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3D Geometry

JEE Mathematics · 279 questions · Page 1 of 28 · Click an option or "Show Solution" to reveal answer

Q1
A plane which passes through the point (3,2,0)(3,2,0) and the line x−41=y−75=z−44{{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4} is :
A x−y+z=1x-y+z=1
B x+y+z=5x+y+z=5
C x+2y−z=1x+2y-z=1
D 2x−y+z=52x-y+z=5
Correct Answer
Option A
Solution

As the point

(3,2,0)\left( {3,2,0} \right)

lies on the given line

x−41=y−75=z−44{{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}

∴\therefore There can be infinite many planes passing through this line.

But here out of the four options only first option is satisfied by the coordinates of both the points

(3, 2, 0)\left( {3,\,2,\,0} \right)

and

(4, 7, 4)\left( {4,\,7,\,4} \right)

∴\therefore

x−y+z=1x - y + z = 1

is the required plane.

Q2
The lines x−21=y−31=z−4−k{{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}} and x−1k=y−42=z−51{{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1} are coplanar if :
A k=3k=3 or −2-2
B k=0k=0 or −1-1
C k=1k=1 or −1-1
D k=0k=0 or −3-3
Correct Answer
Option D
Solution

Coplanar if

∣x2−x1y2−y1z2−z1l1m1n1l2m2n2∣=0\left| \begin{array}{lll}{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \\ {{l_1}} & {{m_1}} & {{n_1}} \\ {{l_2}} & {{m_2}} & {{n_2}} \end{array} \right| = 0

∴\therefore

∣1−1−111−kk21∣=0\left| \begin{array}{lll}1 & { - 1} & { - 1} \\ 1 & 1 & { - k} \\ k & 2 & 1 \end{array} \right| = 0
⇒∣00−121+k−kk+211∣=0\Rightarrow \left| \begin{array}{lll}0 & 0 & { - 1} \\ 2 & {1 + k} & { - k} \\ {k + 2} & 1 & 1 \end{array} \right| = 0
k2+3k=0⇒k(k+3)=0{k^2} + 3k = 0 \Rightarrow k\left( {k + 3} \right) = 0

or

k=0k = 0

or

−3-3
Q3
The radius of the circle in which the sphere x2+y2+z2+2x−2y−4z−19=0{x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0 is cut by the plane x+2y+2z+7=0x+2y+2z+7=0 is
A 44
B 11
C 22
D 33
Correct Answer
Option D
Solution

Center of sphere

=(−1,1,2)= \left( { - 1,1,2} \right)

Radius of sphere

1+1+4+19=5\sqrt {1 + 1 + 4 + 19} = 5

Perpendicular distance from center to the plane

OC=d=∣−1+2+4+71+4+4∣=123=4.OC = d = \left| {{{ - 1 + 2 + 4 + 7} \over {\sqrt {1 + 4 + 4} }}} \right| = {{12} \over 3} = 4.
AC2=AO2−OC2=52−42=9A{C^2} = A{O^2} - O{C^2} = {5^2} - {4^2} = 9
⇒AC=3\Rightarrow AC = 3
Q4
Two systems of rectangular axes have the same origin. If a plane cuts then at distances a,b,ca,b,c and a′,b′,c′a', b', c' from the origin then
A 1a2+1b2+1c2−1a′2−1b′2−1c′2=0{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0
B  1a2+1b2+1c2+1a′2+1b′2+1c′2=0\,{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} + {1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}} = 0
C 1a2+1b2−1c2+1a′2−1b′2−1c′2=0{1 \over {{a^2}}} + {1 \over {{b^2}}} - {1 \over {{c^2}}} + {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0
D 1a2−1b2−1c2+1a′2−1b′2−1c′2=0{1 \over {{a^2}}} - {1 \over {{b^2}}} - {1 \over {{c^2}}} + {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0
Correct Answer
Option A
Solution

Equation of planes be

xa+yb+zc=1  &  xa′+yb′+zc′=1{x \over a} + {y \over b} + {z \over c} = 1\,\,\& \,\,{x \over {a'}} + {y \over {b'}} + {z \over {c'}} = 1

