Three Dimensional Geometry Class 12th Mathematics Part Ii CBSE Solution

Class 12^th Mathematics Part Ii CBSE Solution

Exercise 11.1

1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively,…
2. Find the direction cosines of a line which makes equal angles with the…
3. If a line has the direction ratios -18, 12, -4, then what are its direction…
4. Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.…
5. Find the direction cosines of the sides of the triangle whose vertices are (3,…

Exercise 11.2

1. Show that the three lines with direction cosines 12/13 , -3/13 , -4/13 4/13 ,…
2. Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular…
3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the…
4. Find the equation of the line which passes through the point (1, 2, 3) and is…
5. Find the equation of the line in vector and in cartesian form that passes…
6. Find the cartesian equation of the line which passes through the point (-2, 4,…
7. The Cartesian equation of a line is x-5/3 = y+4/7 = z-6/2 . Write its vector…
8. Find the vector and the cartesian equations of the lines that passes through…
9. . Find the vector and the cartesian equations of the line that passes through…
10. Find the angle between the following pairs of lines:
11. . Find the angle between the following pair of lines: x-2/2 = y-1/5 - z+3/-3…
12. Find the values of p so that the lines 1-x/3 = 7y-14/2p = z-3/2 7-7x/3p =…
13. Show that the lines x-5/7 = y+2/-5 = z/1 x/1 = y/2 = z/3 are perpendicular to…
14. vector r = (i+2 j + k) + lambda (i - j + k) vector r = 2 i - j - k + μ (2 i +…
15. x+1/7 = y+1/-6 = z+1/1 x-3/1 = y-5/-2 = z-7/1 Find the shortest distance…
16. Find the shortest distance between the lines whose vector equations are…
17. vector r = (1-t) i + (t-2) j + (3-2t) k vector r = (s+1) i + (2s-1) j - (2s+1)…

Exercise 11.3

1. z = 2 In each of the following cases, determine the direction cosines of the…
2. x + y + z = 1 In each of the following cases, determine the direction cosines…
3. 2x + 3y - z = 5 In each of the following cases, determine the direction…
4. 5y + 8 = 0 In each of the following cases, determine the direction cosines of…
5. Find the vector equation of a plane which is at a distance of 7 units from the…
6. vector t (i + j - k) = 2 Find the Cartesian equation of the following planes:…
7. Find the Cartesian equation of the following planes:
8. Find the Cartesian equation of the following planes:
9. 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
10. 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
11. 3y + 4z - 6 = 0 In the following cases, find the coordinates of the foot of…
12. x + y + z = 1 In the following cases, find the coordinates of the foot of the…
13. that passes through the point (1, 0, -2) and the normal to the plane is i + j…
14. that passes through the point (1,4, 6) and the normal vector to the plane is…
15. (1, 1, -1), (6, 4, -5), (-4, -2, 3) Find the equations of the planes that…
16. (1, 1, 0), (1, 2, 1), (-2, 2, -1) Find the equations of the planes that passes…
17. Find the intercepts cut off by the plane 2x + y - z = 5.
18. Find the equation of the plane with intercept 3 on the y-axis and parallel to…
19. Find the equation of the plane through the intersection of the planes 3x - y +…
20. Find the vector equation of the plane passing through the intersection of the…
21. Find the equation of the plane through the line of intersection of the planes…
22. Find the angle between the planes whose vector equations are
23. 7x + 5y + 6z + 30 = 0 and 3x - y - 10z + 4 = 0 In the following cases,…
24. 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0 In the following cases, determine…
25. 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
26. 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
27. 4x + 8y + z - 8 = 0 and y + z - 4 = 0 In the following cases, determine…
28. Point Plane (0, 0, 0) 3x - 4y + 12 z = 3 In the following cases, find the…
29. Point Plane (3, - 2, 1) 2x - y + 2z + 3 = 0 In the following cases, find the…
30. Point Plane (2, 3, - 5) x + 2y - 2z = 9 In the following cases, find the…
31. Point Plane (-6, 0, 0) 2x - 3y + 6z - 2 = 0 In the following cases, find the…

