Three Dimensional Geometry- 1 - Free MCQ Practice Test with solutions,

On comparing the given equations with :
In the cartesian form two lines


we get ;

x₁ = -1, y₁ = -1,z₁ = -1, ; a₁ = 7, b₁ = -6, c₁ = 1 and 

x₂ = 3, y₂ = 5, z₂ = 7; a₂ = 1, b₂ = -2, c₂ = 1

Now the shortest distance between the lines is given by :

By definition , The angle θ between the planes A₁x + B₁y + C₁z + D₁ = 0 and A₂ x + B₂ y + C₂ z + D₂ = 0 is given by :

The equation of the plane through the line of intersection of the planes

On comparing the given equations with: 
, we get: 

The distance of a point whose position vector is  from the plane  given by :

is a vector joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂). If  Direction cosines of  are given by : 

In Cartesian coordinate system Shortest distance between the lines

Find the shortest distance between the lines 

On comparing them with :

we get : 

In vector form Distance between two parallel lines  given by :

If θ is the acute angle between

then cosine of the angle between
these two lines is given by :


Here, 


Then, 

We have , 
2x + y + 3z – 2 = 0 and x – 2y + 5 = 0. Let θ be the angle between the planes , then 

From the figure,

If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then , the directions cosines of the line are given by :

Midpoint of the line segment is

Parallel vector to the required line

Hence, the equation of the line is

If a line makes angles 90^∘, 135^∘, 45^∘ with the x, y and z – axes respectively, then the direction cosines of this line is given by :

Let the components of the line PQ on X-axis, Y-axis and Z-axis be d_x,d_y and d_z respectively.
Let the angles made by the line with X-axis, Y-axis and Z-axis be α,β,γ respectively.

Now, the projection of the line on XY-plane will be 
The projection of the line on YZ-plane will be 
The projection of the line on XZ-plane will be 

By comparing with the given data we get the value of k.
∴ k=2 .

In the vector form, equation of a plane which is at a distance d from the origin, and  is the unit vector normal to the plane through the origin is given by : 

We know that if P(x,y,z) then
x = rsinθ ⋅ cosϕ
y = rsinθ ⋅ sinϕ
z = rcosθ
Here P(1,2,3)
∴  1=rsinθ⋅cosϕ  ...(i)
2=rsinθ⋅sinϕ   ...(ii)
3=rcosθ ...(iii)
Square equations (i) & (ii) and add
⇒ 1² + 2² = r²sin²θ ⋅ cos²ϕ + r²sin²θ ⋅ sin²ϕ
= r²sin²θ(cos²ϕ+sin²ϕ)=r²sin²θ.
∴ 5=r²sin²θ.
∴ rsinθ=√5 ...(iv)
(Clearly, θ& ϕ are acute).
Using equation (iii) & (iv)

Using equation (i) & (ii)

Here , D.R’s of normal to the plane are 1, 1 , 1 ,its D.C ‘s are :

On dividing x + y + z = 1 by √3 , we get :
 It is of the form : lx+my+nz = d , therefore , d = 1/√3 .

Vector perpendicular given lines

Option C is correct; the required number is 30.

Total number of face diagonals = 12.

Total unordered pairs of these diagonals = C(12,2) = 66.

Fix one face diagonal. The number of other diagonals is 11.

Count diagonals that are not skew to the fixed diagonal:

They meet the fixed diagonal by sharing an endpoint: there are 4 such diagonals (two through each endpoint of the fixed diagonal).

They meet by crossing on the same face: the other diagonal of that face gives 1.

There is exactly 1 diagonal parallel to the fixed diagonal (the diagonal on the opposite parallel face).

Thus the total non-skew diagonals = 4 + 1 + 1 = 6, so skew diagonals per fixed diagonal = 11 - 6 = 5.

Each skew pair is counted twice when we multiply by 12, so number of unordered skew pairs = (12 × 5) / 2 = 30. Hence option C.