Question

NCERT Solutions Class 12 Maths Chapter 11 Three-dimensional Geometry is concisely made by the mentors of the subject matter. We provide NCERT Solutions to make it easy for students to learn and revise all different concepts. NCERT Solutions Class 12 has in-depth solutions to important concepts.

• Three Dimensional Geometry – 3D geometry has mathematical shapes in 3D space which also involves all 3 coordination. We have three different coordinates which work together namely \(x\)-coordination, \(y\)-coordination, and \(z\)-coordination. In 3d geometry, we can find the location of any point in the space with the help of other coordination. In the application of 3 d geometry, we can perfectly find out location and position.
• Rectangular coordinate system – when three lines are mutually perpendicular to each other. The common point of all three axes meets at the origin.  
• Distance from the origin – the distance between origin and any particular point is found by the Pythagoras theorem. The distance of \(P(x,y,z)\) from origin \((0,0,0)\) is \(\sqrt {x^2 + y^2 + z^2}\)
•  Distance between two different points in space - Distance between 2 points  \(P(x_1+y_1+z_1)\) and \(Q(x_2+y_2+z_2)\) is \(\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\) 
• Division of the line joining 2 points - Let \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\) be \(2\) points. R derives the line segment PQ in ratio internally. Then R has coordinate \({mx_2+nx_1 \over m+n},{my_2+ny_1 \over m+n},{mz_2+nz_1 \over m+n}\)
• Direction Cosines and Direction Ratios of a Line – the direction ratio of a line are calculated with the help of a number proportional to the direction cosine. The direction numbers are represented with symbols such as \(a,b,\) and \(c\). at the unit level the sum of the square of the direction cosines is \(1\).
• Equation of a Line in Space – whenever we make a line on a graphical plane it has a slope which is represented by \(m\). the equation of a line is represented by \(y = mx + c.\)

Our expert team at Sarthaks has made NCERT Solutions Class 12 Maths for students’ complete understanding of different concepts for easy learning and understanding of all kinds of topics.

Answer 1

NCERT Solutions Class 12 Maths Chapter 11 Three dimensional Geometry

1. If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.

Answer:

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

Therefore, the direction cosines of the line are 0, \(- \frac1{\sqrt2}\) and \(\frac1{\sqrt2}\) .

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

Answer:

Let the direction cosines of the line make an angle α with each of the coordinate axes. 

l = cos α, m = cos α, n = cos α

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are \(\pm \frac1{\sqrt3}\), \(\pm \frac1{\sqrt3}\) and \(\pm \frac1{\sqrt3}\).

3. If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

Answer:

If a line has direction ratios of −18, 12, and −4, then its direction cosines are

Thus, the direction cosines are \(- \frac9{11}, \frac6{11}\) and \(\frac{-2}{11}\).

4. Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

Answer:

The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7). 

It is known that the direction ratios of line joining the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), are given by, x₂ − x₁, y₂ − y₁, and z₂ − z₁. 

The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3. 

The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.

It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional. 

Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear.

5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2).

Answer:

The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.

Therefore, the direction cosines of AB are

The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4. 

Therefore, the direction cosines of BC are

The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., −8, −10, and 2. 

Therefore, the direction cosines of AC are

6. Show that the three lines with direction cosines

are mutually perpendicular.

Answer:

Two lines with direction cosines, l₁, m₁, n₁ and l₂, m₂, n₂, are perpendicular to each other, if l₁l₂ + m₁m₂ + n₁n₂ = 0

(i) For the lines with direction cosines,

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines,

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines, \(\frac3{13}, \frac{-4}{13}, \frac{12}{13}\) and \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\), we obtain

Therefore, the lines are perpendicular. 

Thus, all the lines are mutually perpendicular.

7. 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).

Answer:

Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6). 

The direction ratios, a₁, b₁, c₁, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4. 

The direction ratios, a₂, b₂, c₂, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4. 

AB and CD will be perpendicular to each other, if a₁a₂ + b₁b₂+ c₁c₂ = 0 a₁a₂ + b₁b₂+ c₁c₂ = 2 × 3 + 5 × 2 + (− 4) × 4 =

 6 + 10 − 16

= 0

Therefore, AB and CD are perpendicular to each other.

8. Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

Answer:

Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5). 

The directions ratios, a₁, b₁, c₁, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.

