NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.1) Exercise 11.1
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The typical promotion criteria for students in Class 12 is a 33% overall average, as well as a 33% average in both theoretical and practical examinations (if applicable). It may be possible for students with failing grades in one subject to write the compartment for that subject. If a student fails any compartment or two or more subjects, the entire course must be retaken the following year.
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In Chapter 11, Three Dimensional Geometry, students will apply the knowledge they gained in previous chapters regarding Vectors. In addition to the direction Cosines and Ratios of a Line Connecting Two Points, the Equation of Lines and Planes in space under different conditions are also included. Students in Class 12 Mathematics are advised to practise topics such as the Equation of a Line in Space, the angle between Two Lines, and the shortest distance between two lines. To prepare for the Class 12 Mathematics Board Examinations, students should improve their conceptual understanding.
To practise and verify their answers, students should take assistance from the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.1. By reviewing the solutions, students will be able to identify where they made mistakes and improve in those areas. In Exercise 11.1 Class 12, students are required to answer 15+ numerical questions to assist them in clarifying their understanding of the topics. By practising additional questions, students can gain a deeper understanding of the concepts.
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NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.1) Exercise 11.1
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Access NCERT Solutions for Class 12 Chapter 11 – Three Dimensional Geometry
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.1
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In Ex 11.1 Class 12 of the NCERT Solutions Class 12, all topics mentioned in the CBSE Class 12 Mathematics Syllabus are covered. To facilitate a better understanding of the concepts, all major theorems and formulas are explained in detail in the book.
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Q.1 If a line makes angles 90°, 135°, 45° with the x, y and zaxes respectively, find its direction cosines.
Ans
$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{be}\hspace{0.17em}\hspace{0.17em}\mathrm{l},\mathrm{m}\mathrm{and}\mathrm{n}.\mathrm{}\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\\ \mathrm{l}=\mathrm{}\mathrm{cos}90\mathrm{\xb0}=0,\hspace{0.17em}\hspace{0.17em}\\ \mathrm{m}=\mathrm{cos}\mathrm{\hspace{0.17em}}135\mathrm{\xb0}=\frac{1}{\sqrt{2}},\\ \mathrm{n}=\mathrm{cos}\mathrm{\hspace{0.17em}}45\mathrm{\xb0}=\frac{1}{\sqrt{2}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{are}0,\frac{1}{\sqrt{2}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\frac{1}{\sqrt{2}}.\end{array}$
Q.2 Find the direction cosines of a line which makes equal angles with the coordinate axes.
Ans
$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line},\mathrm{which}\mathrm{makes}\mathrm{equal}\mathrm{angle}\mathrm{\alpha}\mathrm{}\\ \mathrm{with}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{coordinate}\mathrm{axes}\mathrm{be}\hspace{0.17em}\hspace{0.17em}\mathrm{l},\mathrm{m}\mathrm{and}\mathrm{n}.\mathrm{}\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\\ \mathrm{l}=\mathrm{}\mathrm{cos\alpha},\hspace{0.17em}\hspace{0.17em}\mathrm{m}=\mathrm{cos}\mathrm{\hspace{0.17em}}\mathrm{\alpha},\mathrm{n}=\mathrm{cos}\mathrm{\hspace{0.17em}}\mathrm{\alpha}\\ \mathrm{Since},\hspace{0.17em}\hspace{0.17em}{\mathrm{l}}^{2}+{\mathrm{m}}^{2}+{\mathrm{n}}^{2}=1\\ \Rightarrow {\mathrm{cos}}^{2}\mathrm{\alpha}+{\mathrm{cos}}^{2}\mathrm{\alpha}+{\mathrm{cos}}^{2}\mathrm{\alpha}=1\\ \Rightarrow 3{\mathrm{cos}}^{2}\mathrm{\alpha}=1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{cos}}^{2}\mathrm{\alpha}=\frac{1}{3}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{cos\alpha}=\pm \frac{1}{\sqrt{3}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line},\mathrm{which}\mathrm{is}\mathrm{equally}\mathrm{inclined}\mathrm{}\\ \mathrm{to}\mathrm{the}\mathrm{coordinate}\mathrm{axes},\mathrm{are}\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}}.\end{array}$
Q.3 If a line has the direction ratios –18, 12, –4 then what are its direction cosines?
