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NCERT Solutions for Class 12 Maths Chapter 11 – Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry are available here to help students understand the means in which different questions should exist solved. Looking around, we tin discover that everything in the real world is in a three-dimensional shape. In the NCERT Solutions for Class 12 Maths of this chapter, students learn about Three-Dimensional Geometry in detail. Students tin easily score total marks in the questions from this chapter, which is included in the second term of Class 12 Maths past solving all the questions nowadays in the NCERT textbook.

The Class 12 Maths NCERT Solutions for Three Dimensional Geometry are very easy to understand. These solutions comprehend all the exercise questions included in the book and are according to the latest update of the CBSE Syllabus for 2021-22. Here, the PDF of the NCERT Solutions for Class 12 Maths Chapter 11 is available which can be downloaded and also referred to in offline mode.

Download PDF of NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry

Access Exercises of Class 12 Maths Affiliate eleven – Three Dimensional Geometry

Practise 11.one Solutions 5 Questions

Exercise 11.2 Solutions 17 Questions

Exercise xi.three Solutions 14 Questions

Miscellaneous Practice On Chapter 11 Solutions 23 Questions

NCERT Solutions for Class 12 Maths Affiliate 11 – Three Dimensional Geometry

EXERCISE 11.i PAGE NO: 467

one. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, notice its management cosines.

Solution:

Let the direction cosines of the line be l, m and n.

Hither let α = ninety°, β = 135° and γ = 45°

So,

l = cos α, chiliad = cos β and northward = cos γ

And so direction cosines are

l = cos 90° = 0

1000 = cos 135°= cos (180° – 45°) = -cos 45° = -i/√2

n = cos 45° = 1/√two

∴ The direction cosines of the line are 0, -1/√two, one/√ii

2. Find the direction cosines of a line which makes equal angles with the coordinate axes.

Solution:

Given:

Angles are equal.

Then let the angles be α, β, γ

Allow the management cosines of the line be 50, one thousand and n

fifty = cos α, m = cos β and northward = cos γ

Hither given α = β = γ (Since, line makes equal angles with the coordinate axes) … (1)

The management cosines are

l = cos α, k = cos β and n = cos γ

We have,

l² + m² + n^two = 1

cos² α + cos²β + cos²γ = 1

From (ane) we have,

cos² α + cos² α + cos² α = 1

3 cosⁱⁱ α = 1

Cos α = ± √(1/3)

∴ The direction cosines are

l = ± √(1/3), grand = ± √(1/three), n = ± √(1/3)

3. If a line has the management ratios –xviii, 12, –four, then what are its direction cosines?

Solution:

Given

Direction ratios as -18, 12, -4

Where, a = -18, b = 12, c = -iv

Let us consider the management ratios of the line as a, b and c

So the direction cosines are

∴ The direction cosines are

-18/22, 12/22, -4/22 => -ix/11, half dozen/11, -2/11

iv. Show that the points (two, 3, 4), (–one, –2, one), (v, eight, 7) are collinear.

Solution:

If the direction ratios of two lines segments are proportional, and then the lines are collinear.

Given:

A(two, 3, four), B(−1, −2, 1), C(5, 8, 7)

Management ratio of line joining A (2, iii, iv) and B (−1, −2, 1), are

(−one−2), (−ii−three), (one−four) = (−3, −5, −three)

Where, a_one = -3, b_i = -5, c₁ = -3

Management ratio of line joining B (−1, −2, 1) and C (5, 8, 7) are

(5− (−1)), (8− (−two)), (7−1) = (6, ten, six)

Where, a₂ = 6, b_two = 10 and c₂ =half-dozen

Now,

∴ A, B, C are collinear.

5. Discover the direction cosines of the sides of the triangle whose vertices are (three, 5, –4), (-1, 1, two) and (–5, –5, –2).

Solution:

Given:

The vertices are (3, 5, –iv), (-ane, 1, ii) and (–5, –5, –ii).

The management cosines of the ii points passing through A(x₁, y₁, z_i) and B(ten₂, y_ii, z₂) is given past (x₂ – x_one), (y_two-y₁), (z₂-z_ane)

Firstly let us observe the direction ratios of AB

Where, A = (3, 5, -4) and B = (-1, i, 2)

Ratio of AB = [(ten₂ – x_one)², (y₂ – y_i)², (z₂ – z₁)²]

= (-one-three), (1-five), (ii-(-iv)) = -4, -4, vi

Then by using the formula,

√[(ten₂ – 10_ane)² + (y₂ – y_i)² + (z₂ – z_one)²]

√[(-4)² + (-4)² + (6)^two] = √(sixteen+16+36)

= √68

= two√17

Now permit united states find the direction cosines of the line AB

By using the formula,

-iv/2√17 , -4/two√17, 6/two√17

Or -2/√17, -2/√17, iii/√17

Similarly,

Allow us find the direction ratios of BC

Where, B = (-1, one, 2) and C = (-v, -5, -two)

Ratio of AB = [(x₂ – 10_i)ⁱⁱ, (y₂ – y₁)², (z_ii – z_one)²]

= (-v+ane), (-5-1), (-2-2) = -4, -6, -4

Then by using the formula,

√[(x₂ – x_one)² + (y₂ – y₁)² + (z₂ – z_i)ⁱⁱ]

√[(-4)^two + (-vi)² + (-4)²] = √(xvi+36+sixteen)

= √68

= 2√17

Now permit united states of america find the direction cosines of the line AB

By using the formula,

-4/ii√17, -6/2√17, -four/2√17

Or -two/√17, -3/√17, -2/√17

Similarly,

Permit us find the direction ratios of CA

Where, C = (-5, -5, -two) and A = (iii, v, -iv)

Ratio of AB = [(10₂ – x₁)², (y_two – y₁)², (z_ii – z_ane)²]

= (3+five), (5+5), (-4+2) = 8, 10, -2

Then by using the formula,

√[(ten₂ – x₁)² + (y_two – y₁)² + (z₂ – z₁)²]

√[(eight)² + (10)^two + (-2)²] = √(64+100+4)

= √168

= ii√42

Now permit us find the direction cosines of the line AB

Past using the formula,

8/2√42, ten/ii√42, -2/2√42

Or 4/√42, v/√42, -1/√42

---

Exercise 11.2 PAGE NO: 477

1. Show that the iii lines with direction cosines

 Are mutually perpendicular.

