}-2 \\hat{j}-3 \\hat{k}\\) be any two adjacent sides of a parallelogram.
\n\\(\\vec{a}\\) + \\(\\vec{b}\\) is a vector along the diagonal
\n<\/p>\n
Question 11.
\nShow that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \\(\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}\\).
\nSolution:
\nLet a, P and y be the angles made by the vector with the axes OX, OY and OZ.
\nLet l, m and n be the direction cosines.
\nSince the vector is equally inclined to the
\naxes, \u03b1 = \u00df = \u03b3
\n\u2234 l = cos\u03b1, m = cos \u03b1, n = cos \u03b1
\nWe have l\u00b2 + m\u00b2 + n\u00b2 = 1
\ni.e, cos\u00b2 \u03b1 + cos\u00b2 \u03b1 + cos\u00b2 \u03b1 = 1
\n\u21d2 3 cos\u00b2 \u03b1 = 1
\ncos\u00b2 \u03b1 = \\(\\frac { 1 }{ 3 }\\)
\ncos \u03b1 = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u2234 l = m = n = \\(\\frac{1}{\\sqrt{3}}\\)
\nHence the direction cosines of the vector equally inclined to the axes are \\(\\frac{1}{\\sqrt{3}}\\), \\(\\frac{1}{\\sqrt{3}}\\), \\(\\frac{1}{\\sqrt{3}}\\).<\/p>\n
<\/p>\n
Question 12.
\nLet \\(\\vec{a}=\\hat{i}+4 \\hat{j}+2 \\hat{k}, \\vec{b}=3 \\hat{i}-2 \\hat{j}+7 \\hat{k}\\) and \\(\\vec{c}=2 \\hat{i}-\\hat{j}+4 \\hat{k}\\) Find a vector \\(\\vec{d}\\) which is perpendicular to both \\(\\vec{a}\\) and \\(\\vec{b}\\) and \\(\\vec{c}\\). \\(\\vec{d}\\) = 15.
\nSolution:
\nLet \\(\\vec{d}\\) = \\(x \\hat{i}+y \\hat{j}+z \\hat{k}\\)
\nSince \\(\\vec{d}\\) is perpendicular to both \\(\\vec{a}\\) and \\(\\vec{b}\\)
\n\\(\\vec{d}\\).\\(\\vec{a}\\) = 0 and \\(\\vec{d}\\).\\(\\vec{b}\\) = 0.
\nx + 4y + 2z = 0 … (1) and
\n3x – 2y + 7z = 0 … (2)
\nGiven \\(\\vec{c}\\).\\(\\vec{d}\\) = 15 … (3)
\n2x – y + 4z = 15 … (3)
\nSolving (1), (2) and (3), we get x = \\(\\frac { 160 }{ 3 }\\)
\ny = \\(\\frac { – 5 }{ 3 }\\) and z = \\(\\frac { -70 }{ 3 }\\)
\n\u2234 \\(\\vec{d}=\\frac{160}{3} \\hat{i}-\\frac{5}{3} \\hat{j}-\\frac{70}{3} \\hat{k}\\)
\n\\(\\vec{d}=\\frac{1}{3}(160 \\hat{i}-5 \\hat{j}-70 \\hat{k})\\)<\/p>\n
Question 13.
\nThe scalar product of the vector \\(\\hat{i}+\\hat{j}+\\hat{k}\\) with a unit vector along the sum of vectors \\(2 \\hat{i}+4 \\hat{j}-5 \\hat{k} \\text { and } \\lambda \\hat{i}+2 \\hat{j}+3 \\hat{k}\\) is equal to one. Find the value of \u03bb.
\nSolution:
\n<\/p>\n
Question 14.
\nIf \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\) are mutually perpendicular vectors of equal magnitudes, show that the vectors \\(\\vec{a}+\\vec{b}+\\vec{c}\\) is equally inclined to \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\).
\nSolution:
\nGiven that \\(|\\vec{a}|=|\\vec{b}|=|\\vec{c}|=\\lambda\\) … (1)
\nSince \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\) are mutually perpendicular
\n
\nHence \\(\\vec{a}+\\vec{b}+\\vec{c}\\) is equally inclined to \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\).<\/p>\n
<\/p>\n
Question 15.