Ans
$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{a}\mathrm{}\mathrm{line}\mathrm{}\mathrm{has}\mathrm{}\mathrm{the}\mathrm{}\mathrm{direction}\mathrm{}\mathrm{ratios}\mathrm{}\u201318,\mathrm{}12,\mathrm{}\u20134,\mathrm{then}\mathrm{direction}\\ \mathrm{cosines}\mathrm{are}:\\ \mathrm{l}=\frac{18}{\sqrt{{(18)}^{2}+{\left(12\right)}^{2}+{(4)}^{2}}}\\ \hspace{0.17em}\hspace{0.17em}=\frac{18}{\sqrt{324+144+16}}\\ \hspace{0.17em}\hspace{0.17em}=\frac{18}{22}\\ \hspace{0.17em}\hspace{0.17em}=\frac{9}{11}\\ \mathrm{m}=\frac{12}{\sqrt{{(18)}^{2}+{\left(12\right)}^{2}+{(4)}^{2}}}\\ \hspace{0.17em}\hspace{0.17em}=\frac{12}{\sqrt{324+144+16}}\\ \hspace{0.17em}\hspace{0.17em}=\frac{12}{22}\\ \hspace{0.17em}\hspace{0.17em}=\frac{6}{11}\\ \mathrm{n}=\frac{4}{\sqrt{{(18)}^{2}+{\left(12\right)}^{2}+{(4)}^{2}}}\\ \hspace{0.17em}\hspace{0.17em}=\frac{4}{\sqrt{324+144+16}}\\ \hspace{0.17em}\hspace{0.17em}=\frac{4}{22}\\ \hspace{0.17em}\hspace{0.17em}=\frac{2}{11}\\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{are}\frac{9}{11},\frac{6}{11},\frac{2}{11}.\end{array}$
Q.4 Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.
Ans
The given points are: A(2, 3, 4), B(–1, –2, 1) and C(5, 8, 7).
Direction ratios of line joining A (2, 3, 4) and B (–1, –2, 1) are:
–1–2, –2 – 3, 1 – 4 i.e., –3, –5, –3.
Direction ratios of line joining B (–1, –2, 1) and C (5, 8, 7) are:
5–(–1), 8 –(–2), 7 –1 i.e., 6, 10, 6.
It can be seen that the direction ratios of BC are –2 times of that of AB.
Therefore, AB is parallel to BC. Since, B is common to both AB and BC, so, points A, B and C are collinear.
Q.5 Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).
Ans
$\begin{array}{l}\mathrm{Let}\mathrm{\xe2\u20ac\u201e}\mathrm{the}\mathrm{\xe2\u20ac\u201e}\mathrm{vertices}\mathrm{\xe2\u20ac\u201e}\mathrm{of}\mathrm{\xe2\u20ac\u201e}\mathrm{triangle}\mathrm{\xe2\u20ac\u201e}\mathrm{ABC}\mathrm{\xe2\u20ac\u201e}\mathrm{be}\mathrm{\xe2\u20ac\u201e}\mathrm{A}\mathrm{\xe2\u20ac\u201e}\left(3,5,4\right),\\ \mathrm{B}\left(1,1,2\right)\mathrm{\xe2\u20ac\u201e}\mathrm{and}\mathrm{\xe2\u20ac\u201e}\mathrm{C}\left(5,5,2\right).\\ \mathrm{The}\mathrm{\xe2\u20ac\u201e}\mathrm{direction}\mathrm{\xe2\u20ac\u201e}\mathrm{ratios}\mathrm{\xe2\u20ac\u201e}\mathrm{of}\mathrm{\xe2\u20ac\u201e}\mathrm{side}\mathrm{\xe2\u20ac\u201e}\mathrm{AB}\xe2\u20ac\u2039\xe2\u20ac\u201e\mathrm{are}:\\ 13,\hspace{0.17em}\hspace{0.17em}15,\hspace{0.17em}\hspace{0.17em}2\left(4\right)\hspace{0.17em}\hspace{0.17em}\mathrm{i}.\mathrm{e}.,4,4,6.