Solution:

Let us consider the direction cosines of 50_one, L_ii and 50₃ exist l_ane, m₁, north₁; l₂, chiliad_ii, north_ii and l_three, m_three, due north₃.

We know that

If l₁, yard₁, n₁ and l_two, m_two, n₂ are the management cosines of two lines;

And θ is the acute bending between the two lines;

Then cos θ = |l₁l_ii + thousand_anem₂ + n₁n₂|

If two lines are perpendicular, then the bending between the two is θ = ninety°

For perpendicular lines, | l₁fifty₂ + m₁m₂ + due north_inorth_two | = cos ninety° = 0, i.e. | l₁l_two + thousand_onem₂ + n₁n_ii | = 0

So, in social club to cheque if the iii lines are mutually perpendicular, we compute | l_i50₂ + g_ione thousand₂ + n₁due north₂ | for all the pairs of the three lines.

Firstly allow us compute, | l₁50₂ + m₁m_two + n₁northward₂ |

So,  L₁⊥ 50₂ …… (1)

Similarly,

Let us compute, | l_two50₃ + one thousand_twoone thousand₃ + n_iidue north_three |

So, 50₂⊥ Fifty₃ ….. (2)

Similarly,

Permit u.s.a. compute, | 50₃l₁ + m₃1000₁ + n₃north₁ |

So, L_one⊥ L₃ ….. (three)

∴ By (1), (2) and (3), the lines are perpendicular.

L_ane, L₂ and L₃ are mutually perpendicular.

two. Show that the line through the points (1, –1, ii), (3, 4, –2) is perpendicular to the line through the points (0, 3, ii) and (iii, 5, 6).

Solution:

Given:

The points (one, –i, ii), (3, four, –2) and (0, three, 2), (3, 5, 6).

Let usa consider AB be the line joining the points, (1, -1, ii) and (3, iv, -2), and CD exist the line through the points (0, 3, 2) and (3, 5, vi).

Now,

The direction ratios, a_i, b_ane, c_i of AB are

(iii – 1), (4 – (-1)), (-two – 2) = 2, 5, -iv.

Similarly,

The management ratios, a_two, b_ii, c₂ of CD are

(3 – 0), (5 – 3), (six – 2) = 3, 2, 4.

And then, AB and CD will be perpendicular to each other, if a_anea_two + b₁b₂ + c_ic₂ = 0

a₁a_ii + b_ib_ii + c₁c_two = two(3) + 5(2) + four(-4)

= 6 + 10 – 16

= 0

∴ AB and CD are perpendicular to each other.

3. Prove that the line through the points (four, 7, eight), (two, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

Solution:

Given:

The points (4, 7, 8), (2, iii, 4) and (–one, –2, 1), (ane, ii, 5).

Let us consider AB be the line joining the points, (iv, 7, eight), (2, iii, four) and CD be the line through the points (–1, –2, 1), (1, 2, 5).

At present,

The direction ratios, a_ane, b₁, c₁ of AB are

(ii – 4), (three – seven), (four – viii) = -2, -4, -4.

The management ratios, a₂, b₂, c₂ of CD are

(1 – (-i)), (2 – (-2)), (5 – 1) = 2, four, 4.

Then AB will exist parallel to CD, if

And then, a₁/a_ii = -ii/2 = -1

b_i/b₂ = -4/4 = -1

c₁/c₂ = -iv/four = -i

∴ We tin can say that,

-i = -one = -1

Hence, AB is parallel to CD where the line through the points (4, seven, 8), (2, 3, 4) is parallel to the line through the points (–one, –ii, ane), (1, 2, 5)

four. Find the equation of the line which passes through the bespeak (one, ii, 3) and is parallel to the vector .

Solution:

5. Notice the equation of the line in vector and in Cartesian form that passes through the point with position vector and is in the direction

Solution:

6. Find the Cartesian equation of the line which passes through the point (–ii, 4, –5) and parallel to the line given past

Solution:

Given:

The points (-ii, 4, -5)

Nosotros know that

The Cartesian equation of a line through a bespeak (x₁, y_ane, z₁) and having direction ratios a, b, c is

7. The Cartesian equation of a line is

. Write its vector form.

Solution:

So when comparison this standard form with the given equation, we get

x_one = 5, y_one = -4, z_ane = half-dozen and

l = three, m = seven, n = 2

8. Find the vector and the Cartesian equations of the lines that passes through the origin and (5, –2, 3).

Solution:

nine. Find the vector and the Cartesian equations of the line that passes through the points (three, –2, –5), (3, –2, half dozen).

Solution:

10. Notice the angle between the following pairs of lines:

Solution:

So,

By (three), we have

11.  Discover the angle between the post-obit pair of lines:

Solution:

12. Find the values of p so that the lines

 are at right angles.

