Three Dimensional Geometry

The plane passes through points , and .
Step 1: Find direction vectors in the plane Step 2: Find the normal vector to the plane

Step 3: Vector to be projected Step 4: Use projection formula

The required plane P contains the point A(3, 7, -7).
The given line passes through the point B(2, 3, -2) and has a direction vector .
Since the plane contains both points A and B, the vector must lie in the plane.
.
The normal vector to the plane, , must be perpendicular to both and as both lie in the plane.


.
The direction ratios of the normal are (1, 1, 1).
The equation of the plane passing through A(3, 7, -7) is .
.
The distance 'd' of this plane from the origin (0, 0, 0) is given by the formula:
.
The question asks for the value of .
.

Let the equation of the required plane be .
The normal vector to the plane is .
The normal vector to the plane is .
The required plane is perpendicular to both and . This means the normal vector of , let's call it , must be perpendicular to both and .
Therefore, is parallel to the cross product .


.
So the direction ratios of the normal to the required plane are (-2, 1, -3).
The equation of the plane is of the form .
The plane passes through the point (-1, 0, -2). We substitute these coordinates to find d.

.
The equation of the plane is .
We are given the equation in the form .
To match the constant term, we multiply our equation by -1:
.
Comparing this with , we get:
.
The value of is .

The correct answer is 125
The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.
∴ Let a point on line be λ


as P′ is foot of perpendicular
(3λ + 5)3 + (2λ – 1)2 + (3λ – 1)3 = 0
⇒ 22λ + 15 – 2 – 3 = 0
⇒λ=−5/11

Mid-point of PP’

⇒ 22(α + β + γ)
= 62 + 23 + 40 = 125

The correct answer is 58
If 2x – 3y = γ + 5 and αx + 5y = β + 1 have infinitely many solutions
then

and 3β + 5γ = –28
So |9α + 3β + 5γ|
= |–30 – 28| = 58

The correct answer is (A) :
Family of Plane’s equation can be given by
(5 + 8λ)x + (8 – 7λ)y + (13 + λ) z – (29 + 20λ) = 0
P₁ passes through (2, 1, 3)
⇒ (10 + 16λ) + (8 – 7λ) + (39 + 3λ) – (29 + 20λ) = 0
⇒ –8λ + 28 = 0

d.r,s of normal to P₁

P₂ passes through (0, 1, 2)
8 – 7λ + 26 + 2λ – (29 + 20λ) = 0
⇒ 5 – 25λ = 0

d.r, s of normal to P₂

Angle between normals

The correct answer is (B) :

Line L is x = y = z



so,



So, Area

The correct answer is (B) :
Given:
Line L:
And plane P: px - qy + z = 5
∵ Line L lies on the plane p
∴ Point (2, -1, -3) will satisfy the equation of plane
So, 2p + q - 3 = 5
⇒ 2p + q = 8 .... (i)
The line is also parallel to the plane.
∴ 3p - 2q - 1 = 0
3p - 2q =1...(ii)
By solving equation (i) and equation (ii),
we get p = 15 ,q = - 22
Therefore , Equation of plane is 15x + 22y + z - 5 = 0
Now , distance of plane from origin is d

The correct answer is (C) : (1, 0, 2)

Family of planes P1 and P2





It passes through the point

The correct answer is 2
Normal to plane



∴ Plane (1 – a) (x – 1) + (y – 1) + z = 0
Distance from(2,1,4) is
i.e




a=2 or -4
Therefore ,

L :lx – y + 3 (1 – l)z = 1, x + 2y – z = 2
and plane containing the line p : 3x – 8y + 7z = 4
Let be the vector parallel to L.
then


∵ R containing L
3(6l – 5) – 8(3 – 2l) + 7 (2l + 1) = 0
18l + 16l + 14l – 15 – 24 + 7 = 0

Let θ be the acute angle between L and y-axis

The correct answer is (B) : 8
∵ PR is perpendicular to given line, so

Now,





Now = 2(Sum of co-ordinate of R)-(Sum of co-ordinates of P)
= 2(7)-11
= 3
a+ = 5+3
= 8

The correct answer is (B) :
, normal vector :
, normal vector
Vector parallel to the line of intersection

Vector normal to is
Angle between line and plane is 30°


Hence,
Direction ratios

The correct answer is 153

Let T(2t – 1, 3t – 2, 2t + 1)

∴ 2(2t – 5) + 3 (3t – 4) + 2 (2t – 6) = 0
17t = 34
t = 2
So T(3, 4, 5)