So the distance from (0, 0, 0) to both the plane is same. ∴\therefore

∣−11a2+1b2+1c2∣=∣−11a′2+1b′2+1c′2∣\left| {{{ - 1} \over {\sqrt {{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}}} }}} \right| = \left| {{{ - 1} \over {\sqrt {{1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}}} }}} \right|

⇒\Rightarrow

1a2+1b2+1c2−1a′2−1b′2−1c′2=0{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {{{a'}^{2}}}} - {1 \over {{{b'}^{2}}}} - {1 \over {{{c'}^{2}}}} = 0
Q5
A line makes the same angle θ\theta , with each of the xx and zz axis. If the angle β \beta \,, which it makes with y-axis, is such that  sin⁡2β=3sin⁡2θ,\,{\sin ^2}\beta = 3{\sin ^2}\theta , then cos⁡2θ{\cos ^2}\theta equals :
A 25{2 \over 5}
B 15{1 \over 5}
C 35{3 \over 5}
D 23{2 \over 3}
Correct Answer
Option C
Solution

Concept : If a line makes the angle

α,β,γ\alpha ,\beta ,\gamma

with x, y, z axis respectively then

cos⁡2α+cos⁡2β+cos⁡2γ=1{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1

$ In this question given that the line makes angle θ with x and z-axis and β with y−axis.

∴ cos2θ+cos2β+cos2θ=1\therefore\: cos^2\theta+cos^2\beta+cos^2\theta=1
⇒ 2cos2θ=1−cos2β\Rightarrow\:2cos^2\theta=1-cos^2\beta
⇒2cos⁡2θ=sin⁡2β\Rightarrow 2{\cos ^2}\theta = {\sin ^2}\beta

But given that

sin2β=3sin2θsin^2\beta=3sin^2\theta

∴\therefore

2cos⁡2θ=3sin⁡2θ2{\cos ^2}\theta = 3{\sin ^2}\theta
⇒2cos⁡2θ=3(1−cos⁡2θ)\Rightarrow 2{\cos ^2}\theta = 3\left( {1 - {{\cos }^2}\theta } \right)
⇒2cos⁡2θ=3−3cos⁡2θ\Rightarrow 2{\cos ^2}\theta = 3 - 3{\cos ^2}\theta
⇒5cos⁡2θ=3\Rightarrow 5{\cos ^2}\theta = 3
⇒cos⁡2θ=35\Rightarrow {\cos ^2}\theta = {3 \over 5}
Q6
If the straight lines x=1+s,y=−3x=1+s,y=-3−λs, - \lambda s, z=1+λsz = 1 + \lambda s and x=t2,y=1+t,z=2−t,x = {t \over 2},y = 1 + t,z = 2 - t, with parameters ss and tt respectively, are co-planar, then λ\lambda equals :
A 00
B −1-1
C −12 - {1 \over 2}
D −2-2
Correct Answer
Option D
Solution

The given lines are

x−1=y+3−λ=z−1λ=s.........(1)x - 1 = {{y + 3} \over { - \lambda }} = {{z - 1} \over \lambda } = s.........\left( 1 \right)

and

2x=y−1=z−2−1=t..........(2)2x = y - 1 = {{z - 2} \over { - 1}} = t..........\left( 2 \right)

The lines are coplanar, if

∣0−(−1)−1−3−2−(−1)1−λλ121−1∣=0\left| \begin{array}{lll}{0 - \left( { - 1} \right)} & { - 1 - 3} & { - 2 - \left( { - 1} \right)} \\ 1 & { - \lambda } & \lambda \\ {{1 \over 2}} & 1 & { - 1} \end{array} \right| = 0
c2→c2+c3;∣1−5110λ120−1∣=0{c_2} \to {c_2} + {c_3};\left| \begin{array}{lll}1 & { - 5} & 1 \\ 1 & 0 & \lambda \\ {{1 \over 2}} & 0 & { - 1} \end{array} \right| = 0
⇒5(−1−λ2)=0⇒λ=−2\Rightarrow 5\left( { - 1 - {\lambda \over 2}} \right) = 0 \Rightarrow \lambda = - 2
Q7
Distance between two parallel planes  2x+y+2z=8\,2x + y + 2z = 8 and 4x+2y+4z+5=04x + 2y + 4z + 5 = 0 is :
A 92{9 \over 2}
B 52{5 \over 2}
C 72{7 \over 2}
D 32{3 \over 2}
Correct Answer
Option C
Solution