Solution:

And so the direction ratios of the lines are

-3, 2p/7, 2 and -3p/7, 1, -five

At present, every bit both the lines are at right angles,

Then, a₁a₂ + b₁b_two + c_anec₂ = 0

(-three) (-3p/vii) + (2p/7) (1) + 2 (-five) = 0

9p/7 + 2p/seven – 10 = 0

(9p+2p)/7 = x

11p/7 = 10

11p = 70

p = 70/11

∴ The value of p is 70/11

xiii. Bear witness that the lines

 are perpendicular to each other.

Solution:

The equations of the given lines are

Two lines with direction ratios is given equally

a_ia₂ + b₁b₂ + c_anec₂ = 0

So the direction ratios of the given lines are vii, -five, 1 and i, 2, 3

i.e., a_ane = seven, b_ane = -5, c₁ = ane and

a_ii = i, b₂ = 2, c_two = 3

Now, Considering

a_anea₂ + b_ib₂ + c₁c₂ = seven × i + (-five) × 2 + 1 × three

= 7 -x + 3

= – 3 + three

= 0

∴ The two lines are perpendicular to each other.

14. Find the shortest distance between the lines

Solution:

Let the states rationalizing the fraction past multiplying the numerator and denominator by √2, we go

∴ The shortest distance is three√2/2

15. Detect the shortest distance betwixt the lines

Solution:

∴ The shortest altitude is 2√29

16. Discover the shortest distance betwixt the lines whose vector equations are

Solution:

Here by comparison the equations we become,

∴ The shortest distance is 3√19

17. Find the shortest altitude betwixt the lines whose vector equations are

Solution:

And,

∴ The shortest distance is 8√29

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EXERCISE 11.3 PAGE NO: 493

1. In each of the following cases, determine the management cosines of the normal to the aeroplane and the distance from the origin.
(a) z = 2

(b) ten + y + z = 1

(c) 2x + 3y – z = 5

(d) 5y + 8 = 0

Solution:

(a) z = 2

Given:

The equation of the plane, z = two or 0x + 0y + z = 2 …. (one)

Direction ratio of the normal (0, 0, i)

Past using the formula,

√[(0)² + (0)² + (1)^two] = √one

= 1

At present,

Divide both the sides of equation (1) by ane, we get

0x/(one) + 0y/(ane) + z/i = 2

So this is of the form lx + my + nz = d

Where, fifty, m, n are the direction cosines and d is the distance

∴ The management cosines are 0, 0, 1

Altitude (d) from the origin is 2 units

(b) x + y + z = 1

Given:

The equation of the airplane, x + y + z = ane…. (i)

Direction ratio of the normal (1, ane, 1)

By using the formula,

√[(1)² + (one)² + (1)ⁱⁱ] = √3

Now,

Split both the sides of equation (1) by √3, we get

x/(√3) + y/(√3) + z/(√3) = one/√3

So this is of the form sixty + my + nz = d

Where, l, k, n are the direction cosines and d is the distance

∴ The direction cosines are i/√3, one/√3, 1/√three

Distance (d) from the origin is i/√three units

(c) 2x + 3y – z = 5

Given:

The equation of the plane, 2x + 3y – z = 5…. (ane)

Management ratio of the normal (2, 3, -one)

By using the formula,

√[(two)² + (3)^two + (-1)ⁱⁱ] = √14

Now,

Carve up both the sides of equation (1) by √14, we become

2x/(√14) + 3y/(√fourteen) – z/(√14) = 5/√14

So this is of the course lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

∴ The management cosines are two/√14, 3/√14, -1/√14

Distance (d) from the origin is 5/√14 units

(d) 5y + 8 = 0

Given:

The equation of the plane, 5y + 8 = 0

-5y = 8 or

0x – 5y + 0z = 8…. (i)

Management ratio of the normal (0, -five, 0)

By using the formula,

√[(0)² + (-5)² + (0)^two] = √25

= 5

Now,

Divide both the sides of equation (1) past v, we get

0x/(5) – 5y/(5) – 0z/(5) = 8/five

Then this is of the form lx + my + nz = d

Where, l, m, north are the direction cosines and d is the distance

∴ The direction cosines are 0, -1, 0

Distance (d) from the origin is 8/v units

2. Observe the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

Solution:

3. Discover the Cartesian equation of the following planes:
(a)

Solution:

Given:

The equation of the aeroplane.

iv. In the following cases, detect the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z – 12 = 0

(b) 3y + 4z – 6 = 0

(c) x + y + z = 1

(d) 5y + 8 = 0

Solution:

(a) 2x + 3y + 4z – 12 = 0

Allow the coordinate of the human foot of ⊥ P from the origin to the given plane be P(x, y, z).

2x + 3y + 4z = 12 …. (1)

Direction ratio are (2, iii, four)

√[(2)² + (iii)² + (4)ⁱⁱ] = √(4 + 9 + 16)

= √29

Now,

Divide both the sides of equation (i) by √29, nosotros get

2x/(√29) + 3y/(√29) + 4z/(√29) = 12/√29

So this is of the form sixty + my + nz = d

Where, l, k, northward are the direction cosines and d is the altitude

∴ The direction cosines are 2/√29, iii/√29, four/√29

Coordinate of the foot (ld, md, nd) =

= [(2/√29) (12/√29), (3/√29) (12/√29), (iv/√29) (12/√29)]

= 24/29, 36/29, 48/29

(b) 3y + 4z – 6 = 0

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x, y, z).