∴Area of
∴Square of

The correct answer is (B) : (1, 2, 1)
P(1,2,1) image in plane x+2y+2z = 16


Q(3,6,5)

Equation of plane = 2(x-0) + 0(y-0) -1(z+1) = 0
2x-z=1
Point lying on plane is (1,2,1)

The correct answer is (D) : 22
Equation of plane through point (2, 3, –5) and perpendicular to planes 2x + y – 5z = 10 and 3x + 5y – 7z = 12 is

∴ Equation of plane is (x – 2) (– 7 + 25) – (y – 3) (– 14 + 15) + (z + 5) · 7 = 0
∴ 18x – y + 7z + 2 = 0
⇒ 18x – y + 7z = – 2
∴ – 18x + y – 7z = 2
On comparing with ax + by + cz = d where d > 0 is
a = – 18, b = 1, c = – 7, d = 2
Therefore a + 7b + c + 20d = 22

Given, through a point parallel to

and through a point parallel to

If lines are coplanar then,

Vector normal to the required plane

Equation of plane

From given option (0, 4, 5) does not lie on the plane.

To find the point P(α, β, γ) where the line intersects the plane, we first express the parametric form of the line:
, , .

The plane contains the lines given by:
as , , , and
and .

We solve the equations for simultaneously to find the plane equation:
Substitute , , into:

Simplifying gives:

The second equation:
, substituting gives:

Simplifying gives:
which is independent of .

The intersection occurs by plugging into plane equations:
satisfies:
Equation of plane through points implies:
\)
.\)

Substitute this value of back:
α = , β = , γ = .

Calculate each:
α = , β = 2 + \frac{38}{77}\) , γ = 3 + \frac{152}{77}\)

α + β + γ = , giving 12\) .

Thus, α + β + γ = 12, verifying it fits the range (12,12).

The correct answer is (B) :
∵ l₁ and l₂ are perpendicular, so

⇒ α = 3
Now angle between l₂ and l₃,

The correct answer is (A) :
Consider the equation of plane,
P : (2x + 3y + z + 20) + λ(x – 3y + 5z – 8) = 0
P : (2 + λ)x + (3 – 3λ)y + (1 + 5λ)z + (20 – 8λ) = 0
∵ Plane P is perpendicular to 2x + 3y + z + 20 = 0
So, 4 + 2λ + 9 – 9λ + 1 + 5λ = 0 ,vso λ=7
P :9x – 18y + 36z – 36 = 0
Or P :x – 2y + 4z = 4
If image of
in plane P is (a, b, c) then

and
Clearly

So, a :b : c = 8 : 5 : – 4

The correct answer is (B) : 93
P₁: x + 3y – z = 6
P₂: –6x + 5y – z = 7
Family of planes passing through line of intersection of P₁ and P₂ is given by x(1 – 6λ) + y(3 + 5λ) + z (–1 – λ) – (6 + 7λ) = 0
It passes through (2, 3, 1/2)
So,



Required plane is
–5x + 8y – 2z – 13 = 0
Or

The correct answer is 137




(x,y,z) ≡ (a-6, β, λ)


⇒ β = 8+ b

⇒ Y = -24+c

⇒ 3a – 4b + 12c = 150….(1)
a + b + c = 5
⇒ 3a + 3b + 3c = 15….(2)
Applying (1) – (2), we get;
-7b + 9c = 135
7b – 9c = -135
7β – 9γ = 7(8 + b) – 9 (–24 + c)
= 56 + 216 + 7b – 9c.
= 56 + 216 – 135 = 137.

The line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector

Equation of line through P(1, 2, 4) and parallel to




⇒ (λ,−4λ+4,−3λ–1)⋅(1,4,−3)=0

Hence
and
So, the correct option is (A):

The correct answer is 9.

eq. of line PM 1x−2​=2y+1​=−1z−3​=λ
any point on line=(λ+2,2λ−1,−λ+3)
for point ' m ’ (λ+2)+2(2λ−1)−(3−λ)=0
λ=

Point m( )

=( ​)

For Image Q(α,β,γ)
​

Q:(3,1,2)

The correct answer is 180.
Any point on L((2λ+1),(−λ−1),(λ+3))
2(2λ+1)+(−λ−1)+3(λ+3)=16
6λ+10=16⇒λ=1
∴P=(3,−2,4)
DR of QR=⟨2λ,−λ,λ+6⟩
DR of L=⟨2,−1,1⟩
4λ+λ+λ+6=0
6λ+6=0⇒λ=−1
Q=(−1,0,2)