The planes are

2x+y+2x−8=0          ...(1)2x + y + 2x - 8 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

and

4x+2y+4z+5=04x + 2y + 4z + 5 = 0

or

2x+y+2z+52=0          ...(2)2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

∴\therefore Distance between

(1)\left( 1 \right)

and

 (2)\,\left( 2 \right)
=∣52+822+12+22∣= \left| {{{{5 \over 2} + 8} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|
=∣2129∣= \left| {{{21} \over {2\sqrt 9 }}} \right|
=72= {7 \over 2}
Q8
If the plane 2ax−3ay+4az+6=02ax-3ay+4az+6=0 passes through the midpoint of the line joining the centres of the spheres x2+y2+z2+6x−8y−2z=13{x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13 and x2+y2+z2−10x+4y−2z=8{x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8 then a equals :
A −1-1
B 11
C −2-2
D 22
Correct Answer
Option C
Solution

Centers of given spheres are

(−3,4,1)\left( { - 3,4,1} \right)

and

(5,−2,1).\left( {5, - 2,1} \right).

Mid point of centers is

(1,1,1).\left( {1,1,1} \right).

Satisfying this in the equation of plane, we get

2a−3a+4a+6=02a - 3a + 4a + 6 = 0
⇒a=−2\Rightarrow a = - 2
Q9
Let S be the set of all real values of λ\lambda such that a plane passing through the points (–λ\lambda 2, 1, 1), (1, –λ\lambda 2, 1) and (1, 1, – λ\lambda 2) also passes through the point (–1, –1, 1). Then S is equal to :
A {1, −-1}
B {3, −- 3}
C {3}\left\{ {\sqrt 3 } \right\}
D {3,−3}\left\{ {\sqrt 3 , - \sqrt 3 } \right\}
Correct Answer
Option D
Solution

All four points are coplanar so

∣1−λ2202−λ2+1022−λ2−1∣=0\left| \begin{array}{lll}{1 - {\lambda ^2}} & 2 & 0 \\ 2 & { - {\lambda ^2} + 1} & 0 \\ 2 & 2 & { - {\lambda ^2} - 1} \end{array} \right| = 0

(λ\lambda2 + 1)2 (3 −- λ\lambda2) = 0 λ\lambda = ±\pm

3\sqrt 3
Q10
The angle between the lines 2x=3y=−z2x=3y=-z and 6x=−y=−4z6x=-y=-4z is :
A 0∘{0^ \circ }
B 90∘{90^ \circ }
C 45∘{45^ \circ }
D 30∘{30^ \circ }
Correct Answer
Option B
Solution

The given lines are

2x=3y=−z2x = 3y = - z

or

          x3=y2=z−6\,\,\,\,\,\,\,\,\,\,{x \over 3} = {y \over 2} = {z \over { - 6}}
   [\,\,\,\left[ {} \right.

Dividing by

66
]\left. {} \right]

and

6x=−y=−4z6x = - y = - 4z

or

         x2=y−12=z−3\,\,\,\,\,\,\,\,\,{x \over 2} = {y \over { - 12}} = {z \over { - 3}}
    [\,\,\,\,\left[ {} \right.

Dividing by

1212
]\left. {} \right]

∴\therefore Angle between two lines is

cos⁡θ=3.2+2.(−12)+(−6).(−3)32+22+(−6)222+(−12)2+(−3)2\cos \theta = {{3.2 + 2.\left( { - 12} \right) + \left( { - 6} \right).\left( { - 3} \right)} \over {\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }}
=6−24+1849157=0⇒θ=90∘= {{6 - 24 + 18} \over {\sqrt {49} \sqrt {157} }} = 0 \Rightarrow \theta = {90^ \circ }
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