The direction ratios, a₂, b₂, c₂, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4. 

AB will be parallel to CD, if \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)

Thus, AB is parallel to CD.

9. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \(3\hat i + 2\hat j - 2\hat k\).

Answer:

It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is

\(\vec a = \hat i + 2\hat j + 3\hat k\)

\(\vec b =3\hat i + 2\hat j - 2\hat k\) 

It is known that the line which passes through point A and parallel to \(\vec b\) is given by

\(\vec r = \vec a + \lambda \vec b, \) where \(\lambda\) is a constant.

This is the required equation of the line.

Answer 2

10. 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 \(\hat i + 2\hat j - \hat k\).

Answer:

It is given that the line passes through the point with position vector

\(\vec a =2\hat i - \hat j + 4\hat k\)   .....(1)

\(\vec b=\hat i + 2\hat j - \hat k\)     ......(2)

It is known that a line through a point with position vector \(\vec a\) and \(\vec b\) parallel to is given by the equation,

\(\vec r = \vec a + \lambda \vec b\)

This is the required equation of the given line in Cartesian form.

11. The Cartesian equation of a line is \(\frac{x-5}{3} = \frac{y + 4}7 = \frac{z -6}2\). Write its vector form.

Answer:

The Cartesian equation of the line is

\(\frac{x-5}{3} = \frac{y + 4}7 = \frac{z -6}2\)   ....(1)

The given line passes through the point (5, −4, 6).

 The position vector of this point is \(\vec a = 5\hat i-4\hat j + 6\hat k\)

Also, the direction ratios of the given line are 3, 7, and 2. 

This means that the line is in the direction of vector, \(\vec b = 3\hat i + 7\hat j + 2\hat k\)

It is known that the line through position vector \(\vec a\) and \(\vec b\) in the direction of the vector is given by the equation,

\(\vec r = \vec a + \lambda \vec b, \lambda \in R\)

⇒ \(\vec r = (5\hat i - 4\hat j + 6\hat k) + \lambda (3\hat i + 7\hat j + 2\hat k)\)

This is the required equation of the given line in vector form.

12. Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

Answer:

The required line passes through the origin. Therefore, its position vector is given by,

\(\vec a = \vec 0\)   .....(1)

The direction ratios of the line through origin and (5, −2, 3) are (5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3 

The line is parallel to the vector given by the equation, \(\vec b = 5\hat i -2\hat j + 3\hat k\)

The equation of the line in vector form through a point with position vector \(\vec a\) and \(\vec b\) parallel to is,

The equation of the line through the point (x₁, y₁, z₁) and direction ratios a, b, c is given by,

Therefore, the equation of the required line in the Cartesian form is

13. Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

Answer:

Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ. 

Since PQ passes through P (3, −2, −5), its position vector is given by, 

\(\vec a = 3\hat i - 2\hat j - 5\hat k\)

The direction ratios of PQ are given by, (3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11 

The equation of the vector in the direction of PQ is

\(\vec b = 0.\hat i - 0.\hat j + 11 \hat k = 11\hat k\)

The equation of PQ in vector form is given by, 

\(\vec r = \vec a + \lambda \vec b, \lambda \in R\)

⇒ \(\vec r = (3\hat i - 2\hat j - 5\hat k) + 11\lambda\hat k\)

The equation of PQ in Cartesian form is

14. Find the angle between the following pairs of lines:

Answer:

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, 

The given lines are parallel to the vectors, \(\vec {b_1} = 3\hat i + 2\hat j + 6\hat k\) and \(\vec{b_2} = \hat i + 2\hat j + 2\hat k\), respectively.

(ii) The given lines are parallel to the vectors, \(\vec {b_1} = \hat i - \hat j - 2\hat k\) and \(\vec{b_2} = 3\hat i - 5\hat j - 4\hat k\), respectively.

15. Find the angle between the following pairs of lines:

Answer:

(i) Let \(\vec {b_1}\) and \(\vec {b_2}\) be the vectors parallel to the pair of lines,

The angle, Q, between the given pair of lines is given by the relation,

(ii) Let \(\vec {b_1}\) and \(\vec {b_2}\) be the vectors parallel to the pair of lines,

If Q is the angle between the given pair of lines, then

16. Show that the lines \(\frac{x - 5}7 = \frac{y + 2}{-5} = \frac z1\) and \(\frac x1 = \frac y2 = \frac z3\) are perpendicular to each other.