Miscellaneous Exercise

1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular…
2. If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually…
3. Find the angle between the lines whose direction ratios are a, b, c and b - c,…
4. Find the equation of a line parallel to x - axis and passing through the…
5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3,…
6. If the lines x-1/3k = y-2/1 = z-3/-5 and x-1/3k = y-2/1 = z-3/-5 are…
7. Find the vector equation of the line passing through (1, 2, 3) and…
8. Find the equation of the plane passing through (a, b, c) and parallel to the…
9. Find the shortest distance between lines vector r = (6 i+2 j+2 k) + lambda (1…
10. Find the coordinates of the point where the line through (5, 1, 6) and (3,…
11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,…
12. Find the coordinates of the point where the line through (3, -4, -5) and (2,…
13. Find the equation of the plane passing through the point (-1, 3, 2) and…
14. If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane vector r…
15. Find the equation of the plane passing through the line of intersection of the…
16. If O be the origin and the coordinates of P be (1, 2, -3), then find the…
17. Find the equation of the plane which contains the line of intersection of the…
18. Find the distance of the point (-1, -5, -10) from the point of intersection of…
19. Find the vector equation of the line passing through (1, 2, 3) and parallel to…
20. Find the vector equation of the line passing through the point (1, 2, - 4) and…
21. Prove that if a plane has the intercepts a, b, c and is at a distance of p…
22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 isA. 2…
23. The planes: 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6 areA. Perpendicular B.…

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Exercise 11.1

Question 1.

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

Answer:

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here α = 90°, β = 135° and γ = 45°

So direction cosines are

l = cos 90° = 0

m = cos 135°= cos (180° - 45°) = -cos 45° = 

n = cos 45° = 

⇒ Direction cosines of the line 

Question 2.

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

Answer:

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here given α = β = γ (line makes equal angles with the coordinate axes) ……….1

Direction Cosines are

⇒ l = cos α, m = cos β and n = cos γ

We have

l² + m ² + n² = 1

cos² α + cos²β + cos²γ = 1

From 1 we have

cos² α + cos² α + cos² α = 1

3 cos² α = 1

The direction cosines are

Question 3.

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

Answer:

If the direction ratios of the line are a, b and c

Then the direction cosines are 

Given direction ratios are – 18, 12 and – 4

⇒ a= -18, b = 12 and c= -4

= 

= 

Direction cosines are

Question 4.

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

Answer:

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

Given A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7)

Direction ratio of line joining A (2,3,4) and B (−1, −2, 1), are

(−1−2), (−2−3), (1−4)

= (−3, −5, −3)

So a₁ = -3, b₁ = -5, c₁ = -3

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

(5− (−1)), (8−(−2)), (7−1)

= (6, 10, 6)

So a₂ = 6, b₂ = 10 and c₂ =6

It is clear that the direction ratios of AB and BC are of same proportions

As

and

Therefore A, B, C are collinear.

Question 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 direction cosines of the two points passing through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is given by

(x₂ – x₁), (y₂-y₁), (z₂-z₁)

And the direction cosines of the line AB is 

Where AB = 

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Exercise 11.2

Question 1.

Show that the three lines with direction cosines  are mutually perpendicular.

Answer:

We know that

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l₁l₂ + m₁m₂ + n₁n₂|

If two lines are perpendicular, then the angle between the two is θ = 90°

⇒ For perpendicular lines, | l₁l₂ + m₁m₂ + n₁n₂ | = cos 90° = 0, i.e.

| l₁l₂ + m₁m₂ + n₁n₂ | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l₁l₂ + m₁m₂ + n₁n₂ | for all the pairs of the three lines.

Now let the direction cosines of L₁, L₂ and L₃ be l₁, m₁, n₁; l₂, m₂, n₂ and l₃, m₃, n₃.

First, consider

⇒ 

⇒ 

⇒ L₁⊥ L₂ ……(i)

Next, consider

⇒ 

⇒ 

⇒ L₂⊥ L₃ …(ii)

Now, consider

⇒ 

⇒ 

⇒ L₁⊥ L₃ …(iii)

∴ By (i), (ii) and (iii), we have

L₁, L₂ and L₃ are mutually perpendicular.

Question 2.

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

Answer:

We know that

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular if the angle between them is θ = 90°, i.e. a₁a₂ + b₁b₂ + c₁c₂ = 0

Also, we know that the direction ratios of the line segment joining (x₁, y₁, z₁) and (x₂, y₂, z₂) is taken as x₂ – x₁, y₂ – y₁, z₂ – z₁ (or x₁ – x₂, y₁ – y₂, z₁ – z₂).

⇒ The direction ratios of the line through the points (1, –1, 2) and (3, 4, –2) is:

a₁ = 3 – 1 = 2, b₁ = 4 – (-1) = 4 + 1 = 5, c₁ = -2 –2 = -4

and the direction ratios of the line through the points (0, 3, 2) and (3, 5, 6) is:
a₂ = 3 – 0 = 3, b₂ = 5 – 3 = 2, c₂ = 6 – 2 = 4

Now, consider

a₁a₂ + b₁b₂ + c₁c₂ = 2 × 3 + 5 × 2 + (-4) × 4 = 6 + 10 + (-16) = 16 + (-16) = 0

⇒ 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).