0x + 3y + 4z = 6 …. (i)

Direction ratio are (0, 3, 4)

√[(0)^two + (3)² + (4)²] = √(0 + 9 + sixteen)

= √25

= 5

Now,

Divide both the sides of equation (ane) by v, we become

0x/(5) + 3y/(5) + 4z/(v) = 6/five

So this is of the grade sixty + my + nz = d

Where, l, m, northward are the management cosines and d is the altitude

∴ The direction cosines are 0/five, 3/five, 4/v

Coordinate of the foot (ld, md, nd) =

= [(0/5) (six/5), (3/v) (half dozen/5), (iv/five) (6/5)]

= 0, 18/25, 24/25

(c) x + y + z = 1

Permit the coordinate of the pes of ⊥ P from the origin to the given aeroplane be P(x, y, z).

x + y + z = 1 …. (i)

Direction ratio are (1, 1, 1)

√[(1)ⁱⁱ + (1)^two + (ane)²] = √(ane + ane + 1)

= √3

Now,

Split up both the sides of equation (1) by √iii, nosotros become

1x/(√three) + 1y/(√3) + 1z/(√iii) = 1/√three

Then this is of the grade lx + my + nz = d

Where, l, thou, due north are the direction cosines and d is the distance

∴ The direction cosines are i/√3, 1/√3, 1/√iii

Coordinate of the pes (ld, md, nd) =

= [(ane/√3) (1/√3), (one/√3) (1/√3), (1/√three) (1/√three)]

= 1/three, 1/3, 1/three

(d) 5y + viii = 0

Let the coordinate of the foot of ⊥ P from the origin to the given airplane be P(x, y, z).

0x – 5y + 0z = eight …. (one)

Direction ratio are (0, -5, 0)

√[(0)^two + (-five)ⁱⁱ + (0)²] = √(0 + 25 + 0)

= √25

= 5

Now,

Divide both the sides of equation (one) by v, we get

0x/(5) – 5y/(5) + 0z/(five) = 8/5

So this is of the class lx + my + nz = d

Where, fifty, m, north are the management cosines and d is the altitude

∴ The management cosines are 0, -1, 0

Coordinate of the pes (ld, md, nd) =

= [(0/5) (8/v), (-5/5) (8/five), (0/v) (8/five)]

= 0, -eight/v, 0

v. Find the vector and Cartesian equations of the planes

(a) that passes through the bespeak (1, 0, –2) and the normal to the plane is

(b) that passes through the point (1,iv, 6) and the normal vector to the plane is

Solution:

x – i – 2y + 8 + z – six = 0

ten – 2y + z + one = 0

ten – 2y + z = -i

∴ The required Cartesian equation of the plane is x – 2y + z = -1

x – 1 – 2y + 8 + z – 6 = 0

x – 2y + z + one = 0

ten – 2y + z = -1

∴ The required Cartesian equation of the plane is x – 2y + z = -1

6. Find the equations of the planes that passes through 3 points.
(a) (1, 1, –1), (6, four, –v), (–four, –2, three)

(b) (1, 1, 0), (1, 2, 1), (–ii, 2, –one)

Solution:

Given:

The points are (i, one, -one), (half dozen, iv, -five), (-4, -2, 3).

Let,

= one(12 – 10) – i(18 – 20) -1 (-12 + sixteen)

= 2 + 2 – 4

= 0

Since, the value of determinant is 0.

∴ The points are collinear as in that location will be infinite planes passing through the given 3 points.

(b) (1, 1, 0), (one, two, 1), (–2, two, –1)

7. Notice the intercepts cutting off by the plane 2x + y – z = 5.

Solution:

Given:

The plane 2x + y – z = v

Let united states of america limited the equation of the aeroplane in intercept course

x/a + y/b + z/c = 1

Where a, b, c are the intercepts cut-off by the plane at x, y and z axes respectively.

2x + y – z = 5 …. (i)

At present dissever both the sides of equation (1) by 5, we get

2x/five + y/5 – z/5 = five/5

2x/5 + y/5 – z/5 = 1

x/(5/2) + y/five + z/(-5) = 1

Hither, a = 5/2, b = five and c = -v

∴ The intercepts cut-off by the plane are v/2, five and -five.

viii. Find the equation of the plane with intercept 3 on the y-centrality and parallel to ZOX plane.

Solution:

We know that the equation of the plane ZOX is y = 0

And then, the equation of plane parallel to ZOX is of the form, y = a

Since the y-intercept of the plane is 3, a = three

∴ The required equation of the plane is y = 3

9. Notice the equation of the aeroplane through the intersection of the planes 3x – y + 2z – 4 = 0 and 10 + y + z – ii = 0 and the point (2, 2, 1).

Solution:

Given:

Equation of the plane passes through the intersection of the plane is given by

(3x – y + 2z – iv) + λ (x + y + z – 2) = 0 and the plane passes through the points (2, 2, 1).

And so, (three × ii – two + 2 × 1 – 4) + λ (two + 2 + 1 – 2) = 0

ii + 3λ = 0

3λ = -2

λ = -2/3 …. (ane)

Upon simplification, the required equation of the plane is given equally

(3x – y + 2z – 4) – 2/3 (x + y + z – 2) = 0

(9x – 3y + 6z – 12 – 2x – 2y – 2z + iv)/3 = 0

7x – 5y + 4z – eight = 0

∴ The required equation of the plane is 7x – 5y + 4z – 8 = 0

10. Detect the vector equation of the airplane passing through the intersection of the planes and through the signal (2, 1, 3).

