NCERT Solutions Miscellaneous Exercise: Three Dimensional Geometry | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Miscellaneous EXERCISE                                                  PAGE NO: 497 
1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points 
(3, 5, –1), (4, 3, –1). 
Solution: 
Let us consider OA to be the line joining the origin (0, 0, 0) and the point A (2, 1, 1). 
And let BC be the line joining the points B (3, 5, -1) and C (4, 3, -1) 
So the direction ratios of OA = (a 1, b 1, c 1) = [(2 – 0), (1 – 0), (1 – 0)] = (2, 1, 1) 
And the direction ratios of BC = (a 2, b 2, c 2) = [(4 – 3), (3 – 5), (-1 + 1)] = (1, -2, 0) 
Given: 
OA is ? to BC 
Now we have to prove that: 
a 1a 2 + b 1b 2 + c 1c 2 = 0 
Let us consider LHS: a 1a 2 + b 1b 2 + c 1c 2 
a 1a 2 + b 1b 2 + c 1c 2 = 2 × 1 + 1 × (-2) + 1 × 0 
= 2 – 2 
= 0 
We know that R.H.S is 0 
So LHS = RHS 
? OA is ? to BC 
Hence proved. 
2. If l 1, m 1, n 1 and l 2, m 2, n 2 are the direction cosines of two mutually perpendicular lines, show that the direction 
cosines of the line perpendicular to both of these are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
Solution: 
Let us consider l, m, n to be the direction cosines of the line perpendicular to each of the given lines. 
Then, ll 1 + mm 1 + nn 1 = 0 … (1) 
And ll 2 + mm 2 + nn 2 = 0 … (2) 
Upon solving (1) and (2) by using cross – multiplication, we get 
Page 2


 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Miscellaneous EXERCISE                                                  PAGE NO: 497 
1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points 
(3, 5, –1), (4, 3, –1). 
Solution: 
Let us consider OA to be the line joining the origin (0, 0, 0) and the point A (2, 1, 1). 
And let BC be the line joining the points B (3, 5, -1) and C (4, 3, -1) 
So the direction ratios of OA = (a 1, b 1, c 1) = [(2 – 0), (1 – 0), (1 – 0)] = (2, 1, 1) 
And the direction ratios of BC = (a 2, b 2, c 2) = [(4 – 3), (3 – 5), (-1 + 1)] = (1, -2, 0) 
Given: 
OA is ? to BC 
Now we have to prove that: 
a 1a 2 + b 1b 2 + c 1c 2 = 0 
Let us consider LHS: a 1a 2 + b 1b 2 + c 1c 2 
a 1a 2 + b 1b 2 + c 1c 2 = 2 × 1 + 1 × (-2) + 1 × 0 
= 2 – 2 
= 0 
We know that R.H.S is 0 
So LHS = RHS 
? OA is ? to BC 
Hence proved. 
2. If l 1, m 1, n 1 and l 2, m 2, n 2 are the direction cosines of two mutually perpendicular lines, show that the direction 
cosines of the line perpendicular to both of these are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
Solution: 
Let us consider l, m, n to be the direction cosines of the line perpendicular to each of the given lines. 
Then, ll 1 + mm 1 + nn 1 = 0 … (1) 
And ll 2 + mm 2 + nn 2 = 0 … (2) 
Upon solving (1) and (2) by using cross – multiplication, we get 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
 
Thus, the direction cosines of the given line are proportional to 
(m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
So, its direction cosines are 
 
 
We know that 
(l 1
2
 + m 1
2
 + n 1
2
) (l 2
2
 + m 2
2
 + n 2
2
) – (l 1l 2 + m 1m 2 + n 1n 2)
2
 
= (m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 … (3) 
It is given that the given lines are perpendicular to each other. 
So, l 1l 2 + m 1m 2 + n 1n 2 = 0 
Also, we have 
l 1
2
 + m 1
2
 + n 1
2
 = 1 
And, l 2
2
 + m 2
2
 + n 2
2
 = 1 
Substituting these values in equation (3), we get 
(m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 = 1 
? = 1 
Hence, the direction cosines of the given line are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
3. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b. 
Solution: 
Page 3