Answer:

The equations of the given lines are \(\frac{x - 5}7 = \frac{y + 2}{-5} = \frac z1\) and \(\frac x1 = \frac y2 = \frac z3\).

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively. 

Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂, are perpendicular to each other, 

if a₁a₂ + b₁ b₂ + c₁c₂ = 0 

∴ 7 × 1 + (−5) × 2 + 1 × 3 

= 7 − 10 + 3 

= 0

Therefore, the given lines are perpendicular to each other.

Answer 3

17. Find the shortest distance between the lines

Answer:

The equations of the given lines are

It is known that the shortest distance between the lines, \(\vec r = \vec {a_1} + \lambda \vec {b_1}\) and \(\vec r = \vec {a_2} + \mu\vec {b_2}\), is given by,

Comparing the given equations, we obtain

'

Substituting all the values in equation (1), we obtain

Therefore, the shortest distance between the two lines is \(\frac{3\sqrt2}{2}\) units.

18. Find the shortest distance between the lines

Answer:

The given lines are

It is known that the shortest distance between the two lines,

Comparing the given equations, we obtain

Substituting all the values in equation (1), we obtain

Since distance is always non-negative, the distance between the given lines is \(2\sqrt{29}\) units.

19. Find the shortest distance between the lines whose vector equations are

Answer:

The given lines are

It is known that the shortest distance between the lines, \(\vec r = \vec {a_1} + \lambda \vec {b_1}\) and \(\vec r = \vec {a_2} + \mu\vec {b_2}\), is given by,

For the given equations,

Substituting all the values in equation (3), we obtain

Therefore, the shortest distance between the lines is \(\frac 8{\sqrt {29}}\) units.

20. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

(a) z = 2

(b) x + y + z = 1

(c) 2x + 3y – z = 5

(d) 5y + 8 = 0

Answer:

(a) z = 2

Given:

The equation of the plane, z = 2 or 0x + 0y + z = 2 …. (1)

Direction ratio of the normal (0, 0, 1)

By using the formula,

√[(0)² + (0)² + (1)²] = √1

= 1

Now,

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

0x/(1) + 0y/(1) + z/1 = 2

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

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

∴ The direction cosines are 0, 0, 1

Distance (d) from the origin is 2 units

(b) x + y + z = 1

Given:

The equation of the plane, x + y + z = 1…. (1)

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

By using the formula,

√[(1)² + (1)² + (1)²] = √3

Now,

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

x/(√3) + y/(√3) + z/(√3) = 1/√3

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

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

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

Distance (d) from the origin is 1/√3 units

(c) 2x + 3y – z = 5

Given:

The equation of the plane, 2x + 3y – z = 5…. (1)

Direction ratio of the normal (2, 3, -1)

By using the formula,

√[(2)² + (3)² + (-1)²] = √14

Now,

Divide both the sides of equation (1) by √14, we get

2x/(√14) + 3y/(√14) – z/(√14) = 5/√14

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

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

∴ The direction cosines are 2/√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…. (1)

Direction ratio of the normal (0, -5, 0)

By using the formula,

√[(0)² + (-5)² + (0)²] = √25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) – 5y/(5) – 0z/(5) = 8/5

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

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

∴ The direction cosines are 0, -1, 0

Distance (d) from the origin is 8/5 units

21. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat i + 5\hat j - 6\hat k\).

Answer:

It is known that the equation of the plane with position vector is given by, \(\vec r. \hat n = d\)

This is the vector equation of the required plane.

Answer 4

22. In the following cases, find 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

Answer:

(a) 2x + 3y + 4z – 12 = 0

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

2x + 3y + 4z = 12 …. (1)

Direction ratio are (2, 3, 4)

√[(2)² + (3)² + (4)²] = √(4 + 9 + 16)

= √29

Now,

Divide both the sides of equation (1) by √29, we get

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

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

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

∴ The direction cosines are 2/√29, 3/√29, 4/√29

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

= [(2/√29) (12/√29), (3/√29) (12/√29), (4/√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 …. (1)

Direction ratio are (0, 3, 4)

√[(0)² + (3)² + (4)²] = √(0 + 9 + 16)

= √25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

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

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

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

∴ The direction cosines are 0/5, 3/5, 4/5

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

= [(0/5) (6/5), (3/5) (6/5), (4/5) (6/5)]