Question 3.

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

Answer:

We know that

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are parallel if the angle between them is θ = 0°, i. e.

⇒ 

Also, we know that the direction ratios of the line segment joining (x₁, y₁, z₁) and (x₂, y₂, z₂) is taken as x₂ – x₁, y₂ – y₁, z₂ – z₁ (or x₁ – x₂, y₁ – y₂, z₁ – z₂).

⇒ The direction ratios of the line through the points (4, 7, 8) and (2, 3, 4) is:

a₁ = 2 – 4 = -2, b₁ = 3 – 7 = -4, c₁ = 4 – 8 = -4

And the direction ratios of the line through the points (– 1, – 2, 1) and (1, 2, 5) is:

a₂ = 1 – (-1) = 1 + 1 = 2, b₂ = 2 – (-2) = 2 + 2 = 4, c₂ = 5 – 1 = 4

Consider 

⇒ 

∴ 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).

Question 4.

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 

Answer:

We know that

Vector equation of a line that passes through a given point whose position vector is  and parallel to a given vector  is .

So, here the position vector of the point (1, 2, 3) is given by  and the parallel vector is .

∴ The vector equation of the required line is:

, where  is a real number.

Question 5.

Find the equation of the line in vector and in cartesian form that passes through the point with position vector  and is in the direction 

Answer:

We know that

Vector equation of a line that passes through a given point whose position vector is  and parallel to a given vector  is .

Here,  and 

⇒ The vector equation of the required line is:

⇒ 

Also, we know that

The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction cosines l, m, n is .

Also, we know that if the direction ratios of the line are a, b, c, then

⇒ 

⇒ The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction ratios a, b, c is .

Here, x₁ = 2, y₁ = -1, z₁ = 4 and a = 1, b = 2, c = -1

⇒ The Cartesian equation of the required line is:

⇒ 

Question 6.

Find the cartesian equation of the line which passes through the point (–2, 4, –5) and parallel to the line given by 

Answer:

We know that

The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction ratios a, b, c is .

Here, The point (x₁, y₁, z₁) is (-2, 4, -5) and the direction ratios are:

a = 3, b = 5, c = 6

⇒ The Cartesian equation of the required line is:

⇒ 

Question 7.

The Cartesian equation of a line is . Write its vector form.

Answer:

We know that

The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction cosines l, m, n is .

Comparing this standard form with the given equation, we get

x₁ = 5, y₁ = -4, z₁ = 6 and l = 3, m = 7, n = 2

⇒ The point through which the line passes has the position vector  and the vector parallel to the line is given by .

Now, ∵ Vector equation of a line that passes through a given point whose position vector is  and parallel to a given vector  is .

∴ The vector equation of the required line is:

⇒ 

Question 8.

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

Answer:

We know that

The vector equation of as line which passes through two points whose position vectors are  and  is .

Here, the position vectors of the two points (0, 0, 0) and (5, -2, 3) are  and , respectively.

So, The vector equation of the required line is:

⇒ 

⇒ 

Now, we know that

Cartesian equation of a line that passes through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is 

So, the Cartesian equation of the line that passes through the origin (0, 0, 0) and (5, – 2, 3) is 

Question 9.

. Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

Answer:

We know that

The vector equation of as line which passes through two points whose position vectors are  and  is .

Here, the position vectors of the two points (3, –2, –5) and (3, –2, 6) are  and , respectively.

So, the vector equation of the required line is:

⇒ 

⇒ 

⇒ 

Now, we also know that

Cartesian equation of a line that passes through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is 

So, the Cartesian equation of the line that passes through the origin (3, -2, -5) and (3, -2, 6) is 

Question 10.

Find the angle between the following pairs of lines:

Answer:

We know that

If θ is the acute angle between  and , then

 ……(i)

(i)  and 

Here  and 

So, from (i), we have

 ……(ii)

⇒ 

⇒ 

And 

Now, 

⇒ 

⇒ By (ii), we have

⇒ 

⇒ 

(ii)  and 

Here,  and 

So, from (i), we have

 …(iii)

⇒ 

⇒ 

And 

Now, 

⇒ 

⇒ By (iii), we have

⇒ 

⇒ 

Question 11.

