 # NCERT Solutions for Class 12 Maths Miscellaneous Exercise chapter 11 3 Dimensional Geometry

The Miscellaneous Exercise of NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry is based on the following topics:

1. Direction Cosines and Direction Ratios of a Line
2. Equation of a Line in Space
3. Angle between Two Lines
4. Shortest Distance between Two Lines
5. Plane
6. Coplanarity of Two Lines
7. Angle between Two Planes
8. Distance of a Point from a Plane
9. Angle between a Line and a Plane

Solve all the problems of this exercise to get thorough with the concepts and topics covered in the entire chapter.

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry Miscellaneous Exercise                       ### Access other exercises of Class 12 Maths Chapter 11

Exercise 11.1 Solutions 5 Questions

Exercise 11.2 Solutions 17 Questions

Exercise 11.3 Solutions 14 Questions

#### Access Answers of Maths NCERT Class 12 Chapter 11 Miscellaneous

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 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 = (a1, b1, c1) ≡ [(2 – 0), (1 – 0), (1 – 0)] ≡ (2, 1, 1)

And the direction ratios of BC = (a2, b2, c2) ≡ [(4 – 3), (3 – 5), (-1 + 1)] ≡ (1, -2, 0)

Given:

OA is ⊥ to BC

Now we have to prove that:

a1a2 + b1b2 + c1c2 = 0

Let us consider LHS: a1a2 + b1b2 + c1c2

a1a2 + b1b2 + c1c2 = 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 l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are (m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1)

Solution:

Let us consider l, m, n be the direction cosines of the line perpendicular to each of the given lines.

Then, ll1 + mm1 + nn1 = 0 … (1)

And ll2 + mm2 + nn2 = 0 … (2)

Upon solving (1) and (2) by using cross – multiplication, we get Thus, the direction cosines of the given line are proportional to

(m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1)

So, its direction cosines are  We know that

(l12 + m12 + n12) (l22 + m22 + n22) – (l1l2 + m1m2 + n1n2)2

= (m1n2 – m2n1)2 + (n1l2 – n2l1)2 + (l1m2 – l2m1)2 … (3)

It is given that the given lines are perpendicular to each other.

So, l1l2 + m1m2 + n1n2 = 0

Also, we have

l12 + m12 + n12 = 1

And, l22 + m22 + n22 = 1

Substituting these values in equation (3), we get

(m1n2 – m2n1)2 + (n1l2 – n2l1)2 + (l1m2 – l2m1)2 = 1

λ = 1

Hence, the direction cosines of the given line are (m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1)

3. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Solution:

Angle between the lines with direction ratios a1, b1, c1 and a2, b2, c2 is given by Given:

a1 = a, b1 = b, c1 = c

a2 = b – c, b2 = c – a, c2 = a – b

Let us substitute the values in the above equation we get, = 0

Cos θ = 0

So, θ = 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 (x1, y1, z1) and parallel to a line with direction ratios a, b, c is Given: the line passes through origin i.e. (0, 0, 0)

x1 = 0, y1 = 0, z1 = 0

Since line is parallel to x – axis,

a = 1, b = 0, c = 0

∴ Equation of Line is given by 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 a1, b1, c1 and a2, b2, c2 is given by So now, a line passing through A (x1, y1, z1) and B (x2, y2, z2) has direction ratios (x1 – x2), (y1 – y2), (z1 – z2)

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)

∴ a1 = 3, b1 = 3, c1 = 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)

∴ a2 = 6, b2 = 6, c2 = 8

Now let us substitute the values in the above equation we get, 6. If the lines and are perpendicular, find the value of k.

Solution:  We get –

x2 = 1, y2 = 2, z2 = 3

And a2 = 3k, b2 = 1, c2 = -5

Since the two lines are perpendicular,

a1a2 + b1b2 + c1c2 = 0

(-3) × 3k + 2k × 1 + 2 × (-5) = 0

-9k + 2k – 10 = 0

-7k = 10

k = -10/7

7

∴ The value of k is -10/7.

