JEE Main Maths Three Dimensional Geometry Previous Year Questions With Solutions

In Cartesian geometry, we basically deal with 2-D geometry and 3-D geometry. In two dimensional geometry, we deal with location of a point in x-y plane whereas in three dimensional geometry, we deal with three orthogonal axes,say, x,y and z axes. In short, 3-D Geometry is nothing but an extension of 2-D geometry in Cartesian geometry. In this article we more concentrate on three dimensional geometry basics and important problems for JEE aspirants. The 3-D geometry is also called Solid Geometry, since it deals with 3-D mathematics of 3d shapes like cubes, cuboids, tetrahedrons, parallelepiped etc. Three-Dimensional Geometry or solid geometry is one of the most interesting and important topics for JEE. This article covers direction cosine, direction ratio, vector equation of a line, cartesian equation of a line, the distance between parallel lines, equation of a plane, etc. Three-Dimensional Geometry questions from the previous years of JEE Main are present in this page, along with the detailed solution for each question. These questions include all the important topics and formulae.

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Three Dimensional Geometry Previous Year Questions With Solutions

Question 1: The projection of any line on co-ordinate axes be, respectively 3, 4, 5 then its length is ______.

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.

Question 2: The angle between two diagonals of a cube will be _____.

Solution:

JEE Previous Year 3D Problems

Let the cube be of side ‘a’ and O (0, 0, 0), D (a, a, a), B (0, a, 0), G (a, 0, a).

Then the equation of OD and BG are x/a = y/a = z/a and x/a = [y − a]/[−a] = z/a, respectively.

Hence, the angle between OD and BG is

cos1a2a2+a23a2×3a2=cos1(13)cos^{−1} \frac{a^2 − a^2 + a^2}{\sqrt{3}a^2 \times \sqrt{3}a^2} = cos^{−1} (\frac{1 }{ 3})

Question 3: The co-ordinates of the foot of perpendicular drawn from point P (1, 0, 3) to the line joining the points A (4, 7, 1) and B (3, 5, 3) is _______.

Solution:

JEE Previous Year 3D Geometry Problems

Let D be the foot of perpendicular drawn from P (1, 0, 3) on the line AB joining (4, 7, 1) and (3, 5, 3).

If D divides AB in ratio λ : 1, then

D=[(3λ+4λ+1),(5λ+7λ+1),(3λ+1λ+1)]D = [(\frac{3 \lambda + 4}{\lambda + 1}), (\frac{5 \lambda + 7}{\lambda + 1}), (\frac{3 \lambda + 1}{\lambda + 1})] …..(i)

Direction ratios of PD are 2λ + 3, 5λ + 7, −2.

Direction ratios of AB are −1, −2, 2 [Because PD⊥AB]

− (2λ + 3) −2 (5λ + 7) − 4 = 0

λ = −7 / 4

Putting the value of λ in (i), we get the point D (5/3, 7/3, 17/3).

Question 4: 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].

Question 5: 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 __________.

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

Question 6: A point moves so that the sum of its distances from the points (4, 0, 0) and (4, 0, 0) remains 10. The locus of the point is _______.

Solution:

(x4)2 +y2 +z2 +(x+4)2 +y2 +z2=10\sqrt{(x − 4)^2 + y^2 + z^2 } + \sqrt{(x + 4)^2 + y^2 + z^2} = 10

2 (x2 + y2 + z2) + 2 [(x4)2 +y2+z2][(x+4)2+y2+z2]\sqrt{[(x − 4)^2 + y^2 + z^2] [(x + 4)^2 + y^2 + z^2]} = 100 − 32 = 68

⇒ (x2 + y2 + z2 − 34)= [(x − 4)2 + y2 + z2] [(x + 4)2 + y2 + z2]

⇒ (x2 + y2 + z2)2 − 68 (x2 + y2 + z2) + (34)= [(x2 + y2 + z2 + 16) − 8x] [(x2 + y2 + z2 + 16) + 8x]

= (x2 + y2 + z2 + 16)2 − 64x2

= (x2 + y2 + z2) + 32 (x2 + y2 + z2) − 64x2 + (16)2

⇒ 9x2 + 25y2 + 25z2 − 225 = 0

Question 7: 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 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)

Question 8: Co-ordinate 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 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.

Question 9: If the plane 2ax − 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres x2 + y2 + z2 + 6x − 8y − 2z = 13 and x2 + y2 + z2 − 10x + 4y − 2z = 8, then a equals ____________.

Solution:

S1 = x2 + y2 + z2 + 6x − 8y − 2z = 13, C1 = (−3, 4, 1)

S2 = x2 + y2 + z2 − 10x + 4y − 2z = 8, C2 = (5, −2, 1)

So, mid point of C1 C2 (say P) = P(532,422,1+12)\frac{5 − 3}{2}, \frac{4 − 2}{2}, \frac{1+1}{2}) = P(1, 1, 1)

Now the plane 2ax − 3ay + 4az + 6 = 0 passes through the point P,

So, 2a (1) − 3a (1) + 4a (1) + 6 = 0 = 2a −3a + 4a + 6 = 0

3a + 6 = 0

3a = −6

⇒a = −2

Question 10: The radius of the sphere x + 2y + 2z = 15 and x2 + y2 + z2 − 2y − 4z = 11 is ___________.

Solution:

Equation of sphere is, x2 + y2 + z2 − 2y − 4z = 11

Centre of sphere = (0, 1, 2) and radius of sphere = 4

Let the centre of the circle be (α, β, γ).

