Test: Introduction To Three Dimensional Geometry- 1 - JEE MCQ

x is 3 units to the right from (0, 4, 5)
y is 4 units upwards from (3, 0, 5)
z is 5 units inwards from (3, 4, 0)
Note that the x-axis has the equation ‘y=0, z=0’, because the axis is horizontal and always has a y and z values of 0 for all x.
Therefore, the distance of point P(3,4,5) from X-axis is the hypotenuse of the triangle formed by the upward and inward movement along y and z-axis, which can be found using Pythagoras’ Theorem:
sqrt(5² + 4² ) = sqrt(41)
The distance is the square root of 41.

The x-axis makes angles 0°, 90° and 90° with x, y and z-axis. Again y-axis makes angles 0°, 90°, 90° with x, y and z-axis. 
∴ direction cosines of x-axes are cos 0°, cos 90°, cos 90° i.e. 1, 0, 0.

Any normal of x−y plane is along z−axis.
normal unit vector to x−y plane, 
n→ = kˆ= 0iˆ+0jˆ+1kˆ
Direction cosines are <0, 0, 1> or
n→ = k =0iˆ+0jˆ−1kˆ
Direction cosines are <0, 0, −1>

There are two different lines that make equal angles with the coordinate axes.
The are y = x line and y = - x line.
The y = x line divides the 1st and 3rd quadrant equally and makes a 45° angle with the positive direction of the x-axis and the y-axis and negative direction of the x-axis and the y-axis.
Again, the line y = - x divides the 2nd and 4th quadrant equally and makes a 45° angle with the positive direction of the x-axis and negative direction of the y-axis & negative direction of the x-axis and positive direction of the y-axis.

Correct Answer : c

Explanation :  l² + m² + n² = 1

(a) 9 + 16 + 25 not equal to 1

(b) y = 4z

4 is not equal to 20

(d) x + y - z = 0

3 + 4 - 5 = 0

=> 2 is not equal to 0

There option c satisfies the equation.

Given that direction cosine of the line(k,k,k)
The value of k = +-1/(3)^½
We know the sum of the squares of the direction cosine is one.
k² + k²+ k² = 1
3k² = 1
k² = +-1/(3)^½

Correct Answer : C

Explanation :  Volume of tetrahedron = k(Area of one face) * (length of perpendicular from the opposite vertex upon it)

1/3 *(Area of one face) * (length of perpendicular from the opposite vertex upon it)

k = 1/3

ABCD is a parallelogram for this we'll show that both the diagonal meet at 0 i.e., 0 is the midpoint of AC and BD
AC = [(4-1)/2, (7-2)/2, (8+1)/2]
      = [3/2, 5/2, 9/2]
BD = [(2+1)/2, (3+2)/2, (4+5)/2]
      = [3/2, 5/2, 9/2]

Given lines are
(x-1)/0 = (y-2)/0 = z/1
and x/1=(y+1)0=z/0
∴ cosθ=0⋅1+0⋅0+1⋅0=0
⇒ θ=90^∘

Given place : x=0 and two points →(−2,3,4) and (1,−2,3)

let say a point (x,y,z) in x=0 place

So, x=(m+n(−2))/m+n

0=(m−2n)/m+n

0 = (m−2n)/m+n  ⇒m=2n

So, m/n = 2/1

⇒ 2:1