7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane Solution:  8. Find the equation of the plane passing through (a, b, c) and parallel to the plane Solution:

The equation of a plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is given as

A (x – x1) + B (y – y1) + C (z – z1) = 0

It is given that, the plane passes through (a, b, c)

So, x1 = a, y1 = b, z1 = c

Since both planes are parallel to each other, their normal will be parallel Direction ratios of normal = (1, 1, 1)

So, A = 1, B =1, C = 1

The Equation of plane in Cartesian form is given as

A (x – x1) + B (y – y1) + C (z – z1) = 0

1(x – a) + 1(y – b) + 1(z – c) = 0

x + y + z – (a + b + c) = 0

x + y + z = a + b + c

∴ The required equation of plane is x + y + z = a + b + c

9. Find the shortest distance between lines and Solution:   10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ – plane.

Solution:  We know that, two vectors are equal if their corresponding components are equal

So,

0 = 5 – 2λ

5 = 2λ

λ = 5/2

y = 1 + 3λ … (5)

And,

z = 6 – 5λ … (6)

Substitute the value of λ in equation (5) and (6), we get –

y = 1 + 3λ

= 1 + 3 × (5/2)

= 1 + (15/2)

= 17/2

And

z = 6 – 5λ

= 6 – 5 × (5/2)

= 6 – (25/2)

= – 13/2

∴ The coordinates of the required point is (0, 17/2, -13/2).

11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX – plane.

Solution:  We know that, two vectors are equal if their corresponding components are equal

So,

x = 5 – 2λ … (5)

0 = 1 + 3λ

-1 = 3λ

λ = -1/3

And,

z = 6 – 5λ … (6)

Substitute the value of λ in equation (5) and (6), we get –

x = 5 – 2λ

= 5 – 2 × (-1/3)

= 5 + (2/3)

= 17/3

And

z = 6 – 5λ

= 6 – 5 × (-1/3)

= 6 + (5/3)

= 23/3

∴ The coordinates of the required point is (17/3, 0, 23/3).

12. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7.

Solution:

We know that the equation of a line passing through two points A (x1, y1, z1) and B (x2, y2, z2) is given as It is given that the line passes through the points A (3, –4, –5) and B (2, –3, 1)

So, x1 = 3, y1 = -4, z1 = -5

And, x2 = 2, y2 = -3, z2 = 1

Then the equation of line is So, x = -k + 3 |, y = k – 4 |, z = 6k – 5 … (1)

Now let (x, y, z) be the coordinates of the point where the line crosses the given plane 2x + y + z + 7 = 0

By substituting the value of x, y, z in equation (1) in the equation of plane, we get

2x + y + z + 7 = 0

2(-k + 3) + (k – 4) + (6k – 5) = 7

5k – 3 = 7

5k = 10

k = 2

Now substitute the value of k in x, y, z we get,

x = – k + 3 = – 2 + 3 = 1

y = k – 4 = 2 – 4 = – 2

z = 6k – 5 = 12 – 5 = 7

∴ The coordinates of the required point are (1, -2, 7).

13. Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution:

We know that the equation of a plane passing through (x1, y1, z1) is given by

A (x – x1) + B (y – y1) + C (z – z1) = 0

Where, A, B, C are the direction ratios of normal to the plane.

It is given that the plane passes through (-1, 3, 2)

So, equation of plane is given by

A (x + 1) + B (y – 3) + C (z – 2) = 0 ……… (1)

Since this plane is perpendicular to the given two planes. So, their normal to the plane would be perpendicular to normal of both planes.

We know that So, required normal is cross product of normal of planes

x + 2y + 3z = 5 and 3x + 3y + z = 0 Hence, the direction ratios are = -7, 8, -3

∴ A = -7, B = 8, C = -3

Substituting the obtained values in equation (1), we get

A (x + 1) + B (y – 3) + C (z – 2) = 0

-7(x + 1) + 8(y – 3) + (-3) (z – 2) = 0

-7x – 7 + 8y – 24 – 3z + 6 = 0

-7x + 8y – 3z – 25 = 0

7x – 8y + 3z + 25 = 0

∴ The equation of the required plane is 7x – 8y + 3z + 25 = 0.