Solution:

The equation of any airplane through the intersection of the planes given in equations (1) and (2) is given by,

11. Find the equation of the aeroplane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

Solution:

Let the equation of the airplane that passes through the ii-given planes

x + y + z = 1 and 2x + 3y + 4z = v is

(10 + y + z – 1) + λ (2x + 3y + 4z – five) = 0

(2λ + one) x + (3λ + ane) y + (4λ + 1) z -1 – 5λ = 0…… (1)

Then the direction ratio of the plane is (2λ + 1, 3λ + 1, 4λ + i)

And management ratio of another plane is (1, -1, 1)

Since, both the planes are ⊥

So by substituting in a₁a_ii + b₁b_ii + c₁c₂ = 0

(2λ + 1 × i) + (3λ + 1 × (-1)) + (4λ + i × ane) = 0

2λ + 1 – 3λ – 1 + 4λ + ane = 0

3λ + one = 0

λ = -i/3

Substitute the value of λ in equation (i) we become,

ten – z + two = 0

∴ The required equation of the plane is x – z + 2 = 0

12. Detect the angle betwixt the planes whose vector equations are

Solution:

13. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, observe the angles betwixt them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + iv = 0

(b) 2x + y + 3z – 2 = 0 and 10 – 2y + 5 = 0

(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – ane = 0

(d) 2x – 2y + 4z + v = 0 and 3x – 3y + 6z – i = 0

(e) 4x + 8y + z – 8 = 0 and y + z – four = 0

Solution:

(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Given:

The equation of the given planes are

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + iv = 0

Two planes are ⊥ if the direction ratio of the normal to the plane is

a₁a₂ + b₁b_ii + c_onec_ii = 0

21 – 5 – 60

-44 ≠ 0

Both the planes are not ⊥ to each other.

Now, ii planes are || to each other if the direction ratio of the normal to the airplane is

∴ The angle is cos^-1 (2/5)

(b) 2x + y + 3z – 2 = 0 and x – 2y + five = 0

Given:

The equation of the given planes are

2x + y + 3z – ii = 0 and x – 2y + 5 = 0

Ii planes are ⊥ if the management ratio of the normal to the plane is

a₁a₂ + b₁b₂ + c_anec₂ = 0

2 × 1 + i × (-2) + 3 × 0

= 0

∴ The given planes are ⊥ to each other.

(c) 2x – 2y + 4z + five = 0 and 3x – 3y + 6z – ane = 0

Given:

The equation of the given planes are

2x – 2y + 4z + v =0 and x – 2y + 5 = 0

We know that, two planes are ⊥ if the management ratio of the normal to the airplane is

a₁a_two + b₁b₂ + c₁c₂ = 0

six + 6 + 24

36 ≠ 0

∴ Both the planes are not ⊥ to each other.

Now let us cheque, both planes are || to each other if the management ratio of the normal to the plane is

∴ The given planes are || to each other.

(d) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – i = 0

Given:

The equation of the given planes are

2x – y + 3z – 1 = 0 and 2x – y + 3z + iii = 0

Nosotros know that, 2 planes are ⊥ if the management ratio of the normal to the plane is

a_ia₂ + b₁b_two + c₁c₂ = 0

2 × ii + (-ane) × (-1) + three × 3

14 ≠ 0

∴ Both the planes are not ⊥ to each other.

Now, let us check two planes are || to each other if the direction ratio of the normal to the aeroplane is

∴ The given planes are || to each other.

(e) 4x + 8y + z – viii = 0 and y + z – 4 = 0

Given:

The equation of the given planes are

4x + 8y + z – 8 = 0 and y + z – 4 = 0

Nosotros know that, two planes are ⊥ if the direction ratio of the normal to the plane is

a_anea₂ + b₁b₂ + c₁c_two = 0

0 + 8 + one

9 ≠ 0

∴ Both the planes are non ⊥ to each other.

Now let us cheque, two planes are || to each other if the management ratio of the normal to the plane is

∴ Both the planes are non || to each other.

At present let usa discover the angle between them which is given as

∴ The bending is 45^o.

14. In the following cases, find the distance of each of the given points from the respective given plane.
Bespeak Plane
(a) (0, 0, 0) 3x – 4y + 12 z = iii

(b) (iii, -2, 1) 2x – y + 2z + 3 = 0

(c) (ii, three, -5) 10 + 2y – 2z = nine

(d) (-6, 0, 0) 2x – 3y + 6z – 2 = 0

Solution:

(a) Signal Plane

(0, 0, 0) 3x – 4y + 12 z = 3

We know that, distance of indicate P(x₁, y_ane, z₁) from the aeroplane Ax + By + Cz – D = 0 is given equally:

Given point is (0, 0, 0) and the plane is 3x – 4y + 12z = iii

= |3/√169|

= 3/13

∴ The altitude is 3/13.

(b) Signal Airplane

(three, -2, 1) 2x – y + 2z + 3 = 0

We know that, altitude of point P(x₁, y₁, z₁) from the plane Ax + By + Cz – D = 0 is given equally:

Given point is (3, -2, 1) and the plane is 2x – y + 2z + 3 = 0

= |13/√ix|

= 13/3

∴ The distance is xiii/3.

(c) Point Plane

(ii, three, -v) x + 2y – 2z = 9

We know that, distance of point P(x_one, y₁, z_i) from the plane Ax + By + Cz – D = 0 is given as:

Given point is (ii, iii, -5) and the plane is x + 2y – 2z = 9

= |9/√9|

= ix/3

= 3

∴ The distance is 3.

(d) Point Plane

(-half-dozen, 0, 0) 2x – 3y + 6z – two = 0

Nosotros know that, distance of betoken P(x₁, y₁, z₁) from the plane Ax + By + Cz – D = 0 is given as:

Given point is (-half-dozen, 0, 0) and the plane is 2x – 3y + 6z – ii = 0

= |14/√49|

= 14/7

= ii

∴ The distance is 2.

---

Miscellaneous EXERCISE PAGE NO: 497

1. Bear witness that the line joining the origin to the point (ii, 1, one) is perpendicular to the line determined by the points (3, v, –1), (4, 3, –one).

Solution:

Let us consider OA be the line joining the origin (0, 0, 0) and the signal A (2, 1, i).

