Maths Quiz on Three Dimensional Geometry for IIT JEE & Engineering Exam

Solution:

Let the two lines be AB and CD having equations

x / 1 = [y + a] / 1 = z / 1 = λ and [x + a] / [2] = y / 1 = z / 1 = μ then

P = (λ, λ − a, λ) and Q = (2μ − a, μ, μ)

So, according to question, [λ − 2μ + a] / [2] = [λ − a − μ] / [1] = [λ − μ] / [2]

μ = a and λ = 3a

Therefore, P = (3a, 2a, 3a) and [(x − 2)² +(y − 3)² + (z − 4)²].

Solution:

Direction cosines of line = (2 / 7, 3 / 7, −6 / 7)

Now, x′ = 1 + [2r /7], y′ = −2+ [3r / 7] and z′ = 3 − [6r / 7]

Therefore, (1 + [2r / 7]) − (−2 + [3r / 7]) + (3 − [6r / 7])

= 5

⇒ r = 1

Solution:

Let the equation of sphere be x² + y² + z² + 2ux + 2vy + 2wz + d = 0

Because it passes through origin (0, 0, 0), d = 0 and also, it passes through (0, 2, 0)

4 + 4v = 0 ⇒ v = −1

Also, it passes through (1, 0, 0)

1 + 2u = 0 ⇒ u = − 1/2

And it passes through (0, 0, 4)

16 + 8w ⇒ w = −2

Centre (−u, −v, −w)=(1 / 2, 1, 2)

Solution:

The required point is the centre of the sphere through the given points.

Let the equation of sphere be x² + y² + z² + 2ux + 2vy + 2wz + d = 0 …..(i)

Sphere (i) is passing through (0, 0, 0), (a, 0, 0), (0, b, 0) and (0, 0, c), hence, d = 0

a² + 2ua = 0 ⇒ u = −a/2

b² + 2vb = 0 ⇒ v = −b/2

c² + 2wc = 0 ⇒ w = −c/2

Therefore, centre of sphere is (a/2, b/2, c/2), which is also the required point.

d) 5√2

Solution:

Let the line segment be AB, then as given AB cos α = 3, AB cos β = 4, AB cos γ = 5

⇒ AB² (cos²α + cos²β + cos²γ) = 3² + 4² + 5²

AB = √[9 + 16 + 25] = 5√2,

where α, β and γ are the angles made by the line with the axes.