14. If the points (1, 1, p) and (–3, 0, 1) be equidistant from the plane , then find the value of p.

Solution:  20 – 12p = ± 8

20 – 12p = 8 or, 20 – 12p = -8

12p = 12 or, 12p = 28

p = 1 or, p = 7/3

∴ The possible values of p are 1 and 7/3.

15. Find the equation of the plane passing through the line of intersection of the planes and and parallel to x-axis.

Solution: Since this plane is parallel to x-axis.

So, the normal vector of the plane (1) will be perpendicular to x-axis.

The direction ratios of Normal (a1, b1, c1) ≡ [(1 – 2λ), (1 – 3λ), (1 +)]

The direction ratios of x–axis (a2, b2, c2) ≡ (1, 0, 0)

Since the two lines are perpendicular,

a1a2 + b1b2 + c1c2 = 0

(1 – 2λ) × 1 + (1 – 3λ) × 0 + (1 + λ) × 0 = 0

(1 – 2λ) = 0

λ = 1/2

Substituting the value of λ in equation (1), we get 16. If O be the origin and the coordinates of P be (1, 2, –3), then find the equation of the plane passing through P and perpendicular to OP.

Solution:

We know that the equation of a plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is given as

A(x – x1) + B(y – y1) + C (z – z1) = 0

It is given that the plane passes through P (1, 2, 3)

So, x1 = 1, y1 = 2, z1 = – 3

Normal vector to plane is = Where O (0, 0, 0), P (1, 2, -3)

So, direction ratios of is = (1 – 0), (2 – 0), (-3 – 0)

= (1, 2, – 3)

Where, A = 1, B = 2, C = -3

Equation of plane in Cartesian form is given as

1(x – 1) + 2(y – 2) – 3(z – (-3)) = 0

x – 1 + 2y – 4 – 3z – 9 = 0

x + 2y – 3z – 14 = 0

∴ The equation of the required plane is x + 2y – 3z – 14 = 0 Solution: Since this plane is perpendicular to the plane So, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).

Direction ratios of Normal of plane (1) = (a1, b1, c1) ≡ [(1 – 2λ), (2 – λ), (3 + λ)]

Direction ratios of Normal of plane (2) = (a2, b2, c2) ≡ (-5, -3, 6)

Since the two lines are perpendicular,

a1a2 + b1b2 + c1c2 = 0

(1 – 2λ) × (-5) + (2 – λ) × (-3) + (3 + λ) × 6 = 0

-5 + 10λ – 6 + 3λ + 18 + 6λ = 0

19λ + 7 = 0

λ = -7/19

By substituting the value of λ in equation (1), we get 18. Find the distance of the point (–1, –5, –10) from the point of intersection of the line Solution: Where,

x = 2, y = -1, z = 2

So, the point of intersection is (2, -1, 2).

Now, the distance between points (x1, y1, z1) and (x2, y2, z2) is given by  ∴ The distance is 13 units.    20. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: and .

Solution:  21. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then Solution:   22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
A. 2 units
B. 4 units
C. 8 units

D. 2/√29 units

Solution:

We know that the distance between two parallel planes Ax + By + Cz = d1 and Ax + By + Cz = d2 is given as It is given that:

First Plane:

2x + 3y + 4z = 4

Let us compare with Ax + By + Cz = d1, we get

A = 2, B = 3, C = 4, d1 = 4

Second Plane:

4x + 6y + 8z = 12 [Divide the equation by 2]

We get,

2x + 3y + 4z = 6

Now comparing with Ax + By + Cz = d1, we get

A = 2, B = 3, C = 4, d2 = 6

So,

Distance between two planes is given as = 2/√29

∴ Option (D) is the correct option.

23. The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
A. Perpendicular
B. Parallel
C. intersect y–axis
D. passes through

Solution:

It is given that:

First Plane:

2x – y + 4z = 5 [Multiply both sides by 2.5]

We get,

5x – 2.5y + 10z = 12.5 … (1)

Given second Plane:

5x – 2.5y + 10z = 6 … (2)

So, It is clear that the direction ratios of normal of both the plane (1) and (2) are same.

∴ Both the given planes are parallel.