The direction ratios of a line joining from centre of the sphere to the centre of the circle are

(α − 0, β − 1, γ − 2) or (α, β − 1, γ − 2)

But, this line is normal at plane x + 2y + 2z = 15

α / 1 = [β − 1]/2 = [γ − 2]/2 = k

α = k, β = 2k + 1, γ = 2k + 2 [Because, centre of circle lies on x + 2y + 2z = 15]

k + 2 (2k + 1) + 2 (2k + 2) = 15

⇒ k = 1

So, the centre of circle = (1, 3, 4)

Therefore, Radius of circle = √[(Radius of sphere)2 − (Length of joining line of centre)2] 42(10)2+(31)2+(42)2=169\sqrt{4^2 − (1 − 0)^2 + (3 − 1)^2 + (4 − 2)^2 } = \sqrt{16-9}

= √7

Question 11: The point at which the line joining the points (2, 3, 1) and (3, 4, 5) intersects the plane 2x + y + z = 7 is _________.

Solution:

Ratio = 2(2)+(3)(1)+(1)(1)72(3)+(4)(1)+(5)(1)7=(5)10\frac{2 (2) + (−3) (1) + (1) (1) − 7} {2 (3) + (−4) (1) + (−5) (1) − 7} = \frac{−(−5)}{− 10} = −(1 / 2)

Hence, x = 2(2)3(1)1\frac{2 (2) − 3 (1)}{1} = 1, y = 3(2)(4)1\frac{−3 (2) − (−4)}{1} = −2 and z = 1(2)(5)1\frac{1 (2) − (−5)}{1} = 7.

Therefore, P (1, −2, 7).

Trick : As (1, 2, 7) and (1, 2, 7) satisfy the equation 2x + y + z = 7, but the point (1, 2, 7) is collinear with (2, 3, 1) and (3, 4, 5).

Note: If a point divides the joining of two points in some particular ratio, then this point must be collinear with the given points.

Question 12: The point where the line x12=y23=z+34\frac{x − 1}{2} = \frac{y − 2}{−3} = \frac{z + 3}{4} meets the plane 2x + 4y − z = 1, is ___________.

Solution:

Let point be (a, b, c), then 2a + 4b − c = 1 …..(i) and a = 2k + 1,b = −3k + 2 and c = 4k − 3,

(where k is constant)

Substituting these values in (i), we get

2 (2k + 1) + 4 (−3k + 2) − (4k − 3) = 1

⇒ k = 1

Hence, the required point is (3, 1, 1).

Trick: The point must satisfy the lines and plane.

Question 13: The equation of the straight line passing through the point (a, b, c) and parallel to the z-axis, is ________.

Solution:

The line through (a,b,c) is xal=ybm=zcn\frac{x − a}{l} = \frac{y − b}{m} = \frac{z -c}{n} ….(i)

Since the line is parallel to z-axis, therefore any point on this line will be of the form (a, b, z1). Also, any point on line (i) is (lr + a, mr + b, nr + c).

Hence, lr + a = a  and  mr + b = b ⇒ l = m = 0

Hence, the line will be xa0=yb0=zc1\frac{x − a}{0} = \frac{y − b}{0} = \frac{z -c}{1}

Question 14: The acute angle between the line joining the points (2, 1, 3), (3, 1, 7) and a line parallel to x13=y4=z+35\frac{x − 1}{3} = \frac{y }{4} = \frac{z + 3}{5} through the point (1, 0, 4) is ____________.

Solution:

Direction ratio of the line joining the point (2, 1, −3), (−3, 1, 7) are (a1, b1, c1) ⇒ (−3 −2, 1 − 1, 7 − (−3))

⇒ (−5, 0, 10)

Direction ratio of the line parallel to line x13=y4=z+35\frac{x − 1}{3} = \frac{y }{4} = \frac{z + 3}{5}  are (a2, b2, c2)

⇒ (3, 4, 5)

Angle between two lines,

cos θ = a1a2+b1b2+c1c2(a12+b12+c12 )×(a22+b22+c22 )\frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{(a_1^2 + b_1^2 + c_1^2 )} \times \sqrt{(a_2^2 + b_2^2 + c_2^2 )}}

cos θ = 5×3+0×4+10×5(25+0+100)×(9+16+25 )\frac{-5 \times 3 + 0 \times 4 + 10 \times 5}{\sqrt{(25 + 0 + 100)} \times \sqrt{(9 + 16 + 25 )}}

cos θ = 352510\frac{35}{25 \sqrt{10}}

⇒ θ = cos−1 (7510\frac{7}{5 \sqrt{10}})

Question 15: If l1, m1, n1 and l2, m2, n2 are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be ______.

Solution:

Let lines are l1x + m1y + n1z + d = 0 ..(i) and

l2x + m2y + n2z + d = 0 …..(ii)

If lx +my + nz + d = 0 is perpendicular to (i) and (ii), then, ll1 + mm1 + nn1 = 0, ll2 + mm2 + nn2 = 0

lm1n2m2n1=mn1l2l1n2=nl1m2l2m1=d\frac{l }{m_1n_2− m_2n_1} = \frac{m}{n_1l_2 − l_1n_2} = \frac{n }{l_1m_2 − l_2m_1} = d

Therefore, direction cosines are (m1n2 − m2n1), (n1l2 − l1n2), (l1m2 − l2m1).

 

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