Practice Start Test

3D Geometry

JEE Mathematics · 279 questions · Page 2 of 28 · Click an option or "Show Solution" to reveal answer

Q11
The plane x+2y−z=4x+2y-z=4 cuts the sphere x2+y2+z2−x+z−2=0{x^2} + {y^2} + {z^2} - x + z - 2 = 0 in a circle of radius
A 33
B 11
C 22
D 2{\sqrt 2 }
Correct Answer
Option B
Solution

Perpendicular distance of center

(12,0,−12)\left( {{1 \over 2},0, - {1 \over 2}} \right)

from

x+2y−2=4x+2y-2=4

is given by

∣12+12−4∣=326{{\left| {{1 \over 2} + {1 \over 2} - 4} \right| = \sqrt {{3 \over 2}} } \over {\sqrt 6 }}

radius of sphere

=14+14+2=52= \sqrt {{1 \over 4} + {1 \over 4} + 2} = \sqrt {{5 \over 2}}

∴\therefore radius of circle

=52−32=1.= \sqrt {{5 \over 2} - {3 \over 2}} = 1.
Q12
The image of the point (−1,3,4)(-1, 3,4) in the plane x−2y=0x-2y=0 is :
A (−173,−193,4)\left( { - {{17} \over 3}, - {{19} \over 3},4} \right)
B (15,11,4)(15,11,4)
C (−173,−193,1)\left( { - {{17} \over 3}, - {{19} \over 3},1} \right)
D None of these
Correct Answer
Option D
Solution
Eqn    E{q^n}\,\,\,\,

of

    PN:\,\,\,\,PN:

-

x+11=y−3−2=z−40=λ{{x + 1} \over 1} = {{y - 3} \over { - 2}} = {{z - 4} \over 0} = \lambda
N(λ−1,−2λ+3−4)N\left( {\lambda - 1, - 2\lambda + 3 - 4} \right)

It lies on

x−2y=0x-2y=0
⇒λ−1+4λ−6=0\Rightarrow \lambda - 1 + 4\lambda - 6 = 0
⇒λ=7/5\Rightarrow \lambda = 7/5
N(25,15,4)N\left( {{2 \over 5},{1 \over 5},4} \right)
NN

is mid point of

PP′PP'

∴\therefore

α−1=45,β+3=25,r+4=8\alpha - 1 = {4 \over 5},\beta + 3 = {2 \over 5},r + 4 = 8
⇒α=95,β=−135,r=4\Rightarrow \alpha = {9 \over 5},\beta = {{ - 13} \over 5},r = 4

∴\therefore Image is

(95,−135,4)\left( {{9 \over 5},{{ - 13} \over 5},4} \right)
Q13
The two lines x=ay+b,z=cy+d;x=ay+b, z=cy+d; and x=a′y+b′,x=a'y+b' , z=c′y+d′z=c'y+d' are perpendicular to each other if :
A aa′+cc′=−1aa'+cc'=-1
B aa′+cc′=1aa'+cc'=1
C aa′+cc′=−1{a \over {a'}} + {c \over {c'}} = - 1
D aa′+cc′=1{a \over {a'}} + {c \over {c'}} = 1
Correct Answer
Option A
Solution

Equation of lines

x−ba=y1=z−dc{{x - b} \over a} = {y \over 1} = {{z - d} \over c}
x−b′a′=y1=z−d′c′{{x - b'} \over {a'}} = {y \over 1} = {{z - d'} \over {c'}}

Line are perpendicular

⇒aa′+1+cc′=0\Rightarrow aa' + 1 + cc' = 0
Q14
If a line makes an angle of π/4\pi /4 with the positive directions of each of xx-axis and yy-axis, then the angle that the line makes with the positive direction of the zz-axis is :
A π4{\pi \over 4}
B π2{\pi \over 2}
C π6{\pi \over 6}
D π3{\pi \over 3}
Correct Answer
Option B
Solution

Let the angle of line makes with the positive direction of

zz

-axis is α\alpha direction cosines of line with the

+ve+ve

directions of

xx

-axis,

yy

-axis, and

zz

-axis is

l,l,
m,m,
nn

respectively. ∴\therefore

l=cos⁡π4,m=cos⁡π4,  n=cos αl = \cos {\pi \over 4},m = \cos {\pi \over 4},\,\,n = cos\,\alpha

as we know that,

l2+m2+n2=1{l^2} + {m^2} + {n^2} = 1

∴\therefore

cos⁡2π4+cos⁡2π4+cos⁡2α=1{\cos ^2}{\pi \over 4} + {\cos ^2}{\pi \over 4} + {\cos ^2}\alpha = 1
⇒12+12+cos⁡2α=1\Rightarrow {1 \over 2} + {1 \over 2} + {\cos ^2}\alpha = 1
⇒cos⁡2α=0⇒α=π2\Rightarrow {\cos ^2}\alpha = 0 \Rightarrow \alpha = {\pi \over 2}

Hence, angle with positive direction of the

zz

-axis is

π2{\pi \over 2}
Q15
If (2,3,5)(2,3,5) is one end of a diameter of the sphere x2+y2+z2−6x−12y−2z+20=0,{x^2} + {y^2} + {z^2} - 6x - 12y - 2z + 20 = 0, then the coordinates of the other end of the diameter are
A (4,3,5)(4, 3, 5)
B (4,3,−3)(4, 3, -3)
C (4,9,−3)(4, 9, -3)
D (4,−3,3)(4, -3, 3)
Correct Answer
Option C
Solution

For given sphere center is

(3,6,1)\left( {3,6,1} \right)

Coordinates of one end of diameter of the sphere are

(2,3,5).\left( {2,3,5} \right).

Let the coordinates of the other end of diameter are

(α,β,γ)\left( {\alpha ,\beta ,\gamma } \right)

∴\therefore

α+22=3,  β+32=6,  γ+52=1{{\alpha + 2} \over 2} = 3,\,\,{{\beta + 3} \over 2} = 6,\,\,{{\gamma + 5} \over 2} = 1
⇒α=4,  β=9  \Rightarrow \alpha = 4,\,\,\beta = 9\,\,

and

  γ=−3\,\,\gamma = - 3

∴\therefore Coordinate of other end of diameter are

(4,9,−3)\left( {4,9, - 3} \right)
Q16
If the straight lines      \,\,\,\,\,      \,\,\,\,\, x−1k=y−22=z−33{{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}      \,\,\,\,\, and     \,\,\,\,\, x−23=y−3k=z−12{{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2} intersects at a point, then the integer kk is equal to
A −5-5
B 55
C 22
D −2-2
Correct Answer
Option A
Solution

The two lines intersect if shortest distance between them is zero

i.e.i.e.
(a→2−a→1).b→1×b→2∣b→1×b→2∣=0{{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0
⇒(a→2−a→1).b→1×b→2=0\Rightarrow \left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{\overrightarrow b _1} \times {\overrightarrow b _2} = 0

where

a→1=i^+2j^+3k^,b→1=ki^+2j^+3k^{\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k,{\overrightarrow b _1} = k\widehat i + 2\widehat j + 3\widehat k
       a→2=2i^+3j^+k^,  b^2=3i^+kj^+2k^\,\,\,\,\,\,\,{\overrightarrow a _2} = 2\widehat i + 3\widehat j + \widehat k,\,\,{\widehat b_2} = 3\widehat i + k\widehat j + 2\widehat k
⇒∣11−2k233k2∣=0\Rightarrow \left| \begin{array}{lll}1 & 1 & { - 2} \\ k & 2 & 3 \\ 3 & k & 2 \end{array} \right| = 0
⇒1(4−3k)−1(2k−9)−2(k2−6)=0\Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0
⇒−2k2−5k+25=0⇒k=−5\Rightarrow - 2{k^2} - 5k + 25 = 0 \Rightarrow k = - 5
    \,\,\,\,

or

    \,\,\,\,
52{5 \over 2}

As

kk

is an integer, therefore

k=−5k=-5
Q17
The line passing through the points (5,1,a)(5,1,a) and (3,b,1)(3, b, 1) crosses the yzyz-plane at the point (0,172,−−132)\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right) . Then
A a=2,a=2, b=8b=8
B a=4,a=4, b=6b=6
C a=6,a=6, b=4b=4
D a=8,a=8, b=2b=2
Correct Answer
Option C
Solution

Equation of line through

(5,1,a)\left( {5,1,a} \right)

and

(3,b,1)\left( {3,b,1} \right)

is

x−5−2=y−1b−1=z−a1−a=λ{{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda

∴\therefore Any point on this line is a

[−2λ+5,(b−1)λ+1,(1−α)λ+a]\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \alpha } \right)\lambda + a} \right]

It crosses

yzyz

plane where

−2λ+5=0{ - 2\lambda + 5 = 0}
λ=52\lambda = {5 \over 2}

∴\therefore

(0,(b−1)52+1,(1−a)52+a)=(0,172,−172)\left( {0,\left( {b - 1} \right){5 \over 2} + 1,\left( {1 - a} \right){5 \over 2} + a} \right) = \left( {0,{{17} \over 2},{{ - 17} \over 2}} \right)
⇒(b−1)52+1=172\Rightarrow \left( {b - 1} \right){5 \over 2} + 1 = {{17} \over 2}

and

(1−a)52+a=−132\left( {1 - a} \right){5 \over 2} + a = - {{13} \over 2}
⇒b=4\Rightarrow b = 4

and

a=6a = 6
Q18
Let the line      \,\,\,\,\, x−23=y−1−5=z+22{{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2} lie in the plane      \,\,\,\,\, x+3y−αz+β=0.x + 3y - \alpha z + \beta = 0. Then (α,β)\left( {\alpha ,\beta } \right) equals
A (−6,7)(-6,7)
B (5,−15)(5,-15)
C (−5,5)(-5,5)
D (6,−17)(6, -17)
Correct Answer
Option A
Solution

As the line

x−23=y−1−5=z+22{{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}

lie in the plane

x+3y−αz+β=0x + 3y - \alpha z + \beta = 0

∴\therefore

Pt(2,1,−2)Pt\left( {2,1, - 2} \right)

lies on the plane i.e.

2+3+2α+β=02 + 3 + 2\alpha + \beta = 0

or

    2α+β+5=0             ....(i)\,\,\,\,2\alpha + \beta + 5 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)

Also normal to plane will be perpendicular to line, ∴\therefore

   3×1−5×3+2×(−α)=0\,\,\,3 \times 1 - 5 \times 3 + 2 \times \left( { - \alpha } \right) = 0
⇒α=−6\Rightarrow \alpha = - 6

From equation

(i)(i)

then,

β=7\beta = 7

∴\therefore

(α,β)=(−6,7)\left( {\alpha ,\beta } \right) = \left( { - 6,7} \right)
Q19
A line ABAB in three-dimensional space makes angles 45∘{45^ \circ } and 120∘{120^ \circ } with the positive xx-axis and the positive yy-axis respectively. If ABAB makes an acute angle θ\theta with the positive zz-axis, then θ\theta equals :
A 45∘{45^ \circ }
B 60∘{60^ \circ }
C 75∘{75^ \circ }
D 30∘{30^ \circ }
Correct Answer
Option B
Solution

Direction cosines of the line :

ℓ=cos⁡45∘=12,m=cos⁡120∘=−12,π=cos⁡θ\ell = \cos {45^ \circ } = {1 \over {\sqrt 2 }},m = \cos {120^ \circ } = {{ - 1} \over 2},\pi = \cos \theta

where θ\theta is the angle, which line makes with positive

zz

-axis. Now

ℓ2+m2+n2=1{\ell ^2} + {m^2} + {n^2} = 1
⇒12+14+cos⁡2θ=1,  cos⁡2θ=14\Rightarrow {1 \over 2} + {1 \over 4} + {\cos ^2}\theta = 1,\,\,{\cos ^2}\theta = {1 \over 4}
⇒cos⁡θ=12    (θ\Rightarrow \cos \theta = {1 \over 2}\,\,\,\,\left( \theta \right.

being acute)

⇒0=π3\Rightarrow 0 = {\pi \over 3}
Q20
Statement-1 : The point A(3,1,6)A(3, 1, 6) is the mirror image of the point B(1,3,4)B(1, 3, 4) in the plane x−y+z=5.x-y+z=5. Statement-2 : The plane x−y+z=5x-y+z=5 bisects the line segment joining A(3,1,6)A(3, 1, 6) and B(1,3,4).B(1, 3, 4).
A Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B Statement - 1 is true, Statement - 2 is false.
C Statement - 1 is false , Statement - 2 is true.
D Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
Correct Answer
Option A
Solution
A(3,1,6);  B=(1,3,4)A\left( {3,1,6} \right);\,\,B = \left( {1,3,4} \right)

Mid-point of

AB=(2,2,5)AB = \left( {2,2,5} \right)

lies on the plane. and d.r's of

AB=(2,−2,2)AB=(2,-2,2)

d.r's of normal to plane

=(1,−1,1).= \left( {1, - 1,1} \right).

Direction ratio of

ABAB

and normal to the plane are proportional therefore,

ABAB

is perpendicular to the plane ∴\therefore

AA

is image of

BB

Statement-

22

is correct but it is not correct explanation.

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