NCERT Solutions For Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2
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| Textbook | NCERT |
| Board | CBSE |
| Category | NCERT Solutions |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 11 |
| Exercise | Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 |
| Number of Questions Solved | 17 |
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2
NCERT TEXTBOOK EXERCISES
Ex 11.2 Class 12 Maths Question 1.
Show that the three lines with direction cosines:
$\frac { 12 }{ 13 } ,\frac { -3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 4 }{ 13 } ,\frac { 12 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 12 }{ 13 }$
are mutually perpendicular.
Solution:
Let the lines be L1,L2 and L3.
∴ For lines L1 and L2
Ex 11.2 Class 12 Maths Question 2.
Show that the line through the points (1,-1,2) (3,4, -2) is perpendicular to the line through the points (0,3,2) and (3,5,6).
Solution:
Let A, B be the points (1, -1, 2), (3, 4, -2) respectively Direction ratios of AB are 2,5, -4
Let C, D be the points (0, 3, 2) and (3, 5, 6) respectively Direction ratios of CD are 3, 2,4 AB is Perpendicular to CD if
Ex 11.2 Class 12 Maths Question 3.
Show that the line through the points (4,7,8) (2,3,4) is parallel to the line through the points (-1,-2,1) and (1,2,5).
Solution:
Let the points be A(4,7,8), B (2,3,4), C (-1,-2,1) andD(1,2,5).
Now direction ratios of AB are
Ex 11.2 Class 12 Maths Question 4.
Find the equation of the line which passes through the point (1,2,3) and is parallel to the vector $3\hat { i } +2\hat { j } -2\hat { k }$
Solution:
Equation of the line passing through the point
Ex 11.2 Class 12 Maths Question 5.
Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat { i } -\hat { j } +4\hat { k }$ and is in the direction $\hat { i } +2\hat { j } -\hat { k }$.
Solution:
The vector equation of a line passing through a point with position vector $\overrightarrow { a }$ and parallel to the
Ex 11.2 Class 12 Maths Question 6.
Find the cartesian equation of the line which passes through the point (-2,4, -5) and parallel to the line is given by $\frac { x+3 }{ 3 } =\frac { y-4 }{ 5 } =\frac { z+8 }{ 6 }$
Solution:
The cartesian equation of the line passing through the point (-2,4, -5) and parallel to the
Ex 11.2 Class 12 Maths Question 7.
The cartesian equation of a line is
$\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 }$
write its vector form.
Solution:
The cartesian equation of the line is
$\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 }$
Clearly (i) passes through the point (5, – 4, 6) and has 3,7,2 as its direction ratios.
=> Line (i) passes through the point A with
Ex 11.2 Class 12 Maths Question 8.
Find the vector and the cartesian equations of the lines that passes through the origin and (5,-2,3).
Solution:
The line passes through point
Undefined control sequence \therefore
Direction ratios of the line passing through the
Ex 11.2 Class 12 Maths Question 9.
Find the vector and cartesian equations of the line that passes through the points (3, -2, -5), (3,-2,6).
Solution:
The PQ passes through the point P(3, -2, -5)
Ex 11.2 Class 12 Maths Question 10.
Find the angle between the following pair of lines
(i) $\overrightarrow { r } =2\hat { i } -5\hat { j } +\hat { k } +\lambda (3\hat { i } +2\hat { j } +6\hat { k } )$
$and\quad \overrightarrow { r } =7\hat { i } -6\hat { j } +\mu (\hat { i } +2\hat { j } +2\hat { k } )$
(ii) $\overrightarrow { r } =3\hat { i } +\hat { j } -2\hat { k } +\lambda (\hat { i } -\hat { j } -2\hat { k } )$
$\overrightarrow { r } =2\hat { i } -\hat { j } -56\hat { k } +\mu (3\hat { i } -5\hat { j } -4\hat { k } )$
Solution:
(i) Let θ be the angle between the given lines.
The given lines are parallel to the vectors
Ex 11.2 Class 12 Maths Question 11.
Find the angle between the following pair of lines
(i) $\frac { x-2 }{ 2 } =\frac { y-1 }{ 5 } =\frac { z+3 }{ -3 } and\frac { x+2 }{ -1 } =\frac { y-4 }{ 8 } =\frac { z-5 }{ 4 }$
(ii) $\frac { x }{ 2 } =\frac { y }{ 2 } =\frac { z }{ 1 } and\frac { x-5 }{ 4 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ 8 }$
Solution:
Given
(i) $\frac { x-2 }{ 2 } =\frac { y-1 }{ 5 } =\frac { z+3 }{ -3 } and\frac { x+2 }{ -1 } =\frac { y-4 }{ 8 } =\frac { z-5 }{ 4 }$
(ii) $\frac { x }{ 2 } =\frac { y }{ 2 } =\frac { z }{ 1 } and\frac { x-5 }{ 4 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ 8 }$
Ex 11.2 Class 12 Maths Question 12.
Find the values of p so that the lines
$\frac { 1-x }{ 3 } =\frac { 7y-14 }{ 2p } =\frac { z-3 }{ 2 } and\frac { 7-7x }{ 3p } =\frac { y-5 }{ 1 } =\frac { 6-z }{ 5 }$ are at right angles
Solution:
The given equation are not in the standard form
The equation of given lines is
Ex 11.2 Class 12 Maths Question 13.
Show that the lines $\frac { x-5 }{ 7 } =\frac { y+2 }{ -5 } =\frac { z }{ 1 } and\frac { x }{ 1 } =\frac { y }{ 2 } =\frac { z }{ 3 }$ are perpendicular to each other
Solution:
Given lines
$\frac { x-5 }{ 7 } =\frac { y+2 }{ -5 } =\frac { z }{ 1 }$ …(i)
$\frac { x }{ 1 } =\frac { y }{ 2 } =\frac { z }{ 3 }$ …(ii)
Ex 11.2 Class 12 Maths Question 14.
Find the shortest distance between the lines
$\overrightarrow { r } =(\hat { i } +2\hat { j } +\hat { k } )+\lambda (\hat { i } -\hat { j } +\hat { k } )$ and
$\overrightarrow { r } =(2\hat { i } -\hat { j } -\hat { k } )+\mu (2\hat { i } +\hat { j } +2\hat { k } )$
Solution:
The shortest distance between the lines
Ex 11.2 Class 12 Maths Question 15.
Find the shortest distance between the lines
$\frac { x+1 }{ 7 } =\frac { y+1 }{ -6 } =\frac { z+1 }{ 1 } and\frac { x-3 }{ 1 } =\frac { y-5 }{ -2 } =\frac { z-7 }{ 1 }$
Solution:
Shortest distance between the lines
Ex 11.2 Class 12 Maths Question 16.
Find the distance between die lines whose vector equations are:
$\overrightarrow { r } =(\hat { i } +2\hat { j } +3\hat { k) } +\lambda (\hat { i } -3\hat { j } +2\hat { k } )$ and
$\overrightarrow { r } =(4\hat { i } +5\hat { j } +6\hat { k) } +\mu (2\hat { i } +3\hat { j } +\hat { k } )$
Solution:
Comparing the given equations with
Ex 11.2 Class 12 Maths Question 17.
Find the shortest distance between the lines whose vector equations are
$\overrightarrow { r } =(1-t)\hat { i } +(t-2)\hat { j } +(3-2t)\hat { k }$ and
$\overrightarrow { r } =(s+1)\hat { i } +(2s-1)\hat { j } -(2s+1)\hat { k }$
Solution:
Comparing these equation with
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 PDF
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