= 0, 18/25, 24/25

(c) x + y + z = 1

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

x + y + z = 1 …. (1)

Direction ratio are (1, 1, 1)

√[(1)² + (1)² + (1)²] = √(1 + 1 + 1)

= √3

Now,

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

1x/(√3) + 1y/(√3) + 1z/(√3) = 1/√3

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

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

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

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

= [(1/√3) (1/√3), (1/√3) (1/√3), (1/√3) (1/√3)]

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

(d) 5y + 8 = 0

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

0x – 5y + 0z = 8 …. (1)

Direction ratio are (0, -5, 0)

√[(0)² + (-5)² + (0)²] = √(0 + 25 + 0)

= √25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) – 5y/(5) + 0z/(5) = 8/5

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

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

∴ The direction cosines are 0, -1, 0

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

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

= 0, -8/5, 0

23. Find the equations of the planes that passes through three points. 

(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3) 

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

Answer:

(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3).

Since A, B, C are collinear points, there will be infinite number of planes passing through the given points. 

(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).

Therefore, a plane will pass through the points A, B, and C. 

It is known that the equation of the plane through the points, \((x_1, y_1,z_1) , (x_2, y_2, z_2)\), and \((x_3, y_3,z_3)\), is

This is the Cartesian equation of the required plane.

24. Find the intercepts cut off by the plane 2x + y - z = 5.

Answer:

2x + y - z = 5   .....(1)

Dividing both sides of equation (1) by 5, we obtain

It is known that the equation of a plane in intercept form is \(\frac xa + \frac yb + \frac zc = 1\), where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively. 

Therefore, for the given equation,

a = \(\frac 52\), b = 5 and c = -5

Thus, the intercepts cut off by the plane are \(\frac 52\), 5 and -5.

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

Answer:

The equation of the plane ZOX is y = 0 

Any plane parallel to it is of the form, y = a 

Since the y-intercept of the plane is 3, 

a = 3 

Thus, the equation of the required plane is y = 3

26. Find the equation of the plane through the intersection of the planes 3x - y + 2z - 4 = 0 and x + y + z - 2 = 0 and the point (2, 2, 1).

Answer:

The equation of any plane through the intersection of the planes, 

3x − y + 2z − 4 = 0 and x + y + z − 2 = 0, is

(3x − y + 2z − 4) + α(x + y + z - 2) = 0, where \(\alpha \in R\)   .....(1)

The plane passes through the point (2, 2, 1). 

Therefore, this point will satisfy equation (1).

Substituting \(\alpha = \frac23\) in equation (1), we obtain

This is the required equation of the plane.

Answer 5

27. Find the vector equation of the plane passing through the intersection of the planes \(\vec r.(2\hat i + 2\hat j - 3\hat k) = 7\), \(\vec r.(2\hat i + 5\hat j + 3\hat k) = 9\) and through the point (2, 1, 3).

Answer:

\(\vec r.(2\hat i + 2\hat j - 3\hat k) = 7\) and \(\vec r.(2\hat i + 5\hat j + 3\hat k) = 9\) 

The equations of the planes are

⇒ \(\vec r.(2\hat i + 2\hat j - 3\hat k) -7= 0\)    ....(1)

\(\vec r.(2\hat i + 5\hat j + 3\hat k) - 9=0\) 

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

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

\(\vec r=2\hat i + 2\hat j + 3\hat k\) 

Substituting in equation (3), we obtain

Substituting \(\lambda = \frac{10}9\) in equation (3), we obtain

This is the vector equation of the required plane.

28. Find the equation of the plane 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.

Answer:

The equation of the plane through the intersection of the planes, x + y + z = 1 and 2x + 3y  4z = 5 is

The direction ratios, a₁, b₁, c₁, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1).

The plane in equation (1) is perpendicular to x - y + z = 0

Its direction ratios, a₂, b₂, c₂, are 1, −1, and 1. 

Since the planes are perpendicular,

Substituting \(\lambda = - \frac13\) in equation (1), we obtain

This is the required equation of the plane.

29. Find the angle between the planes whose vector equations are \(\vec r.(2\hat i + 2\hat j - 3\hat k) = 5\)  and \(\vec r.(3\hat i - 3\hat j + 5\hat k) = 3\).