And let BC be the line joining the points B (3, 5, −1) and C (4, three, −1)

So the direction ratios of OA = (a₁, b₁, c₁) ≡ [(2 – 0), (1 – 0), (i – 0)] ≡ (2, ane, 1)

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

Given:

OA is ⊥ to BC

Now we have to show that:

a₁a₂ + b₁b_ii + c_onec₂ = 0

Let the states consider LHS: a₁a_two + b₁b_ii + c₁c_ii

a_anea₂ + b₁b_two + c_onec_two = 2 × i + 1 × (−2) + i × 0

= 2 – 2

= 0

Nosotros know that R.H.S is 0

So LHS = RHS

∴ OA is ⊥ to BC

Hence proved.

ii. If l_i, grand₁, n₁ and l₂, one thousand₂, n₂ are the management cosines of two mutually perpendicular lines, testify that the management cosines of the line perpendicular to both of these are (thou₁due north₂ – m_twon₁), (n_anel₂ – north_twofifty₁), (50_i1000_ii – l_iim_i)

Solution:

Permit u.s.a. consider l, m, n be the direction cosines of the line perpendicular to each of the given lines.

And then, ll_i + mm₁ + nn₁ = 0 … (ane)

And ll_ii + mm_ii + nn_two = 0 … (2)

Upon solving (ane) and (2) past using cantankerous – multiplication, we become

Thus, the management cosines of the given line are proportional to

(m₁n₂ – yard_iidue north₁), (north_anel₂ – n₂l₁), (l₁m₂ – l₂chiliad₁)

And so, its management cosines are

We know that

(l₁ ⁱⁱ + m₁ ⁱⁱ + n_one ²) (l₂ ² + thousand_ii ² + n₂ ^two) – (l₁l₂ + m₁m_ii + n₁n₂)ⁱⁱ

= (1000₁n_ii – m₂northward₁)² + (n₁l₂ – north₂l_ane)² + (fifty_ione thousand₂ – l₂m_i)² … (3)

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

So, 50₁l_ii + k₁m_ii + n₁n_two = 0

Also, we have

l₁ ² + m_one ⁱⁱ + n_one ² = 1

And, 50_two ² + m₂ ² + n_ii ^two = 1

Substituting these values in equation (3), nosotros get

(thousand_in₂ – m_twon_ane)ⁱⁱ + (north₁l_ii – n_ii50₁)² + (l_iyard₂ – l₂k₁)^two = i

λ = one

Hence, the direction cosines of the given line are (m₁n₂ – yard_twon_ane), (n₁l₂ – n₂l_one), (l_oneyard₂ – l₂m₁)

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

Solution:

Bending betwixt the lines with direction ratios a₁, b₁, c₁ and a_ii, b₂, c₂ is given by

Given:

a_ane = a, b₁ = b, c_ane = c

a_ii = b – c, b₂ = c – a, c₂ = a – b

Allow us substitute the values in the higher up equation we get,

= 0

Cos θ = 0

And then, θ = xc° [Since, cos 90 = 0]

Hence, Angle betwixt the given pair of lines is 90°.

4. Find the equation of a line parallel to 10 – centrality and passing through the origin.

Solution:

We know that, equation of a line passing through (10₁, y₁, z₁) and parallel to a line with direction ratios a, b, c is

Given: the line passes through origin i.eastward. (0, 0, 0)

x₁ = 0, y₁ = 0, z₁ = 0

Since line is parallel to x – axis,

a = one, b = 0, c = 0

∴ Equation of Line is given by

5. If the coordinates of the points A, B, C, D be (one, 2, 3), (4, 5, seven), (–4, 3, –6) and (2, 9, ii) respectively, then find the angle betwixt the lines AB and CD.

Solution:

Nosotros know that the angle between the lines with direction ratios a₁, b_one, c₁ and a_ii, b_two, c_ii is given by

And then at present, a line passing through A (x₁, y₁, z₁) and B (10₂, y₂, z₂) has direction ratios (10₁ – 10_two), (y₁ – y_two), (z₁ – z₂)

The management ratios of line joining the points A (one, 2, three) and B (4, five, 7)

= (4 – one), (v – 2), (seven – 3)

= (3, 3, iv)

∴ a₁ = 3, b₁ = 3, c_one = 4

The direction ratios of line joining the points C (-four, 3, -6) and B (2, 9, two)

= (2 – (-4)), (nine – iii), (2-(-6))

= (6, vi, 8)

∴ a₂ = 6, b_ii = 6, c₂ = viii

At present let us substitute the values in the above equation we get,

6. If the lines

 and
 are perpendicular, find the value of 1000.

Solution:

We become –

x₂ = 1, y₂ = two, z₂ = 3

And a₂ = 3k, b_two = one, c₂ = -five

Since the two lines are perpendicular,

a₁a₂ + b_oneb_ii + c₁c₂ = 0

(-3) × 3k + 2k × i + 2 × (-five) = 0

-9k + 2k – ten = 0

-7k = ten

thousand = -ten/7

∴ The value of thousand is -ten/7.

7. Find the vector equation of the line passing through (i, 2, iii) and perpendicular to the plane

Solution:

viii. Find the equation of the aeroplane passing through (a, b, c) and parallel to the aeroplane

Solution:

The equation of a aeroplane passing through (10₁, y₁, z₁) and perpendicular to a line with direction ratios A, B, C is given as

A (x – ten₁) + B (y – y₁) + C (z – z₁) = 0

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

And so, 10_i = a, y₁ = b, z_one = c

Since both planes are parallel to each other, their normal will be parallel

Direction ratios of normal = (1, 1, 1)

So, A = one, B =1, C = i

The Equation of plane in Cartesian form is given equally

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

1(x – a) + ane(y – b) + 1(z – c) = 0

x + y + z – (a + b + c) = 0

x + y + z = a + b + c

∴ The required equation of airplane is 10 + y + z = a + b + c

9. Find the shortest altitude betwixt lines
 and

Solution:

10. Find the coordinates of the point where the line through (five, i, half-dozen) and (iii, iv,1) crosses the YZ – plane.

