Test: Three Dimensional Geometry- 1 - JEE MCQ

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

For A′ & C′, there are five diagonals which are skew i.e. the diagonals which do not pass through vertices A′ & C′ (not intersecting with A′C′) and neither parallel to diagonal A′C′.
Those diagonals are D′A, DB, D′C, CB′ & AB′.
We can see that for one of the diagonal, there are 5 options possible for the diagonals which are skew.
Also, there are total 12 face diagonals.
So, total number of skew diagonals possible are 12×5=60.
But we also see that for the diagonal A′C′, we get AB′ as skew diagonal and also when we take AB′ as diagonal, we get A′C′ as skew diagonal i.e. total number of skew diagonals possible are repeated twice.
Hence, required number of skew diagonals = 60/2 = 30.