Answer:

The equations of the given planes are \(\vec r.(2\hat i + 2\hat j - 3\hat k) = 5\)  and \(\vec r.(3\hat i - 3\hat j + 5\hat k) = 3\) 

It is known that if \(\vec {n_1}\) and \(\vec {n_2}\) are \(\vec r.\vec {n_1} = d_1\) and \(\vec r.\vec{n_2} = d_2\) 

normal to the planes, , then the angle between them, Q, is given by,

30. 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).

Answer:

Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1). 

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

The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and (−1 + 1) = 0 

OA is perpendicular to BC, if a₁a₂ + b₁b₂ + c₁c₂ = 0 

a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 (−2) + 1 ×0 = 2 − 2 = 0 

Thus, OA is perpendicular to BC.

31. Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b.

Answer:

The angle Q between the lines with direction cosines, a, b, c and b − c, c − a, a − b, is given by,

Thus, the angle between the lines is 90°.

32. Find the equation of a line parallel to x-axis and passing through the origin.

Answer:

The line parallel to x-axis and passing through the origin is x-axis itself. 

Let A be a point on x-axis. 

Therefore, the coordinates of A are given by (a, 0, 0), where a \(\in\) R. 

Direction ratios of OA are (a − 0) = a, 0, 0 The equation of OA is given by,

Thus, the equation of line parallel to x-axis and passing through origin is

\(\frac x1 = \frac y 0 = \frac z0\)

33. 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.

Answer:

The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (−4, 3, −6), and (2, 9, 2) respectively. 

The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4 The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8

It can be seen that, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac12\)

Therefore, AB is parallel to CD. 

Thus, the angle between AB and CD is either 0° or 180°.

34. If the lines \(\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z-3}2\) and \(\frac{x- 1}{3k} = \frac{y - 1}1 = \frac{z - 6}{-5}\) are perpendicular, find the value of k.

Answer:

The direction of ratios of the lines, \(\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z-3}2\) and \(\frac{x- 1}{3k} = \frac{y - 1}1 = \frac{z - 6}{-5}\) , are −3, 2k, 2 and 3k, 1, −5 respectively.

It is known that two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂, are perpendicular, if a₁a₂ + b₁b₂ + c₁c₂ = 0

Therefore, for \(k = -\frac{10}7\), the given lines are perpendicular to each other.

Answer 6

35. Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane \(\vec r.(\hat i + 2\hat j - 5\hat k) + 9 = 0\) 

Answer:

The position vector of the point (1, 2, 3) is \(\vec{r_1} = \hat i + 2\hat j + 3\hat k\) 

The direction ratios of the normal to the plane, \(\vec r.(\hat i + 2\hat j - 5\hat k) + 9 = 0\) are 1, 2, and −5 and the normal vector is \(\vec N = \hat i + 2\hat j - 5\hat k\)

The equation of a line passing through a point and perpendicular to the given plane is given by, \(\vec l = \vec r + \lambda \vec N, \lambda \in R\)

⇒ \(\vec l = (\hat i + 2\hat j + 3\hat k) + \lambda (\hat i + 2\hat j - 5\hat k)\) 

36. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.

Answer:

It is known that the equation of the line passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), is

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k). 

The equation of YZ-plane is x = 0 

Since the line passes through YZ-plane, 

5 − 2k = 0

Therefore, the required point is \(\left(0, \frac{17}2, \frac{-13}2\right)\).

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

Answer:

It is known that the equation of the line passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), is 

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k). 

Since the line passes through ZX-plane,

Therefore, the required point is \(\left(\frac{17}{3}, 0, \frac{23}{3}\right)\).

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

Answer:

It is known that the equation of the line through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), is

Since the line passes through the points, (3, −4, −5) and (2, −3, 1), its equation is given by,

Therefore, any point on the line is of the form (3 − k, k − 4, 6k − 5). 

This point lies on the plane, 2x + y + z = 7 

∴ 2 (3 − k) + (k − 4) + (6k − 5) = 7 

⇒ 5k - 3 = 7

⇒ k = 2

Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5) i.e., (1, −2, 7).

39. 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.

Answer:

The equation of the plane passing through the point (−1, 3, 2) is a (x + 1) + b (y − 3) + c (z − 2) = 0 … (1) 

where, a, b, c are the direction ratios of normal to the plane. 