. Find the angle between the following pair of lines:

Answer:

We know that

If and  are the equations of two lines, then the acute angle between the two lines is given by

cos θ = | l₁l₂ + m₁m₂ + n₁n₂ | ……(i)

(i)  and 

Here, a₁ = 2, b₁ = 5, c₁ = -3 and a₂ = -1, b₂ = 8, c₂ = 4

Now,  ……(ii)

Here, 

And 

So, from (ii), we have

⇒ 

And 

∴ From (i), we have

⇒ 

⇒ 

(ii)  and 

Here, a₁ = 2, b₁ = 2, c₁ = 1 and a₂ = 4, b₂ = 1, c₂ = 8

Here, 

And 

So, from (ii), we have

⇒ 

And 

∴ From (i), we have

⇒ 

⇒ 

Question 12.

Find the values of p so that the lines  are at right angles.

Answer:

For any two lines to be at right angles, the angle between them should be θ = 90°.

⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0, where a₁, b₁, c₁ and a₂, b₂, c₂ are the direction ratios of two lines.

The standard form of a pair of Cartesian lines is:

 and  …(i)

Now, first we rewrite the given equations according to the standard form, i.e.

 and , i.e.

 and  …(ii)

Now, comparing (i) and (ii), we get

 and 

Now, as both the lines are at right angles,

so a₁a₂ + b₁b₂ + c₁c₂ = 0

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 11p = 70

⇒ 

Question 13.

Show that the lines  are perpendicular to each other.

Answer:

We know that

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular if the angle between them is θ = 90°, i.e. a₁a₂ + b₁b₂ + c₁c₂ = 0

Also, here the direction ratios are:

a₁ = 7, b₁ = -5, c₁ = 1 and a₂ = 1, b₂ = 2, c₂ = 3

Now, Consider

a₁a₂ + b₁b₂ + c₁c₂ = 7 × 1 + (-5) × 2 + 1 × 3 = 7 -10 + 3 = - 3 + 3 = 0

∴ The two lines are perpendicular to each other.

Question 14.

Find the shortest distance between the lines

Answer:

We know that

Shortest distance between two lines  and  is

 …(i)

Here, ,  and

, 

Now, 

 …(ii)

Now, 

⇒ 

 ……….(iii)

 ……….(iv)

Now, 

……….(v)

Now, using (i), we have

The shortest distance between the two lines, d  [From (iv) and (v)]

⇒ 

Rationalizing the fraction by multiplying the numerator and denominator by √2,

⇒ 

Question 15.

Find the shortest distance between the lines

Answer:

We know that

_{Shortest distance between the lines:}

 and  is

 …(i)

The standard form of a pair of Cartesian lines is:

 and 

And the given equations are:  and 

_{Comparing the given equations with the standard form, we get}

_x1 = -1, y₁ = -1, z₁ = -1; x₂ = 3, y₂ = 5, z₂ = 7

_a1 = 7, b₁ = -6, c₁ = 1; a₂ = 1, b₂ = -2, c₂ = 1

_{Now, consider}

_⇒

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Next, consider

⇒ 

⇒ 

⇒ 

⇒ 

_{⇒From (i), we have}

⇒ 

Question 16.

Find the shortest distance between the lines whose vector equations are

Answer:

We know that

Shortest distance between two lines  and  is

 ……….(i)

Here, ,  and

, 

Now, 

 ……….(ii)

Now, 

⇒ 

 ……….(iii)

 ……….(iv)

Now, 

 ……….(v)

Now, using (i), we have

The shortest distance between the two lines, 

Question 17.

Find the shortest distance between the lines whose vector equations are

Answer:

Firstly, consider

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

So, we need to find the shortest distance between  and .

Now, We know that

Shortest distance between two lines  and  is

 …(i)

Here,  and

⇒ 

Now, 

 …(ii)

Now, 

⇒ 

 ……….(iii)

 ……….(iv)

Now, 

 ……….(v)

Now, using (i), we have

The shortest distance between the two lines, 

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Exercise 11.3

Question 1.

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

z = 2

Answer:

The eq. of the plane

z = 2

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

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosines = 0,0,1

Distance(d) = 2

Question 2.

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

x + y + z = 1

Answer:

The eq. of the plane

x + y + z = 1

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

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Question 3.

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

2x + 3y – z = 5

Answer:

The eq. of the plane

2x + 3y-z = 5

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

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Question 4.

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

5y + 8 = 0

Answer:

The eq. of the plane

0x-5y + 0z = 8

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

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosine = 0, -1, 0

Question 5.

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

Answer:

Vector eq. of the plane with position vector  is

Question 6.

Find the Cartesian equation of the following planes:

Answer:

(Letbe the position vector of P(x,y,z)

Hence,

So, Cartesian eq. is

x + y - z = 2

Question 7.

