Three Dimensional 3D Geometry- 1 - Free MCQ Practice Test with solutions,

If a line makes the angles α,β,γ,
Then,  cos² α+cos² β+cos2 γ=1
It is given that, α+β=90°
⇒  α=β−90°
⇒  cosα=cos(90o−β) 
⇒  cosα=sinβ
⇒  cos² α = sin² β=1−cos² β
⇒  cos²α+cos²β=1
As,   cos² α+cos² β+cos^2γ=1
⇒  1+cos² γ=1
⇒  cos² γ=0
hence,  γ=π/2=90°

A(3,4,−1) and  B(2,−1,5)

Vector AB is normal to the required plane 
⇒ Directions of normal (−1,−5,6)
∴ Equation ,−x−5y+6z=k
Point (2,−3,1) passes through the plane,
∴ −2+15+6=k⇒k=19
∴ −x−5y+6z = 19
x+5y−6z+19 = 0

Let (x,y,z) be the point.
Given sum of the squares of distance from point to the axes is 36. 
⇒(x²+y²)+(y²+z²)+(z²+x²)=36
⇒2(x²+y²+z²)=36⇒x²+y²+z²=18
So the distance of the point from the origin is =3(2)^1/2

The normals to the planes x+2y+2z=5 and 3x+3y+2z=8 are their respective unit vectors ie


Since the required plane is perpendicular to the planes 
x+2y+2z=5 and 3x+3y+2z=8,
 its normal would be perpendicular to the normals to the planes
The cross product of  is normal to the required plane

The equation of the required plane would be

-2x+4y-3z=2+12-6 ie 2x-4y+3z-8 = 0

α(x − α) + β(y − β) + γ(z − γ) = 0
α(1 − α) + β (2 − β) + γ ( 3 − γ) = 0
α + 2β + 3γ = α² + β² + γ²
α² + β² + γ² − α − 2β − 3γ = 0
x² + y² + z² − x − 2y− 3z = 0

line AB = (x-2)/3 = (y+1)/-2 = (z-3)/-1 = λ
(x,y,z)=(3λ+2,−2λ−1,−λ+3)
3x−2y−z=9
3(3λ+2)−2(−2λ−1)−3+λ=9
9λ+6+4λ+2−3+λ=9
14λ=4
λ=2/7
C(x,y,z)=(207, -117, 19/7)
A(2,-1,3)
C is MP of A and B
C= (A+B)/2,    B= (2C-A)
B=(26/7, −22/7+1, 38/7−3)
B=(26/7,−15/7,17/7)

 =AC = −ai^ + ck^
AB = −ai^ + bj^
 Area of △ABC= ½|AB × AC∣ 
|AB × AC∣ =

−(bc)i^− (ac)j^ − (ab)k^
∣AB × AC∣ = (b²c² + a²c² + a²b²)^1/2
Area = 1/2(a²b² + b²c² + c²a²)^1/2

 Let P(x,y,z) be any point on the locus, then the distances from the six faces are ∣x+1∣, ∣x−1∣, ∣y−1∣, ∣z−1∣
According to the given condition, we have
∣x+1∣² + ∣x−1∣² + ∣y+1∣² + ∣y−1∣² + ∣z+1∣²+ ∣z−1∣²=10
⇒ 2(x²+y²+z²)=10−6=4
⇒ x² + y² + z² = 2

Let equation of plane is 
x/α + y/β + z/γ = 1.............(1)
Plane passes through point (a,b,c)
a/α + b/β + c/γ = 1................(2)
Again plane meets the coordinate points. 
Coordinates of points are (α,0,0)  
Coordinates of point B are (0, β, 0) and 
coordinates of point C are (0, 0, γ)  
∴ Equation of planes, parallel to coordinate axis and passing through  
Point A is x = α …..(3)  
Point B is y = β …..(4)  
Point C is z = γ …..(5)  
∴ Locus of the point of intersection is 
a/x + b/y + c/z = 1

2x - y +3z + 4 = 0
x - 3y + z = 0
x + 2y + z + 1 = 0
x = 12/5
y = -1/5
z= -3
this point also satisfied by
ax - y + z + 2 = 0
a(12 / 5) - (-1/5) + (-3) = 0
⇒ 12a/5 + 1/5 -3 = 0
a= -2

Let the point of intersection is (λ, 2λ, 3λ). 
Clearly, λ = 3μ + 1, 2λ = 2 – μ 
Solving, we get λ = 1, μ = 0 
Hence, the point of intersection is (1, 2, 3).
Therefore, (1+k)/3 = (2-1)/2 = (3-2)/h
= (1+k)/3 = 1/2 = 1/h
⇒ h = 2   k = 1/2

We have four coplanar points.
The three vectors connecting two of them at a time are thus coplanar.
⇒{(−x, y, 0) (-x, 0, z) (1−x, 1, 1)} = 0
⇒−x (−z) −y (−x−z+xz)+0=0
⇒xz + xy + yz = xyz
⇒1/y + 1/z + 1/x = 1

Let x₁, y₁, z₁ be any point on the plane

The co-ordinates of any point on L1 in terms of parameter r1 are given by,x=y+a=z=r1
⇒x=r1,y=r1 - a,z=r1....(1)
Similarly the co-ordinates of any point on L2 in terms of parameter 2r2 are given by 
⇒x+a=2y=2z=2r2
⇒ x=2r2−a,y=r2, z=r2...(2)
Let 'A' be a point on L1 and 'B' be a point on L2
​Using (1) and (2), the direction ratios of AB are 
2r2−a−r1,r2−r1+a,r2−r1
​If the above line is same as the line whose direction cosines are proportional to (2,1,2) as given in the question, then  
(2r2−r1−a)/2= (r2−r1+a)/1 = (r2−r1)/2
Solving the first two of the above equation, we get r1=3a
Again solving the last two, we get r2=a
Using these values in (1) and (2), we get the coordinates of points as 
 
(3a,2a,3a) and (a,a,a).

(x-x1)a1 = (y-y1)b1 = (z-z1)c1 &
(x-x2)a2 = (y-y2)b2 = (z-z2)c2 are coplanar if

k² + 3k = 0
k = 0 or -3

A(a,0,0),B(0,b,0),C(0,0,c) are the points on coordinate axis and centriod (α,β,γ).
According to centroid formula,
α= a/3
​a=3α
b=3β
c=3γ
Equation of plane is,x/a+y/b+z/c =1
x/α+y/β+z/γ=3