Solution:

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

Then,

0 = 5 – 2λ

five = 2λ

λ = 5/2

y = ane + 3λ … (5)

And,

z = six – 5λ … (6)

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

y = 1 + 3λ

= one + iii × (five/two)

= 1 + (xv/two)

= 17/ii

And

z = half dozen – 5λ

= half dozen – 5 × (five/ii)

= 6 – (25/2)

= – xiii/2

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

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

Solution:

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

Then,

x = five – 2λ … (v)

0 = 1 + 3λ

-1 = 3λ

λ = -1/3

And,

z = 6 – 5λ … (6)

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

ten = 5 – 2λ

= 5 – 2 × (-1/3)

= 5 + (two/3)

= 17/iii

And

z = 6 – 5λ

= 6 – five × (-i/3)

= 6 + (5/3)

= 23/3

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

12. Find the coordinates of the point where the line through (3, –4, –v) 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 (x_two, y₂, z₂) is given as

It is given that the line passes through the points A (3, –4, –5) and B (two, –3, 1)

And so, x_one = iii, y₁ = -4, z₁ = -v

And, x_two = two, y_two = -3, z_ii = 1

So the equation of line is

And then, x = -g + iii |, y = k – four |, z = 6k – 5 … (1)

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

By substituting the value of 10, y, z in equation (ane) in the equation of plane, we go

2x + y + z + seven = 0

two(-grand + iii) + (k – 4) + (6k – 5) = vii

5k – iii = 7

5k = 10

grand = two

At present substitute the value of k in 10, y, z we get,

x = – yard + 3 = – 2 + 3 = 1

y = k – 4 = 2 – 4 = – 2

z = 6k – 5 = 12 – 5 = 7

∴ The coordinates of the required point are (one, -two, 7).

13. Find the equation of the plane passing through the indicate (–1, three, 2) and perpendicular to each of the planes x + 2y + 3z = v 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 – y_one) + C (z – z₁) = 0

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

Information technology is given that the plane passes through (-one, three, 2)

And so, equation of plane is given past

A (10 + 1) + B (y – 3) + C (z – 2) = 0 ……… (i)

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

We know that

Then, required normal is cross product of normal of planes

10 + 2y + 3z = 5 and 3x + 3y + z = 0

Hence, the direction ratios are = -7, eight, -3

∴ A = -7, B = 8, C = -3

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

A (x + 1) + B (y – 3) + C (z – 2) = 0

-seven(x + i) + 8(y – 3) + (-3) (z – 2) = 0

-7x – seven + 8y – 24 – 3z + half dozen = 0

-7x + 8y – 3z – 25 = 0

7x – 8y + 3z + 25 = 0

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

xiv. If the points (one, 1, p) and (–3, 0, 1) be equidistant from the plane

, then notice the value of p.

Solution:

20 – 12p =± 8

20 – 12p = 8 or, 20 – 12p = -8

12p = 12 or, 12p = 28

p = i or, p = seven/3

∴ The possible values of p are 1 and seven/iii.

15. Find the equation of the aeroplane passing through the line of intersection of the planes  and and parallel to 10-axis.

Solution:

Since this airplane is parallel to ten-centrality.

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

The direction ratios of Normal (a_1, b_1, c₁) ≡ [(i – 2λ), (one – 3λ), (1 +)]

The direction ratios of x–axis (a_2, b_2, c_ii) ≡ (1, 0, 0)

Since the two lines are perpendicular,

a_onea_two + b_aneb₂ + c₁c₂ = 0

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

(1 – 2λ) = 0

λ = 1/2

Substituting the value of λ in equation (1), we become

xvi. If O be the origin and the coordinates of P exist (1, 2, –three), then observe 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 management 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)

Then, x₁ = 1, y₁ = 2, z₁ = – three

Normal vector to plane is =

Where O (0, 0, 0), P (i, 2, -3)

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

= (1, 2, – 3)

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

Equation of airplane in Cartesian grade is given every bit

ane(x – ane) + 2(y – ii) – 3(z – (-iii)) = 0

x – i + 2y – four – 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

Then, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).

Direction ratios of Normal of plane (1) = (a_1, b_i, c₁) ≡ [(1 – 2λ), (ii – λ), (3 + λ)]

Direction ratios of Normal of plane (ii) = (a_2, b_2, c_two) ≡ (-5, -iii, 6)

Since the two lines are perpendicular,

a_anea₂ + b_ib₂ + c₁c₂ = 0

(1 – 2λ) × (-5) + (2 – λ) × (-three) + (3 + λ) × half dozen = 0

-v + 10λ – 6 + 3λ + 18 + 6λ = 0

19λ + seven = 0

λ = -7/19

By substituting the value of λ in equation (1), we go

18. Find the distance of the point (–1, –5, –10) from the point of intersection of the line

Solution:

Where,

ten = 2, y = -1, z = 2

And so, the bespeak of intersection is (two, -1, ii).

Now, the distance between points (x_ane, y_ane, z_i) and (10₂, y₂, z₂) is given by

∴ The distance is 13 units.

xx. Find the vector equation of the line passing through the signal (ane, two, – 4) and perpendicular to the ii lines:
 and.