 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Miscellaneous EXERCISE                                                  PAGE NO: 497 
1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points 
(3, 5, –1), (4, 3, –1). 
Solution: 
Let us consider OA to be the line joining the origin (0, 0, 0) and the point A (2, 1, 1). 
And let BC be the line joining the points B (3, 5, -1) and C (4, 3, -1) 
So the direction ratios of OA = (a 1, b 1, c 1) = [(2 – 0), (1 – 0), (1 – 0)] = (2, 1, 1) 
And the direction ratios of BC = (a 2, b 2, c 2) = [(4 – 3), (3 – 5), (-1 + 1)] = (1, -2, 0) 
Given: 
OA is ? to BC 
Now we have to prove that: 
a 1a 2 + b 1b 2 + c 1c 2 = 0 
Let us consider LHS: a 1a 2 + b 1b 2 + c 1c 2 
a 1a 2 + b 1b 2 + c 1c 2 = 2 × 1 + 1 × (-2) + 1 × 0 
= 2 – 2 
= 0 
We know that R.H.S is 0 
So LHS = RHS 
? OA is ? to BC 
Hence proved. 
2. If l 1, m 1, n 1 and l 2, m 2, n 2 are the direction cosines of two mutually perpendicular lines, show that the direction 
cosines of the line perpendicular to both of these are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
Solution: 
Let us consider l, m, n to be the direction cosines of the line perpendicular to each of the given lines. 
Then, ll 1 + mm 1 + nn 1 = 0 … (1) 
And ll 2 + mm 2 + nn 2 = 0 … (2) 
Upon solving (1) and (2) by using cross – multiplication, we get 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
 
Thus, the direction cosines of the given line are proportional to 
(m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
So, its direction cosines are 
 
 
We know that 
(l 1
2
 + m 1
2
 + n 1
2
) (l 2
2
 + m 2
2
 + n 2
2
) – (l 1l 2 + m 1m 2 + n 1n 2)
2
 
= (m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 … (3) 
It is given that the given lines are perpendicular to each other. 
So, l 1l 2 + m 1m 2 + n 1n 2 = 0 
Also, we have 
l 1
2
 + m 1
2
 + n 1
2
 = 1 
And, l 2
2
 + m 2
2
 + n 2
2
 = 1 
Substituting these values in equation (3), we get 
(m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 = 1 
? = 1 
Hence, the direction cosines of the given line are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
3. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b. 
Solution: 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Angle between the lines with direction ratios a 1, b 1, c 1 and a 2, b 2, c 2 is given by 
 
Given: 
a 1 = a, b 1 = b, c 1 = c 
a 2 = b – c, b 2 = c – a, c 2 = a – b 
Let us substitute the values in the above equation. We get, 
 
= 0 
Cos ? = 0 
So, ? = 90° [Since, cos 90 = 0] 
Hence, Angle between the given pair of lines is 90°. 
4. Find the equation of a line parallel to x – axis and passing through the origin. 
Solution: 
We know that, equation of a line passing through (x 1, y 1, z 1) and parallel to a line with direction ratios a, b, c is 
 
Given: the line passes through origin i.e. (0, 0, 0) 
x 1 = 0, y 1 = 0, z 1 = 0 
Since line is parallel to x – axis, 
a = 1, b = 0, c = 0 
? Equation of Line is given by 
Page 4