Find the Cartesian equation of the following planes:

Answer:

(Letbe the position vector of P(x,y,z)

.

.

So, Cartesian eq. is

2x + 3y - 4z = 1

Question 8.

Find the Cartesian equation of the following planes:

Answer:

Letbe the position vector of P(x,y,z)

Hence, 

So, Cartesian eq. is

(s-2t)x + (3-t)y + (2s + t)z=15

Question 9.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

Answer:

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

2x + 3y + 4z = 12

Direction ratio (2,3,4)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Question 10.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

Answer:

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

0x + 3y + 4z = 6

Direction ratio (0,3,4)

.

= 5

his is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd)

Question 11.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z – 6 = 0

Answer:

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

x + y + z = 1

Direction ratio (1,1,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Question 12.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1

Answer:

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

0x – 5y + 0z = 8

Direction ratio (0,-5,0)

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd) 

Question 13.

Find the vector and cartesian equations of the planes

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

Answer:

Let the position vector of the point

Normal⊥to the plane

Vector eq. of the plane,

x – 1 + y – z – 2 = 0

x + y – z – 3 = 0

Required Cartesian eq. of the plane

x + y – z = 3

Question 14.

Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is 

Answer:

Let the position vector of the point

Vector eq. of the plane,

x – 1 – 2y + 8 + z – 6 = 0

x – 2y + z + 1 = 0

Required Cartesian eq. of the plane

x – 2y + z = - 1

Question 15.

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

(1, 1, –1), (6, 4, –5), (–4, –2, 3)

Answer:

The given points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).

Let,

= 1(12 - 10) – 1(18 - 20) -1 (-12 + 16)

= 2 + 2 – 4

= 0

Since, the value of determinant is 0.

Therefore, these points are collinear as there will be infinite planes passing through the given 3 points.

Question 16.

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

(1, 1, 0), (1, 2, 1), (–2, 2, –1)

Answer:

The given points are (1, 1, 0), (1, 2, 1), (-2, 2, -1).

Let,

= 1(-2 - 2) – 1(-1 + 2)

= -4 – 1

= -5 ≠ 0

There passes a unique plane from the given 3 points.

Equation of the plane passes through the points, (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃), i.e.,

⇒ (x - 1)(-2) – (y - 1)(3) + 3z = 0

⇒ -2x + 2 – 3y + 3 + 3z = 0

⇒ 2x + 3y – 3z = 5

This is the required eq. of the plane.

Question 17.

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

Answer:

We know that, the eq. of the plane in intercept form

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

⇒ 2x + y – z = 5 (i)

Dividing both side of (i)eq. by 5, we get

Thus, the intercepts cut-off by the plane are 5/2, 5 and -5.

Question 18.

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

Answer:

We know that the eq. of the plane ZOX is

y = 0

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

Hence, the required eq. of the plane is

y = 3

Question 19.

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:

Eq. of the plane passes through the intersection of the plane is given by

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

∵ Plane passes through the points (2,2,1)

(3 × 2 – 2 + 2 × 1 – 4) + λ(2 + 2 + 1 – 2) = 0

2 + 3λ = 0

3λ = -2

 (i)

Hence, the required eq. of the plane

7x – 5y + 4z – 8 = 0

This is the required eq. of the plane.

Question 20.

Find the vector equation of the plane passing through the intersection of the planes  and through the point (2, 1, 3).

Answer:

Let the vector eq. of the plane passing through the intersection of the planes

Here, 

∵ Plane passes through points (2,1,3)

4 + 4λ + 2 + 5λ - 9 + 9λ - 7 - 9λ = 0

9 λ = 10

This is the required vector eq. of the plane.

Question 21.

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:

Let the eq. of the plane that passes through the two-given plane x + y + z = 1 and 2x + 3y + 4z = 5 is

(x + y + z – 1) + λ(2x + 3y + 4z – 5) = 0

(2λ + 1)x + (3λ + 1)y + (4λ + 1)z -1 - 5λ = 0 (i)

Direction ratio of the plane (2λ + 1, 3λ + 1, 4λ + 1)

and Direction ratio of another plane (1, -1, 1)

∵ Both are ⊥ hence

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

2λ + 1 - 3λ - 1 + 4λ + 1 = 0

Put the value of λ in (i)eq., we get

x – z + 2 = 0

This is the required eq. of the plane.

Question 22.

Find the angle between the planes whose vector equations are 

Answer:

The eq. of the given planes are

If n₁ and n₂ are normal to the planes, 

Angle between two planes

Question 23.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

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

Answer:

The eq. of the given planes are

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

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

a₁a₂ + b₁b₂ + c₁c₂ = 0

21 – 5 – 60

-44 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

∴ Both the planes are not || to each other.