It is known that two planes, \(a_1x + b_1y + c_1z + d_1 = 0\) and \(a_2x + b_2y + c_2z + d_2 = 0\) are perpendicular, if \(a_1a_2 + b_1b_2 + c_1c_2 = 0\)

Plane (1) is perpendicular to the plane, x + 2y + 3z = 5

Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0

From equations (2) and (3), we obtain

Substituting the values of a, b, and c in equation (1), we obtain

This is the required equation of the plane.

40. Find the equation of the plane passing through the line of intersection of the planes \(\vec r.(\hat i + \hat j + \hat k) = 1 \) and \(\vec r. (2\hat i + 3\hat j - \hat k) + 4 = 0\) and parallel to x-axis.

Answer:

The given planes are

The equation of any plane passing through the line of intersection of these planes is

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis. 

The direction ratios of x-axis are 1, 0, and 0.

Substituting \(\lambda =-\frac12\) in equation (1), we obtain

Therefore, its Cartesian equation is y − 3z + 6 = 0 

This is the equation of the required plane.

41. 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.

Answer:

The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively. 

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3 

It is known that the equation of the plane passing through the point (x₁, y₁ z₁) is \(a(x- x_1) + b(y - y_1) + c(z - z_1) = 0\) where, a, b, and c are the direction ratios of normal. 

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3). 

Thus, the equation of the required plane is

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

⇒ x + 2y - 3z - 14 = 0

Answer 7

42. Find the equation of the plane which contains the line of intersection of the planes

and which is perpendicular to the plane.

Answer:

The equations of the given planes are

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

The plane in equation (3) is perpendicular to the plane,

\(\vec r.(5\hat i + 3\hat j - 6\hat k) + 8 = 0\)

Substituting \(\lambda = \frac7{19}\) in equation (3), we obtain

This is the vector equation of the required plane. 

The Cartesian equation of this plane can be obtained by substituting \(\vec r = x\hat i + y\hat j + z \hat k\) in equation (3).

43. Find the distance of the point (−1, −5, −10) from the point of intersection of the line \(\vec r = 2\hat i - \hat j + 2\hat k + \lambda(3\hat i + 4\hat j + 2\hat k)\) and the plane \(\vec r.(\hat i - \hat j + \hat k) = 5\).

Answer:

The equation of the given line is

\(\vec r = 2\hat i - \hat j + 2\hat k + \lambda(3\hat i + 4\hat j + 2\hat k)\)     ...(1)

The equation of the given plane is

\(\vec r.(\hat i - \hat j + \hat k) = 5\)     .....(2)

Substituting the value of \(\vec r\) from equation (1) in equation (2), we obtain

Substituting this value in equation (1), we obtain the equation of the line as

\(\vec r = 2\hat i - \hat j + 2\hat k\)

This means that the position vector of the point of intersection of the line and the plane is

\(\vec r = 2\hat i - \hat j + 2\hat k\)

This shows that the point of intersection of the given line and plane is given by the coordinates, (2, −1, 2). 

The point is (−1, −5, −10). 

The distance d between the points, (2, −1, 2) and (−1, −5, −10), is

44. Prove that if a plane has the intercepts a, b, c and is at a distance of P units from the origin, then \(\frac1{a^2} + \frac1{b^2} + \frac1{c^2} = \frac1{p^2}\).

Answer:

The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

\(\frac xa + \frac yb + \frac z c = 1\)   .....(1)

The distance (p) of the plane from the origin is given by,

45. 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) \(\frac2{\sqrt{29}} \) units

Answer:

The equations of the planes are

It can be seen that the given planes are parallel. It is known that the distance between two parallel planes, ax + by + cz = d₁ and ax + by + cz = d₂, is given by,

Thus, the distance between the lines is \(\frac2{\sqrt{29}}\) units. 

Hence, the correct answer is D.

46. The planes: 2x − y + 4z = 5 and 5x − 2.5y + 10z = 6 are \(\left(0, 0, \frac54\right)\)

(A) Perpendicular 

(B) Parallel 

(C) intersect y-axis 

(D) passes through

Answer:

The equations of the planes are 

2x − y + 4z = 5 …(1) 

5x − 2.5y + 10z = 6 …(2)

It can be seen that,

Therefore, the given planes are parallel. 

Hence, the correct answer is B.