 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Miscellaneous EXERCISE                                                  PAGE NO: 497 
1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points 
(3, 5, –1), (4, 3, –1). 
Solution: 
Let us consider OA to be the line joining the origin (0, 0, 0) and the point A (2, 1, 1). 
And let BC be the line joining the points B (3, 5, -1) and C (4, 3, -1) 
So the direction ratios of OA = (a 1, b 1, c 1) = [(2 – 0), (1 – 0), (1 – 0)] = (2, 1, 1) 
And the direction ratios of BC = (a 2, b 2, c 2) = [(4 – 3), (3 – 5), (-1 + 1)] = (1, -2, 0) 
Given: 
OA is ? to BC 
Now we have to prove that: 
a 1a 2 + b 1b 2 + c 1c 2 = 0 
Let us consider LHS: a 1a 2 + b 1b 2 + c 1c 2 
a 1a 2 + b 1b 2 + c 1c 2 = 2 × 1 + 1 × (-2) + 1 × 0 
= 2 – 2 
= 0 
We know that R.H.S is 0 
So LHS = RHS 
? OA is ? to BC 
Hence proved. 
2. If l 1, m 1, n 1 and l 2, m 2, n 2 are the direction cosines of two mutually perpendicular lines, show that the direction 
cosines of the line perpendicular to both of these are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
Solution: 
Let us consider l, m, n to be the direction cosines of the line perpendicular to each of the given lines. 
Then, ll 1 + mm 1 + nn 1 = 0 … (1) 
And ll 2 + mm 2 + nn 2 = 0 … (2) 
Upon solving (1) and (2) by using cross – multiplication, we get 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
 
Thus, the direction cosines of the given line are proportional to 
(m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
So, its direction cosines are 
 
 
We know that 
(l 1
2
 + m 1
2
 + n 1
2
) (l 2
2
 + m 2
2
 + n 2
2
) – (l 1l 2 + m 1m 2 + n 1n 2)
2
 
= (m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 … (3) 
It is given that the given lines are perpendicular to each other. 
So, l 1l 2 + m 1m 2 + n 1n 2 = 0 
Also, we have 
l 1
2
 + m 1
2
 + n 1
2
 = 1 
And, l 2
2
 + m 2
2
 + n 2
2
 = 1 
Substituting these values in equation (3), we get 
(m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 = 1 
? = 1 
Hence, the direction cosines of the given line are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
3. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b. 
Solution: 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Angle between the lines with direction ratios a 1, b 1, c 1 and a 2, b 2, c 2 is given by 
 
Given: 
a 1 = a, b 1 = b, c 1 = c 
a 2 = b – c, b 2 = c – a, c 2 = a – b 
Let us substitute the values in the above equation. We get, 
 
= 0 
Cos ? = 0 
So, ? = 90° [Since, cos 90 = 0] 
Hence, Angle between the given pair of lines is 90°. 
4. Find the equation of a line parallel to x – axis and passing through the origin. 
Solution: 
We know that, equation of a line passing through (x 1, y 1, z 1) and parallel to a line with direction ratios a, b, c is 
 
Given: the line passes through origin i.e. (0, 0, 0) 
x 1 = 0, y 1 = 0, z 1 = 0 
Since line is parallel to x – axis, 
a = 1, b = 0, c = 0 
? Equation of Line is given by 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
 
5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (–4, 3, –6) and (2, 9, 2), respectively, then find 
the angle between the lines AB and CD. 
Solution: 
We know that the angle between the lines with direction ratios a 1, b 1, c 1 and a 2, b 2, c 2 is given by 
 
So now, a line passing through A (x 1, y 1, z 1) and B (x 2, y 2, z 2) has direction ratios (x 1 – x 2), (y 1 – y 2), (z 1 – z 2) 
The direction ratios of line joining the points A (1, 2, 3) and B (4, 5, 7) 
= (4 – 1), (5 – 2), (7 – 3) 
= (3, 3, 4) 
? a 1 = 3, b 1 = 3, c 1 = 4 
The direction ratios of line joining the points C (-4, 3, -6) and B (2, 9, 2) 
= (2 – (-4)), (9 – 3), (2-(-6)) 
= (6, 6, 8) 
? a 2 = 6, b 2 = 6, c 2 = 8 
Now let us substitute the values in the above equation. We get, 
Page 5