The angle between them is given by

Question 24.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

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

Answer:

The eq. of the given planes are

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

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

a₁a₂ + b₁b₂ + c₁c₂ = 0

2 × 1 + 1 × (-2) + 3 × 0

= 0

Thus, the given planes are ⊥ to each other.

Question 25.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Answer:

The eq. of the given planes are

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

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

a₁a₂ + b₁b₂ + c₁c₂ = 0

6 + 6 + 24

36 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

Thus, the given planes are || to each other.

Question 26.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Answer:

The eq. of the given planes are

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

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

a₁a₂ + b₁b₂ + c₁c₂ = 0

2 × 2 + (-1) × (-1) + 3 × 3

14 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

Thus, the given planes are || to each other.

Question 27.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

4x + 8y + z – 8 = 0 and y + z – 4 = 0

Answer:

The eq. of the given planes are

4x + 8y + z – 8 = 0 and y + z – 4 = 0

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

a₁a₂ + b₁b₂ + c₁c₂ = 0

0 + 8 + 1

9 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

∴ Both the planes are not || to each other.

The angle between them is given by

= 45⁰

Question 28.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(0, 0, 0) 3x – 4y + 12 z = 3

Answer:

Distance of point P(x₁,y₁,z₁) from the plane Ax + By + Cz – D = 0 is

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

Question 29.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(3, – 2, 1) 2x – y + 2z + 3 = 0

Answer:

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

Question 30.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(2, 3, – 5) x + 2y – 2z = 9

Answer:

Given point is (2,3,-5) and the plane is x + 2y – 2z = 9

= 3

Question 31.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(–6, 0, 0) 2x – 3y + 6z – 2 = 0

Answer:

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

= 2

---

Miscellaneous Exercise

Question 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).

Answer:

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

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

Direction ratios of OA = (a₁, b₁, c₁) ≡ [(2 - 0), (1 - 0), (1 - 0)] ≡ (2,1,1)

Direction ratios of BC = (a₂, b₂, c₂) ≡ [(4 - 3), (3 - 5), (-1 + 1)]

≡ (1, -2, 0)

Given-

OA is ⊥ to BC

∴ we have to prove that -

a₁a₂ + b₁b₂ + c₁c₂ = 0

L.H.S = a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 × (−2) + 1 × 0 = 2 - 2 = 0

R.H.S = 0

Thus, L.H.S = R.H.S ….PROVED

Hence OA is ⊥ to BC.

Question 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)

Answer:

Let 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)

On solving (1) and (2) by 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

where .

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. Therefore,

l₁l₂ + m₁m₂ + n₁n₂ = 0

Also, we have

l₁² + m₁² + n₁² = 1

and, l₂² + m₂² + n₂² = 1

Putting these values in (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)

Question 3.

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

Answer:

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

So,

= 0

∴ cosθ = 0

So, θ = 90°

Hence, Angle between the given pair of Lines is 90°.

Question 4.

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

Answer:

Equation of a line passing through (x₁, y₁, z₁) and parallel to a line with direction ratios a,b,c is

Since 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 –

Question 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.

Answer:

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

A line passing through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) has direction ratios (x₁ - x₂), (y₁ - y₂), (z₁ - z₂)

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

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,

 ∴ cosθ = 1

So, θ = 0°

Hence, Angle between the lines AB and CD is 0°.

Question 6.

If the lines  and  are perpendicular, find the value of k.

Answer:

Two lines  and 

are perpendicular to each other if

a₁a₂ + b₁b₂ + c₁c₂ = 0

Given -

comparing with

we get -

x₁ =1, y₁ = 2, z₁ = 3

& a₁ = - 3, b₁ = 2k, c₁ = 2

Similarly,

comparing with

we get -

x₂ = 1, y₂ = 2, z₂ = 3

& 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

Hence, the value of k is -10/7.

Question 7.

Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane .

Answer:

The vector equation of a line passing through a point with position vector  and parallel to vector  is

⇒ 

Given, the line passes through (1,2,3)

So, 

Finding normal of plane

Comparing with ,

Since line is perpendicular to plane, the line will be parallel of the plane

∴ 

Hence,

This is the required vector equation of line.

Question 8.

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

Answer:

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

A(x - x₁) + B(y - y₁) + C(z - z₁) = 0

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

= Direction ratios of normal of 

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

∴ A = 1, B =1, C = 1

Thus,

Equation of plane in cartesian form is

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

Thus, x + y + z = a + b + c is the required equation of plane.