\\ \mathrm{So},\mathrm{\xe2\u20ac\u201e}\mathrm{the}\mathrm{\xe2\u20ac\u201e}\mathrm{direction}\mathrm{\xe2\u20ac\u201e}\mathrm{cosines}\mathrm{\xe2\u20ac\u201e}\mathrm{of}\mathrm{\xe2\u20ac\u201e}\mathrm{AB}\mathrm{\xe2\u20ac\u201e}\mathrm{are}:\\ \hspace{0.17em}\hspace{0.17em}\frac{4}{\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}+{6}^{2}}},\frac{4}{\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}+{6}^{2}}},\frac{6}{\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}+{6}^{2}}}\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\frac{4}{2\sqrt{17}},\frac{4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\\ \Rightarrow \frac{2}{\sqrt{17}},\frac{2}{\sqrt{17}},\frac{3}{\sqrt{17}}\\ \mathrm{The}\mathrm{\xe2\u20ac\u201e}\mathrm{direction}\mathrm{\xe2\u20ac\u201e}\mathrm{ratios}\mathrm{\xe2\u20ac\u201e}\mathrm{of}\mathrm{\xe2\u20ac\u201e}\mathrm{side}\mathrm{\xe2\u20ac\u201e}\mathrm{BC}\mathrm{\xe2\u20ac\u201e}\mathrm{are}:\\ 5\left(1\right),\hspace{0.17em}\hspace{0.17em}51,\mathrm{\hspace{0.17em}}22\hspace{0.17em}\hspace{0.17em}\mathrm{i}.\mathrm{e}.,4,\mathrm{\hspace{0.17em}}6,\mathrm{\hspace{0.17em}}4.\\ \mathrm{So},\mathrm{\xe2\u20ac\u201e}\mathrm{the}\mathrm{\xe2\u20ac\u201e}\mathrm{direction}\mathrm{\xe2\u20ac\u201e}\mathrm{cosines}\mathrm{\xe2\u20ac\u201e}\mathrm{of}\mathrm{\xe2\u20ac\u201e}\mathrm{AB}\mathrm{\xe2\u20ac\u201e}\mathrm{are}:\\ \hspace{0.17em}\hspace{0.17em}\frac{\mathrm{\hspace{0.17em}}4}{\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}+{6}^{2}}},\frac{\mathrm{\hspace{0.17em}}6}{\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}+{6}^{2}}},\frac{\mathrm{\hspace{0.17em}}4}{\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}+{6}^{2}}}\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{\hspace{0.17em}}4}{2\sqrt{17}},\frac{\mathrm{\hspace{0.17em}}6}{2\sqrt{17}},\frac{\mathrm{\hspace{0.17em}}4}{2\sqrt{17}}\\ \Rightarrow \frac{2}{\sqrt{17}},\frac{3}{\sqrt{17}},\frac{2}{\sqrt{17}}\\ \mathrm{The}\mathrm{\xe2\u20ac\u201e}\mathrm{direction}\mathrm{\xe2\u20ac\u201e}\mathrm{ratios}\mathrm{\xe2\u20ac\u201e}\mathrm{of}\mathrm{\xe2\u20ac\u201e}\mathrm{side}\mathrm{\xe2\u20ac\u201e}\mathrm{AC}\mathrm{\xe2\u20ac\u201e}\mathrm{are}:\\ 53,\hspace{0.17em}\hspace{0.17em}55,\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}2\left(4\right)\hspace{0.17em}\hspace{0.17em}\mathrm{i}.\mathrm{e}.,8,\hspace{0.17em}\hspace{0.17em}10,\hspace{0.17em}\hspace{0.17em}2.\\ \mathrm{So},\mathrm{\xe2\u20ac\u201e}\mathrm{the}\mathrm{\xe2\u20ac\u201e}\mathrm{direction}\mathrm{\xe2\u20ac\u201e}\mathrm{cosines}\mathrm{\xe2\u20ac\u201e}\mathrm{of}\mathrm{\xe2\u20ac\u201e}\mathrm{AB}\mathrm{\xe2\u20ac\u201e}\mathrm{are}:\\ \hspace{0.17em}\hspace{0.17em}\frac{8}{\sqrt{{\left(8\right)}^{2}+{\left(10\right)}^{2}+{2}^{2}}},\frac{10}{\sqrt{{\left(8\right)}^{2}+{\left(10\right)}^{2}+{2}^{2}}},\frac{2}{\sqrt{{\left(8\right)}^{2}+{\left(10\right)}^{2}+{2}^{2}}}\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{\hspace{0.17em}}8}{2\sqrt{42}},\frac{10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\\ \Rightarrow \mathrm{\hspace{0.17em}}\frac{\mathrm{\hspace{0.17em}}4}{\sqrt{42}},\frac{5}{\sqrt{42}},\frac{1}{\sqrt{42}}\end{array}$
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