Solution:

21. Bear witness that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, and so

Solution:

22. Distance between the two planes: 2x + 3y + 4z = iv 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:

Offset Plane:

2x + 3y + 4z = four

Allow us compare with Ax + Past + Cz = d_i, we get

A = 2, B = 3, C = 4, d_ane = 4

Second Airplane:

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

We get,

2x + 3y + 4z = 6

Now comparison with Ax + By + Cz = d₁, we go

A = 2, B = iii, C = 4, d_two = 6

So,

Distance betwixt two planes is given as

= 2/√29

∴ Pick (D) is the right option.

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

Solution:

It is given that:

Start Plane:

2x – y + 4z = 5 [Multiply both sides past ii.5]

We go,

5x – 2.5y + 10z = 12.v … (1)

Given second Plane:

5x – 2.5y + 10z = six … (2)

Then,

Information technology is articulate that the direction ratios of normal of both the plane (1) and (2) are same.

∴ Both the given planes are parallel.

---

The major concepts of Maths covered in Affiliate 11- Three Dimensional Geometry of NCERT Solutions for Class 12 are:

11.i Introduction

xi.ii Management Cosines and Management Ratios of a Line

11.2.1 Relation betwixt the direction cosines of a line

eleven.2.two Direction cosines of a line passing through ii points

eleven.iii Equation of a Line in Space

xi.3.1 Equation of a line through a given signal and parallel to a given vector b.

11.iii.2 Equation of a line passing through two given points

11.4 Angle between Two Lines

11.5 Shortest Distance between Two Lines

11.v.1 Distance between 2 skew lines

eleven.5.ii Distance betwixt parallel lines

11.6 Plane

11.half dozen.1 Equation of a aeroplane in normal class

11.6.two Equation of a plane perpendicular to a given vector, passing through a given point

eleven.six.3 Equation of a plane passing through three non-collinear points

11.half dozen.4 Intercept form of the equation of a aeroplane

11.six.5 Plane passing through the intersection of two given planes

eleven.7 Coplanarity of Two Lines

11.8 Angle between Two Planes

eleven.nine Distance of a Indicate from a Airplane

11.10 Angle between a Line and a Plane

NCERT Solutions for Grade 12 Maths Affiliate 11- Three Dimensional Geometry

The chapter Three Dimensional Geometry belongs to the unit of measurement Vectors and 3 – Dimensional Geometry of term – Ii every bit per latest CBSE Syllabus 2021-22. It adds upwards to 14 marks of the total marks. In that location are 3 exercises along with a miscellaneous exercise in this chapter to aid students understand the concepts related to 3 Dimensional Geometry clearly. Some of the topics discussed in Chapter eleven of NCERT Solutions for Class 12 Maths are as follows:

1. Direction cosines of a line are the cosines of the angles fabricated by the line with the positive directions of the coordinate axes.
2. If l, g, n are the management cosines of a line, then fifty^two + chiliadⁱⁱ + due north^two = 1
3. Management ratios of a line are the numbers which are proportional to the direction cosines of a line.
4. Skew lines are lines in infinite which are neither parallel nor intersecting. They lie in different planes.
5. Angle between skew lines is the angle between two intersecting lines drawn from any signal (preferably through the origin) parallel to each of the skew lines.
6. If 50_i , m₁, n₁ and l_ii, m_ii, n₂ are the management cosines of two lines; and θ is the acute bending between the two lines; then cosθ = |50₁l₂ + m_anechiliad₂ + n_in₂|

These are a few topics that are discussed in chapter 3 dimensional geometry. Students can utilize the NCERT Solutions for Course 12 Maths Chapter 11 for any quick reference to cover circuitous topics. As well, to know more near the affiliate, refer to the NCERT Textbook and NCERT Solutions of Grade 12 Maths.

Key Features of NCERT Solutions for Grade 12 Maths Chapter 11- Three Dimensional Geometry

Studying the Iii Dimensional Geometry of Course 12 using the NCERT Solutions enables the students to sympathise the post-obit:

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, coplanar and skew lines, the shortest distance between ii lines. Cartesian and vector equation of a plane. The angle between (i) ii lines, (ii) two planes, (3) a line and a plane. Distance of a point from a plane.

Oftentimes Asked Questions on NCERT Solutions for Course 12 Maths Chapter 11

What is three dimensional geometry in NCERT Solutions for Class 12 Maths Chapter 11?

It is a branch of Mathematics that includes the study of lines, points, and solid shapes in iii dimensional coordinate systems. It will introduce students to the concept of z-coordinate along with ten and y coordinates to decide the exact location of a point in the three dimensional coordinate planes.

What are the important topics covered in Chapter 11 of NCERT Solutions for Class 12 Maths?

The important topics covered in NCERT Solutions for Class 12 Maths Chapter 11 are:
11.1 Introduction
11.2 Direction Cosines and Management Ratios of a Line
11.3 Equation of a Line in Space
eleven.4 Angle between Two Lines
11.5 Shortest Altitude between Two Lines
eleven.6 Plane
11.seven Coplanarity of 2 Lines
xi.8 Angle between Two Planes
11.9 Distance of a Point from a Plane
eleven.10 Bending between a Line and a Plane

Is NCERT Solutions for Grade 12 Maths Chapter 11 the best report material for the students during revision?

Yes, the NCERT Solutions for Class 12 Maths Chapter 11 is the best study material that helps students to revise complex concepts effortlessly. Each solution provides a logical explanation to make learning easier for the students. The in-house team of experts at BYJU'South take framed the stepwise solutions to encourage the analytical thinking arroyo amidst students. These solutions can also exist compared in club to become an idea most the other methods, which can be used to solve the textbook bug.

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Source: https://byjus.com/ncert-solutions-class-12-maths/chapter-11-3-dimensional-geometry/

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