Question 9.

Find the shortest distance between lines

 and .

Answer:

Shortest distance between lines with vector equations

 and  is

⇒ 

Given -

Comparing with  , we get -

 & 

Similarly,

Comparing with , we get -

 & 

Now,

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Magnitude of  = 

= √144

= 12

Also,

⇒ 

= -80 + (-16) + (-12)

= -108

Shortest Distance  = 9

Hence, the shortest distance between the given two lines is 9.

Question 10.

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

Answer:

The vector equation of a line passing through two points with position vectors  &  is

⇒ 

The position vector of point A(5,1,6) is given as -

⇒ 

The position vector of point B(3,4,1) is given as -

⇒ 

⇒ 

∴ ...(1)

Let the coordinates of the point where the line crosses the YZ plane be (0,y,z)

So,  …(2)

Since point lies in line, it will satisfy its equation,

Putting (2) in (1)

⇒ 

Two vectors are equal if their corresponding components are equal

So,

0 = 5 - 2λ…(3)

y = 1 + 3λ…(4)

and, z = 6 - 5λ…(5)

From equation (3), we get -

λ = 5/2

Substitute the value of λ in equation (4) and (5), 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

Therefore, the coordinates of the required point is

(0, 17/2, -13/2).

Question 11.

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

Answer:

The vector equation of a line passing through two points with position vectors  &  is

⇒ 

The position vector of point A(5,1,6) is given as -

.

The position vector of point B(3,4,1) is given as -

⇒ 

⇒ 

∴  ...(1)

Let the coordinates of the point where the line crosses the ZX plane be (0,y)

So,  …(2)

Since point lies in line, it will satisfy its equation,

Putting (2) in (1)

Two vectors are equal if their corresponding components are equal

So,

x = 5 - 2λ …(3)

0 = 1 + 3λ …(4)

and, z = 6 - 5λ …(5)

From equation (4), we get -

λ = -1/3

Substitute the value of λ in equation (3) and (5), 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

Therefore, the coordinates of the required point is

(17/3, 0, 23/3).

Question 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.

Answer:

The equation of a line passing through two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is

⇒ 

Given the line passes through the points A(3, –4, –5) and

B(2, –3, 1)

∴ x₁ = 3, y₁ = -4, z₁ = -5

and, x₂ = 2, y₂ = -3, z₂ = 1

So, the equation of line is

⇒ 

⇒ 

So,

x = -k + 3 | y = k - 4 | z = 6k - 5 …(1)

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

Putting the value of x,y,z from (1) in the equation of plane,

2x + y + z + 7 = 0

⇒ 2(-k + 3) + (k - 4) + (6k - 5) = 7

⇒ 5k - 3 = 7

⇒ 5k = 10

∴ k = 2

Putting the value of k in x, y, z

x = - k + 3 = - 2 + 3 = 1

y = k - 4 = 2 - 4 = - 2

z = 6k - 5 = 12 - 5 = 7

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

Question 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.

Answer:

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.

Now the plane passes through (-1,3,2)

So, equation of plane is

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 normals of both planes.

we know that -

 is perpendicular to both  & 

So, required normal is cross product of normals of planes

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

Required Normal 

Hence, direction ratios = -7, 8, -3

∴ A = -7, B = 8, C = -3

Putting above values in (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

Therefore, equation of the required plane is 7x - 8y + 3z + 25 = 0.

Question 14.

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

Answer:

The distance of a point with position vector  from the plane  is .

The position vector of point (1,1,p) is given as -

⇒ 

The position vector of point (-3,0,1) is given as -

⇒ 

It is given that the points (1,1,p) and (-3,0,1) are equidistant from the plane 

∴ 

⇒ 20 - 12p =  8

⇒ 20 - 12p = 8 or, 20 - 12p = -8

⇒ 12p = 12 or, 12p = 28

∴ p = 1 or, p = 7/3

us, the possible values of p are 1 and 7/3.

Question 15.

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

Answer:

The equation of any plane through the line of intersection of the planes  and  is given by -

.

So, the equation of any plane through the line of intersection of the given planes is

.

.

∴ . …(1)

Since this plane is parallel to x-axis.

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

Direction ratios of Normal (a_1, b_1, c₁)≡ [(1 - 2λ), (1 - 3λ), (1 +)]

Direction ratios of x–axis (a_2, b_2, 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

Putting the value of λ in (1), we get -

⇒ 

⇒ 

Hence, the equation of the required plane is 

Question 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.

Answer:

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

A(x - x₁) + B(y - y₁) + C(z - z₁) = 0

The plane passes through P(1,2,3)

So, x₁ = 1, y₁ = 2, z₁ = - 3

Normal vector to plane = 

where O(0,0,0), P(1,2, - 3)

Direction ratios of  = (1 - 0), (2 - 0), (-3 - 0)

= (1,2, - 3)

∴ A = 1, B = 2, C = -3

Equation of plane in cartesian form is

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

⇒ x - 1 + 2y - 4 - 3z - 9 = 0

⇒ x + 2y - 3z - 14 = 0

Question 17.

Find the equation of the plane which contains the line of intersection of the planes  and . And which is perpendicular to the plane .

Answer:

The equation of any plane through the line of intersection of the planes  and  is given by -

⇒ 

So, the equation of any plane through the line of intersection of the given planes is

⇒ 

⇒ 

∴ …(1)

Since this plane is perpendicular to the plane

…(2)

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) = (a_1, b_1, c₁)

≡ [(1 - 2λ), (2 - λ), (3 + λ)]

Direction ratios of Normal of plane (2) = (a_2, b_2, 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

Putting the value of λ in (1), we get -

⇒ 

⇒ 

⇒ 

Hence, the equation of the required plane is 

Question 18.

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

Answer:

Given -

The equation of line is

and the equation of the plane is

To find the intersection of line and plane, putting value of  from equation of line into equation of plane, we get -

⇒ 

⇒ 

⇒ (2 + 3λ) × 1 + (-1 + 4λ) × (-1) + (2 + 2λ) × 1 = 5

⇒ 2 + 3λ + 1 - 4λ + 2 + 2λ = 5

⇒ λ = 0

So, the equation of line is

Let the point of intersection be (x,y,z)

So, 

∴ 

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

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

Now, the distance between points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by -

 units

Distance between the points A(2, -1, 2) and B(-1, -5, -10) is given by -

= 13 units

Question 19.

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes s and  .

Answer:

The vector equation of a line passing through a point with position vector  and parallel to a vector  is

⇒ 

Given, the line passes through (1,2,3)

So, 

Given, line is parallel to both planes

∴ Line is perpendicular to normal of both planes.

i.e  is perpendicular to normal of both planes.

we know that -

 is perpendicular to both  & 

So,  is cross product of normals of planes

 and 

Required Normal 

Thus, 

Now, putting the value of  &  in formula

Therefore, the equation of the line is

Question 20.

Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

 and .

Answer:

The vector equation of a line passing through a point with position vector  and parallel to a vector  is

⇒ 

Given, the line passes through (1, 2, -4)

So, 

Given, line is parallel to both planes

∴ Line is perpendicular to normal of both planes.

i.e  is perpendicular to normal of both planes.

we know that -

 is perpendicular to both  & 

So,  is cross product of normals of planes

 and 

Required Normal 

Thus, 

Now, putting the value of  &  in formula

Therefore, the equation of the line is

Question 21.

Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then 

Answer:

Distance of the point (x₁,y₁,z₁) from the plane Ax + By + Cz = D is

⇒ 

The equation of a plane having intercepts a, b, c on the x-, y-, z- axis respectively is

⇒ 

Comparing with Ax + By + Cz = D, we get -

A = 1/a, B = 1/b, C = 1/c, D = 1

Given, the plane is at a distance of 'p' units from the origin.

So, The point is O(0,0,0)

∴ x₁ = 0, y₁ = 0, z₁ = 0

Now,

Distance 

Substituting all values, we get -

⇒ 

squaring both sides, we get -

⇒ 

Hence Proved.

Question 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.  units

Answer:

Distance between two parallel planes Ax + By + Cz = d₁ and Ax + By + Cz = d₂ is

⇒ 

Given -

First Plane is

2x + 3y + 4z = 4

Comparing with Ax + By + Cz = d₁, we get -

A = 2, B = 3, C = 4, d₁ = 4

Second Plane is

4x + 6y + 8z = 12

After Dividing by 2,

2x + 3y + 4z = 6

Comparing with Ax + By + Cz = d₁, we get -

A = 2, B = 3, C = 4, d₂ = 6

So,

Distance between two planes

= 2/√29

Hence, (D) is the correct option.

Question 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

Answer:

Given -

First Plane is

2x – y + 4z = 5

Multiply both sides by 2.5, we get -

5x - 2.5y + 10z = 12.5 …(1)

Second Plane is

5x – 2.5y + 10z = 6 …(2)

Clearly, the direction ratios of normals of both the plane (1) and (2) are same.

Hence, Both the given planes are parallel.