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3D Geometry

JEE Mathematics · 279 questions · Page 3 of 28 · Click an option or "Show Solution" to reveal answer

Q21
Statement - 1 : The point A(1,0,7)A(1,0,7) is the mirror image of the point B(1,6,3)B(1,6,3) in the line : x1=y−12=z−23{x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3} Statement - 2 : The line x1=y−12=z−23{x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3} bisects the line segment joining A(1,0,7)A(1,0,7) and B(1,6,3)B(1, 6, 3)
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

The directions ratios of the line segment joining points

A(1,0,7)  A\left( {1,0,7} \right)\,\,

and

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

are

0,6,−4.0,6, - 4.

The direction ratios of the given line are

1,2,3.1,2,3.

Clearly

1×0+2×6+3×(−4)=01 \times 0 + 2 \times 6 + 3 \times \left( { - 4} \right) = 0

So, the given line is perpendicular to line

AB.AB.

Also, the mid point of

AA

and

BB

is

(1,3,5)\left( {1,3,5} \right)

which lies on the given line. So, the image of

BB

in the given line is

AA

, because the given line is the perpendicular bisector of line segment joining points

AA

and

BB

, But statement -

22

is not a correct explanation for statement

−1.-1.
Q22
A equation of a plane parallel to the plane x−2y+2z−5=0x-2y+2z-5=0 and at a unit distance from the origin is :
A x−2y+2z−3=0x-2y+2z-3=0
B x−2y+2z+1=0x-2y+2z+1=0
C x−2y+2z−1=0x-2y+2z-1=0
D x−2y+2z+5=0x-2y+2z+5=0
Correct Answer
Option A
Solution

Given equation of a plane is

x−2y+2z−5=0x - 2y + 2z - 5 = 0

So, Equation of parallel plane is given by

x−2y+2z+d=0x - 2y + 2z + d = 0

Now, it is given that distance from origin to the parallel planes is

1.1.

∴\therefore

∣d12+22+23∣=1⇒d=±3\left| {{d \over {\sqrt {{1^2} + {2^2} + {2^3}} }}} \right| = 1 \Rightarrow d = \pm 3

So equation of required plane

x−2y+2z±3=0x - 2y + 2z \pm 3 = 0
Q23
If the line x−12=y+13=z−14{{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4} and x−31=y−k2=z1{{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1} intersect, then kk is equal to :
A −1-1
B 29{2 \over 9}
C 92{9 \over 2}
D 00
Correct Answer
Option C
Solution

Given lines in vector form are

r→=(i^−j→+k→)+λ(2i→+3j→+4j→)\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)

and

r→=(3i^+kj^)+μ(i^+2j^+k^)\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)

These will intersect if shortest distance between them

=0=0

i.e.

(a→2−a→1).b1→×b→2=0\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}} \times {\overrightarrow b _2} = 0
⇒∣3−1k+1−1234121∣=0\Rightarrow \left| \begin{array}{lll}{3 - 1} & {k + 1} & { - 1} \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} \right| = 0
⇒2(−5)−(k+1)(−2)−1(1)=0\Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0
⇒k=9/2\Rightarrow k = 9/2
Q24
The image of the line x−13=y−31=z−4−5 {{x - 1} \over 3} = {{y - 3} \over 1} = {{z - 4} \over { - 5}}\, in the plane 2x−y+z+3=02x-y+z+3=0 is the line :
A x−33=y+51=z−2−5{{x - 3} \over 3} = {{y + 5} \over 1} = {{z - 2} \over { - 5}}
B x−3−3=y+5−1=z−25 {{x - 3} \over { - 3}} = {{y + 5} \over { - 1}} = {{z - 2} \over 5}\,
C x+33=y−51=z−2−5 {{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}\,
D x+3−3=y−5−1=z+25{{x + 3} \over { - 3}} = {{y - 5} \over { - 1}} = {{z + 2} \over 5}
Correct Answer
Option C
Solution
a−12=b−3−1=c−41=λ(let){{a - 1} \over 2} = {{b - 3} \over { - 1}} = {{c - 4} \over 1} = \lambda \left( {let} \right)
⇒a=2λ+1,  b=3−λ,  c=4+λ\Rightarrow a = 2\lambda + 1,\,\,b = 3 - \lambda ,\,\,c = 4 + \lambda
p=(a+12,b+32,c+42)p = \left( {{{a + 1} \over 2},{{b + 3} \over 2},{{c + 4} \over 2}} \right)
=(λ+1,6−λ2,λ+82)= \left( {\lambda + 1,{{6 - \lambda } \over 2},{{\lambda + 8} \over 2}} \right)

∴\therefore

2(λ+1)−6−λ2+λ+82+3=02\left( {\lambda + 1} \right) - {{6 - \lambda } \over 2} + {{\lambda + 8} \over 2} + 3 = 0
3λ+6=0⇒λ=−23\lambda + 6 = 0 \Rightarrow \lambda = - 2
a=−3,b=5,c=2a = - 3,b = 5,c = 2

Required line is

x+33=y−51=z−2−5{{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}
Q25
The angle between the lines whose direction cosines satisfy the equations l+m+n=0l+m+n=0 and l2=m2+n2{l^2} = {m^2} + {n^2} is :
A π6{\pi \over 6}
B π2{\pi \over 2}
C π3{\pi \over 3}
D π4{\pi \over 4}
Correct Answer
Option C
Solution

Given

l+m+n=0l + m + n = 0

and

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

Now,

(−m−n)2=m2+n2{\left( { - m - n} \right)^2} = {m^2} + {n^2}
⇒mn=0⇒m=0  \Rightarrow mn = 0 \Rightarrow m = 0\,\,

or

  n=0\,\,n = 0

If

m=0m=0

then

l=−nl=-n

We know

l2+m2+n2=1⇒n=±12{l^2} + {m^2} + {n^2} = 1 \Rightarrow n = \pm {1 \over {\sqrt 2 }}

i.e.

(l1,m1,n1)=(−12,0,12)\left( {{l_1},{m_1},{n_1}} \right) = \left( { - {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }}} \right)

If

n=0n=0

then

l=−ml=-m
l2+m2+n2=1   ⇒2m2=1{l^2} + {m^2} + {n^2} = 1\,\,\, \Rightarrow 2{m^2} = 1
⇒m=±12\Rightarrow m = \pm {1 \over {\sqrt 2 }}

Let

m=12⇒l=−12m = {1 \over {\sqrt 2 }} \Rightarrow l = - {1 \over {\sqrt 2 }}

and

n=0n=0
(l2,m2,n2)=(−12,12,0)\left( {{l_2},{m_2},{n_2}} \right) = \left( { - {1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }},0} \right)

∴\therefore

cos⁡θ=12⇒θ=π3\cos \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 3}
Q26
The distance of the point (1,0,2)(1, 0, 2) from the point of intersection of the line x−23=y+14=z−212{{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}} and the plane x−y+z=16,x - y + z = 16, is :
A 3213\sqrt {21}
B 1313
C 2142\sqrt {14}
D 88
Correct Answer
Option B
Solution

General point on given line

≡P(3r+2,4r−1,12r+2)\equiv P\left( {3r + 2,4r - 1,12r + 2} \right)

Point

PP

must satisfy equation of plane

(3r+2)−(4r−1)+(12r+2)=16\left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16
11r+5=1611r + 5 = 16
r=1r=1
P(3×1+2,4×1−1,12×1+2)=P(5,3,14)P\left( {3 \times 1 + 2,4 \times 1 - 1,12 \times 1 + 2} \right) = P\left( {5,3,14} \right)

Distance between

PP

and

(1,0,2)(1,0,2)
D=(5−1)2+32+(14−2)2=13D = \sqrt {{{\left( {5 - 1} \right)}^2} + {3^2} + {{\left( {14 - 2} \right)}^2}} = 13
Q27
The equation of the plane containing the line 2x−5y+z=3;x+y+4z=5,2x-5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1,x+3y+6z=1, is :
A x+3y+6z=7x+3y+6z=7
B 2x+6y+12z=−132x+6y+12z=-13
C 2x+6y+12z=132x+6y+12z=13
D x+3y+6z=−7x+3y+6z=-7
Correct Answer
Option A
Solution

Equation of the plane containing the lines

2x−5y+z=32x - 5y + z = 3

and

x+y+4z=5x + y + 4z = 5

is

2x−5y+z−3+λ(x+y+4z−5)=02x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right) = 0
⇒(2+λ)x+(−5+λ)y+(1+4λ)z+\Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z +
               (−3−5λ)=0....(i)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 - 5\lambda } \right) = 0....\left( i \right)

Since the plane

(i)(i)

parallel to the given plane

x+3y+6z=1x + 3y + 6z = 1

∴\therefore

   2+λ1=−5+λ3=1+4λ6\,\,\,{{2 + \lambda } \over 1} = {{ - 5 + \lambda } \over 3} = {{1 + 4\lambda } \over 6}
                ⇒λ=−112\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \lambda = - {{11} \over 2}

Hence equation of the required plane is

(2−112)x+(−5−112)y+(1−442)z+\left( {2 - {{11} \over 2}} \right)x + \left( { - 5 - {{11} \over 2}} \right)y + \left( {1 - {{44} \over 2}} \right)z +
                      (−3+552)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 + {{55} \over 2}} \right) = 0
⇒x+3y+6z=7\Rightarrow x + 3y + 6z = 7
Q28
The distance of the point (1,−5,9)(1,-5,9) from the plane x−y+z=5x-y+z=5 measured along the line x=y=zx=y=z is :
A 103{{10} \over {\sqrt 3 }}
B 203{20 \over 3}
C 3103\sqrt {10}
D 10310\sqrt {3}
Correct Answer
Option D
Solution
eqn  e{q^n}\,\,

of

  PO:  \,\,PO:\,\,
x−11=y+51=z−91=λ{{x - 1} \over 1} = {{y + 5} \over 1} = {{z - 9} \over 1} = \lambda
⇒x=λ+1;  y=λ−5;  z=λ+9\Rightarrow x = \lambda + 1;\,\,y = \lambda - 5;\,\,z = \lambda + 9

Putting these in

eqne{q^n}

of plane :-

λ+1−λ+5+λ+9=5\lambda + 1 - \lambda + 5 + \lambda + 9 = 5
⇒λ=−10\Rightarrow \lambda = - 10
⇒O\Rightarrow O

is

(−9,−15,−1)\left( { - 9, - 15, - 1} \right)

⇒\Rightarrow

   \,\,\,

Distance

OP=103.OP = 10\sqrt 3 .
Q29
ABC is a triangle in a plane with vertices A(2, 3, 5), B(−1, 3, 2) and C(λ\lambda , 5, μ\mu ). If the median through A is equally inclined to the coordinate axes, then the value of (λ\lambda 3 + μ\mu 3 + 5) is :
A 1130
B 1348
C 676
D 1077
Correct Answer
Option B
Solution

DR's of AD are

λ−12−2,5+32−3,μ+22−5{{\lambda - 1} \over 2} - 2,{{5 + 3} \over 2} - 3,{{\mu + 2} \over 2} - 5

i.e.

λ−52,  1,  μ−82{{\lambda - 5} \over 2},\,\,1,\,\,{{\mu - 8} \over 2}

As medium is making equal angles with coordinate axes, ∴\therefore

λ−52=1=μ−82{{\lambda - 5} \over 2} = 1 = {{\mu - 8} \over 2}

⇒\Rightarrow λ\lambda = 7, μ\mu = 10 ∴\therefore λ\lambda3 + μ\mu3 + 5 = 73 + 103 + 5 = 1348

Q30
If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of Δ\Delta ABC is :
A 1x2+1y2+1z2=1{1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1
B 1x2+1y2+1z2=3{1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 3
C 1x2+1y2+1z2=19{1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = {1 \over 9}
D 1x2+1y2+1z2=9{1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 9
Correct Answer
Option A
Solution

Suppose centroid be (h, k,

ll

) ∴\therefore x −- intp == 3h, y −- intp == 3k, z −- intp == 3

ll

Equation

x3h+y3k+z3l=1{x \over {3h}} + {y \over {3k}} + {z \over {3l}} = 1

∴\therefore Distance from (0, 0, 0)

∣−119h2+19k2+19l2∣=3\left| {{{ - 1} \over {\sqrt {{1 \over {9{h^2}}} + {1 \over {9{k^2}}} + {1 \over {9{l^2}}}} }}} \right| = 3

⇒\Rightarrow

1x2+1y2+1z2=1{1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1
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