Chapter 11: Three Dimensional Geometry

Any point in the given line is

x – 3/ 2 = y – 3/1 = z/1 = λ

x = 2λ + 3, y = λ + 3 and z = λ

Let it be the coordinates of P

So, the direction ratios of OP are (2λ + 3 – 0), (λ + 3 – 0) and (λ – 0) ⇒ 2λ + 3, λ + 3, λ

But the direction ratios of the line PQ are 2, 1, 1

Now, we know that

λ² + 3λ + 3 = 4λ² + 9 + 12λ (On squaring on both sides)

3λ² + 9λ + 6 = 0

λ² + 3λ + 2 = 0

(λ + 1)( λ + 2) = 0

λ = -1, -2

So, the direction are:

Thus, the required equation of planes are

x/1 = y/2 = z/-1 and x/-1 = y/1 = z/-2

Given equations are,

l + m + n = 0 ….. (i)

l² + m² – n² = 0 …. (ii)

From equation (i), we have n = – (l + m)

Putting the value of n is equation (ii), we get

l² + m² + [-(l + m)]² = 0

l² + m² – l² – m² – 2lm = 0

-2lm = 0

lm = 0 ⇒ (- m – n)m = 0 [Since, l = – m – n]

(m + n)m = 0 ⇒ m = 0 or m = -n

⇒ l = 0 or l = -n

Now, the direction cosines of the two lines are

0, -n, n and -n, 0, n ⇒ 0, -1, 1 and -1, 0, 1

Thus, the required angle π/3.

Given that l, m, n and l + dl, m + dm, n + dn are the direction cosines of a variable line in two positions

l² + m² + n² = 1 ….. (i) and

(l + dl)² + (m + dm)² + (n + dn)² = 1 …. (ii)

Given, O (0, 0, 0) and A(a, b, c)

So, the direction ratios of OA = a – 0, b – 0, c – 0 = a, b, c

And, the direction cosines of line OA

Now, the direction ratios of the normal to the plane are (a, b, c).

We know that, the equation of the plan passing through the point A(a, b, c) is

a(x – a) + b(y – b) + c(z – c) = 0

ax – a² + by – b² + cz – c² = 0

ax + by + cz = a² + b² + c²

Thus, the required equation of the plane is ax + by + cz = a² + b² + c²

Let’s take OX, OY, OZ and ox, oy, oz to be two rectangular systems.

And, the equations of two planes are

Long Answer (L.A.)

Given,

is the equation of the line.

Now, the coordinates of any point Q on the line are

x = -2λ + 4, y = 6λ and z = -3λ + 1 and the given point is P(2, 3, -8)

The direction ratios of PQ are -2λ + 4 – 2, 6λ – 3, -3λ + 1 + 8 i.e. -2λ + 2, 6λ – 3, -3λ + 9

And the direction ratios of the given line are -2, 6, -3.

If PQ ⊥ line, then

-2(-2λ + 2) + 6(6λ – 3) – 3(-3λ + 9) = 0

4λ – 4 + 36λ – 18 + 9λ – 27 = 0

49λ – 49 = 0

λ = 1

Now, the foot of the perpendicular is -2(1) + 4, 6(1), -3(1) + 1 i.e. 2, 6, -2

Hence, the distance PQ is

Thus, the required coordinates of the foot of perpendicular are 2, 6, -2 and the required distance is 3√5 units.

Given plane is 2x – 2y + 4z + 5 = 0 and point (1, 3/2, 2)

The direction ratios of the normal to the plane are 2, -2, 4

So, the equation of the line passing through (1, 3/2, 2) and direction ratios are equal to the direction ratios of the normal to the plane i.e. 2, -2, 4 is

Now, any point in the plane is 2λ + 1, -2λ + 3/2, 4λ + 2

Since, the point lies in the plane, then

2(2λ + 1) – 2(-2λ + 3/2) + 4(4λ + 2) + 5 = 0

4λ + 2 + 4λ – 3 + 16λ + 8 + 5 = 0

24λ + 12 = 0 λ = ½

So, the coordinates of the point in the plane are

2(-1/2) + 1, -2(-1/2) + 3/2, 4(-1/2) + 2 = 0, 5/2, 0

Thus, the foot of the perpendicular is (0, 5/2, 0) and the required length

Given point is (3, 0, 1) and the equation of planes are

x + 2y = 0 …. (i) and 3y – z = 0 …. (ii)

Equation of any line l passing through (3, 0, 1) is

l: (x – 3)/a = (y – 0)/b = (z – 1)/c

Now, the direction ratios of the normal to plane (i) and plane (ii) are (1, 2, 0) and (0, 3, 1).

As the line is parallel to both the planes, we have

1.a + 2.b + 0.c = 0 ⇒ a + 2b + 0c = 0 and

0.a + 3.b – 1.c = 0 ⇒ 0a + 3b – c = 0

We know that, equation of the plane passing through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) with its normal’s direction ratios is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 …. (i)

Now, if the plane is passing through two points (2, 1, -1) and (-1, 3, 4) then

a(x₂ -x₁) + b(y₂ – y₁) + c(z₂ – z₁) = 0

a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0

-3a + 2b + 5c = 0 …. (ii)

As the required plane is perpendicular to the given plane x – 2y + 4z = 10, then

1.a – 2.b + 4.c = 10 …. (iii)

On solving (ii) and (iii) we get,

So, a = 18λ, b = 17λ and c = 4λ

Thus, the required plane is

18λ (x – 2) + 17λ (y – 1) + 4λ (z + 1) = 0

18x – 36 + 17y – 17 + 4z + 4 = 0

⇒ 18x + 17y + 4z – 49 = 0