 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Miscellaneous EXERCISE                                                  PAGE NO: 497 
1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points 
(3, 5, –1), (4, 3, –1). 
Solution: 
Let us consider OA to be the line joining the origin (0, 0, 0) and the point A (2, 1, 1). 
And let BC be the line joining the points B (3, 5, -1) and C (4, 3, -1) 
So the direction ratios of OA = (a 1, b 1, c 1) = [(2 – 0), (1 – 0), (1 – 0)] = (2, 1, 1) 
And the direction ratios of BC = (a 2, b 2, c 2) = [(4 – 3), (3 – 5), (-1 + 1)] = (1, -2, 0) 
Given: 
OA is ? to BC 
Now we have to prove that: 
a 1a 2 + b 1b 2 + c 1c 2 = 0 
Let us consider LHS: a 1a 2 + b 1b 2 + c 1c 2 
a 1a 2 + b 1b 2 + c 1c 2 = 2 × 1 + 1 × (-2) + 1 × 0 
= 2 – 2 
= 0 
We know that R.H.S is 0 
So LHS = RHS 
? OA is ? to BC 
Hence proved. 
2. If l 1, m 1, n 1 and l 2, m 2, n 2 are the direction cosines of two mutually perpendicular lines, show that the direction 
cosines of the line perpendicular to both of these are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
Solution: 
Let us consider l, m, n to be the direction cosines of the line perpendicular to each of the given lines. 
Then, ll 1 + mm 1 + nn 1 = 0 … (1) 
And ll 2 + mm 2 + nn 2 = 0 … (2) 
Upon solving (1) and (2) by using cross – multiplication, we get 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
 
Thus, the direction cosines of the given line are proportional to 
(m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
So, its direction cosines are 
 
 
We know that 
(l 1
2
 + m 1
2
 + n 1
2
) (l 2
2
 + m 2
2
 + n 2
2
) – (l 1l 2 + m 1m 2 + n 1n 2)
2
 
= (m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 … (3) 
It is given that the given lines are perpendicular to each other. 
So, l 1l 2 + m 1m 2 + n 1n 2 = 0 
Also, we have 
l 1
2
 + m 1
2
 + n 1
2
 = 1 
And, l 2
2
 + m 2
2
 + n 2
2
 = 1 
Substituting these values in equation (3), we get 
(m 1n 2 – m 2n 1)
2
 + (n 1l 2 – n 2l 1)
2
 + (l 1m 2 – l 2m 1)
2
 = 1 
? = 1 
Hence, the direction cosines of the given line are (m 1n 2 – m 2n 1), (n 1l 2 – n 2l 1), (l 1m 2 – l 2m 1) 
3. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b. 
Solution: 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
Angle between the lines with direction ratios a 1, b 1, c 1 and a 2, b 2, c 2 is given by 
 
Given: 
a 1 = a, b 1 = b, c 1 = c 
a 2 = b – c, b 2 = c – a, c 2 = a – b 
Let us substitute the values in the above equation. We get, 
 
= 0 
Cos ? = 0 
So, ? = 90° [Since, cos 90 = 0] 
Hence, Angle between the given pair of lines is 90°. 
4. Find the equation of a line parallel to x – axis and passing through the origin. 
Solution: 
We know that, equation of a line passing through (x 1, y 1, z 1) and parallel to a line with direction ratios a, b, c is 
 
Given: the line passes through origin i.e. (0, 0, 0) 
x 1 = 0, y 1 = 0, z 1 = 0 
Since line is parallel to x – axis, 
a = 1, b = 0, c = 0 
? Equation of Line is given by 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
 
5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (–4, 3, –6) and (2, 9, 2), respectively, then find 
the angle between the lines AB and CD. 
Solution: 
We know that the angle between the lines with direction ratios a 1, b 1, c 1 and a 2, b 2, c 2 is given by 
 
So now, a line passing through A (x 1, y 1, z 1) and B (x 2, y 2, z 2) has direction ratios (x 1 – x 2), (y 1 – y 2), (z 1 – z 2) 
The direction ratios of line joining the points A (1, 2, 3) and B (4, 5, 7) 
= (4 – 1), (5 – 2), (7 – 3) 
= (3, 3, 4) 
? a 1 = 3, b 1 = 3, c 1 = 4 
The direction ratios of line joining the points C (-4, 3, -6) and B (2, 9, 2) 
= (2 – (-4)), (9 – 3), (2-(-6)) 
= (6, 6, 8) 
? a 2 = 6, b 2 = 6, c 2 = 8 
Now let us substitute the values in the above equation. We get, 
 
 
 
 
 NCERT Solutions for Class 12 Maths Chapter 11 – 
Three Dimensional Geometry  
 
6. If the lines  
 and  
 are perpendicular, find the value of k. 
Solution: