JEE Maths Quiz on Three Dimensional Geometry
JEE Maths Quiz on Three Dimensional Geometry : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Three Dimensional Geometry.
Quiz on Three Dimensional Geometry
Q1. A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points of intersection are given by __________.
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}].
Q2. The distance of the point (1, 2, 3) from the plane x − y + z = 5 measured parallel to the line x / 2 = y / 3 = z / −6, is __________.
a) 1
b) 2
c) 3
d) 4
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
Q3. The centre of the sphere passes through four points (0, 0, 0), (0, 2, 0), (1, 0, 0) and (0, 0, 4) is ________.
Solution:
Let the equation of sphere be x^{2} + y^{2} + z^{2} + 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)
Q4. Coordinate of a point equidistant from the points (0,0,0), (a, 0, 0), (0, b, 0), (0, 0, c) is ___________.
Solution:
The required point is the centre of the sphere through the given points.
Let the equation of sphere be x^{2} + y^{2} + z^{2} + 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
a^{2 }+ 2ua = 0 ⇒ u = −a/2
b^{2} + 2vb = 0 ⇒ v = −b/2
c^{2 }+ 2wc = 0 ⇒ w = −c/2
Therefore, centre of sphere is (a/2, b/2, c/2), which is also the required point.
Q5. The projection of any line on coordinate axes be, respectively 3, 4, 5 then its length is ______.
a) 2√2
b) 4√2
c) 3√2
d) 5√2
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
⇒ AB^{2 }(cos^{2}α + cos^{2}β + cos^{2}γ) = 3^{2} + 4^{2} + 5^{2}
AB = √[9 + 16 + 25] = 5√2,
where α, β and γ are the angles made by the line with the axes.
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