\nProve that \\((\\vec{a}+\\vec{b}) \\cdot(\\vec{a}+\\vec{b})=|\\vec{a}|^{2}+|\\vec{b}|^{2}\\), if and only if \\(\\vec{a}\\), \\(\\vec{b}\\) are perpendicular, given a \u2260 \\(\\vec{0}\\), b \u2260 \\(\\vec{0}\\)
\nSolution:
\n\\((\\vec{a}+\\vec{b}) \\cdot(\\vec{a}+\\vec{b})=\\vec{a} \\cdot \\vec{a}+\\vec{a} \\cdot \\vec{b}+\\vec{b} \\cdot \\vec{a}+\\vec{b} \\cdot \\vec{b}\\)
\n\\((\\vec{a}+\\vec{b}) \\cdot(\\vec{a}+\\vec{b})=|\\vec{a}|^{2}+|\\vec{b}|^{2}+2(\\vec{a} \\cdot \\vec{b}) \\ldots \\ldots\\) … (1)
\nCase 1
\nLet
\nFrom (1), we get \\(\\vec{a}\\), \\(\\vec{b}\\) = 0
\nHence \\(\\vec{a}\\) \u22a5 \\(\\vec{b}\\) since \\(\\vec{a}\\) and \\(\\vec{b}\\) are nonzero vectors.<\/p>\n
Case 2
\nLet \\(\\vec{a}\\) \u22a5 \\(\\vec{b}\\). Then \\(\\vec{a}\\).\\(\\vec{b}\\) = 0
\n(1) \u2192 \\((\\vec{a}+\\vec{b}) \\cdot(\\vec{a}+\\vec{b})=|\\vec{a}|^{2}+|\\vec{b}|^{2}\\) + 0
\n\\((\\vec{a}+\\vec{b}) \\cdot(\\vec{a}+\\vec{b})=|\\vec{a}|^{2}+|\\vec{b}|^{2}\\)<\/p>\n
Question 16.
\nIf \u03b8 is the angle between two vectors \\(\\vec{a}\\) and \\(\\vec{b}\\), then \\(\\vec{a}\\).\\(\\vec{b}\\) \u2265 0 only when
\na. 0 < \u03b8 < \\(\\frac { \u03c0 }{ 2 }\\)
\nb. 0 \u2264 \u03b8 \u2264 \\(\\frac { \u03c0 }{ 2 }\\)
\nc. 0 < \u03b8 < \u03c0
\nd. 0 \u2264 \u03b8 \u2264 \u03c0
\nSolution:
\n<\/p>\n
Question 17.
\nLet \\(\\vec{a}\\) and \\(\\vec{b}\\) be two unit vectors and \u03b8 is the angle between them. Then \\(\\vec{a}\\) + \\(\\vec{b}\\) is a unit vector if
\na. \u03b8 = \\(\\frac { \u03c0 }{ 4 }\\)
\nb. \u03b8 = \\(\\frac { \u03c0 }{ 3 }\\)
\nc. \u03b8 = \\(\\frac { \u03c0 }{ 2 }\\)
\nd. \u03b8 = \\(\\frac { 2\u03c0 }{ 3 }\\)
\nSolution:
\n<\/p>\n
Question 18.
\nThe value of
\n\\(\\hat{i} \\cdot(\\hat{j} \\times \\hat{k})+\\hat{j} \\cdot(\\hat{i} \\times \\hat{k})+\\hat{k} \\cdot(\\hat{i} \\times \\hat{j}) \\text { is }\\)
\na. 0
\nb. -1
\nc. 1
\nd. 3
\nSolution:
\nc. 1
\n\\(\\hat{i} \\cdot(\\hat{j} \\times \\hat{k})+\\hat{j} \\cdot(\\hat{i} \\times \\hat{k})+\\hat{k} \\cdot(\\hat{i} \\times \\hat{j}) \\text { is }\\)
\n= \\(\\hat{i} \\cdot \\hat{i}-\\hat{j} \\cdot \\hat{j}+\\hat{k} \\hat{k}\\)
\n= 1 – 1 + 1 = 1<\/p>\n
<\/p>\n
Question 19.
\nIf \u03b8 is the angle between any two vectors \\(\\vec{a}\\) and \\(\\vec{b}\\), then \\(|\\vec{a} \\cdot \\vec{b}|=|\\vec{a} \\times \\vec{b}|\\), when \u03b8 is equal to
\na. 0
\nb. \\(\\frac { \u03c0 }{ 4 }\\)
\nc. \\(\\frac { \u03c0 }{ 2 }\\)
\nd. \u03c0
\nSolution:
\nb. \\(\\frac { \u03c0 }{ 4 }\\)
\n<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 10 Vector Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-10-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 10 Vector Miscellaneous Exercise Question 1. Write down a unit vector in XY – plane, making an angle of 30\u00b0 with the positive direction …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Miscellaneous Exercise<\/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 Miscellaneous Exercise - 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