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3D Geometry

JEE Mathematics · 279 questions · Page 26 of 28 · Click an option or "Show Solution" to reveal answer

Q251
If the angel θ\theta between the line x+11=y−12=z−22{{x + 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2} and the plane 2x−y+λ  z+4=02x - y + \sqrt \lambda \,\,z + 4 = 0 is such that sin⁡  θ=13\sin \,\,\theta = {1 \over 3} then value of λ\lambda is :
A 53{5 \over 3}
B −35{-3 \over 5}
C 34{3 \over 4}
D −43{-4 \over 3}
Correct Answer
Option A
Solution

If θ\theta is the angle between line and plane then

(π2−0)\left( {{\pi \over 2} - 0} \right)

is the angle between line and normal to plane given by

cos⁡(π2−0)=(i^+2j^+2k^).(2i^−j^+λk^)34+1+λ\cos \left( {{\pi \over 2} - 0} \right) = {{\left( {\widehat i + 2\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j + \sqrt \lambda \widehat k} \right)} \over {3\sqrt {4 + 1 + \lambda } }}
cos⁡(π2−θ)=2−2+2λ3×5+λ\cos \left( {{\pi \over 2} - \theta } \right) = {{2 - 2 + 2\sqrt \lambda } \over {3 \times \sqrt 5 + \lambda }}
⇒sin⁡θ=2λ35+λ=13\Rightarrow \sin \theta = {{2\sqrt \lambda } \over {3\sqrt 5 + \lambda }} = {1 \over 3}
⇒4λ=5+λ⇒λ=53\Rightarrow 4\lambda = 5 + \lambda \Rightarrow \lambda = {5 \over 3}
Q252
If the angle between the line x=y−12=z−3λx = {{y - 1} \over 2} = {{z - 3} \over \lambda } and the plane x+2y+3z=4x+2y+3z=4 is cos⁡−1(514),{\cos ^{ - 1}}\left( {\sqrt {{5 \over {14}}} } \right), then λ\lambda equals :
A 32{3 \over 2}
B 25{2 \over 5}
C 53{5 \over 3}
D 23{2 \over 3}
Correct Answer
Option D
Solution

If θ\theta be the angle between the given line and plane, then

sin⁡θ=1×1+2×2+λ×312+22+λ2.12+22+32\sin \theta = {{1 \times 1 + 2 \times 2 + \lambda \times 3} \over {\sqrt {{1^2} + {2^2} + {\lambda ^2}} .\sqrt {{1^2} + {2^2} + {3^2}} }}
=5+3λ14.5+λ2= {{5 + 3\lambda } \over {\sqrt {14} .\sqrt {5 + {\lambda ^2}} }}

But it is given that

θ=cos⁡−1514⇒sin⁡θ=314\theta = {\cos ^{ - 1}}\sqrt {{5 \over {14}}} \Rightarrow \sin \theta = {3 \over {\sqrt {14} }}

∴\therefore

5+3λ145+λ2=314⇒λ=23{{5 + 3\lambda } \over {\sqrt {14} \sqrt {5 + {\lambda ^2}} }} = {3 \over {\sqrt {14} }} \Rightarrow \lambda = {2 \over 3}
Q253
A line with direction cosines proportional to 2,1,22,1,2 meets each of the lines x=y+a=zx=y+a=z and x+a=2y=2zx+a=2y=2z . The co-ordinates of each of the points of intersection are given by :
A (2a,3a,3a),(2a,a,a)\left( {2a,3a,3a} \right),\left( {2a,a,a} \right)
B (3a,2a,3a),(a,a,a)\left( {3a,2a,3a} \right),\left( {a,a,a} \right)
C (3a,2a,3a),(a,a,2a)\left( {3a,2a,3a} \right),\left( {a,a,2a} \right)
D (3a,3a,3a),(a,a,a)\left( {3a,3a,3a} \right),\left( {a,a,a} \right)
Correct Answer
Option B
Solution

Let a point on the line

x=y+a=zx = y + a = z

is

(λ,λ−a,λ)\left( {\lambda ,\lambda - a,\lambda } \right)

and a point on the line

x+a=2y=2zx + a = 2y = 2z

is

(μ−a,μ2,μ2),\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),

then direction ratio of the line joining these points are

λ−μ+a,  λ−a−μ2,  λ−μ2\lambda - \mu + a,\,\,\lambda - a - {\mu \over 2},\,\,\lambda - {\mu \over 2}

If it represents the required line, then

λ−μ+a2{{\lambda - \mu + a} \over 2}
=λ−a−μ21= {{\lambda - a - {\mu \over 2}} \over 1}
=λ−μ22= {{\lambda - {\mu \over 2}} \over 2}

on solving we get

λ=3a, μ=2a\lambda = 3a,\,\mu = 2a

∴\therefore The required points of intersection are

(3a,3a−a,3a)\left( {3a,3a - a,3a} \right)

and

(2a−a,2a2,2a2)\left( {2a - a,{{2a} \over 2},{{2a} \over 2}} \right)

or

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

and

(a,a,a)\left( {a,a,a} \right)
Q254
Let LL be the line of intersection of the planes 2x+3y+z=12x+3y+z=1 and x+3y+2z=2.x+3y+2z=2. If LL makes an angle α\alpha with the positive xx-axis, then cos α\alpha equals
A 11
B 12{1 \over {\sqrt 2 }}
C 13{1 \over {\sqrt 3 }}
D 12{1 \over 2}
Correct Answer
Option C
Solution

Let the direction cosines of line

LL

be

l,m,n,l,m,n,

then

2l+3m+n=02l+3m+n=0
       ....(i)\,\,\,\,\,\,\,....\left( i \right)

and

l+3m+2n=0          ....(ii)l + 3m + 2n = 0\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)

on solving equation

(i)(i)

and

(ii),(ii),

we get

l6−3=m1−4=n6−3{l \over {6 - 3}} = {m \over {1 - 4}} = {n \over {6 - 3}}
          ⇒l3=m−3=n3\,\,\,\,\,\,\,\,\,\, \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3}

Now

⇒l3=m−3=n3=l2+m2+n232+(−3)2+32\Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3} = {{\sqrt {{l^2} + {m^2} + {n^2}} } \over {\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }}

As

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

∴\therefore

l3=m−3=n3=127{l \over 3} = {m \over { - 3}} = {n \over 3} = {1 \over {\sqrt {27} }}
⇒l=327=13,  m=−13,n=13\Rightarrow l = {3 \over {\sqrt {27} }} = {1 \over {\sqrt 3 }},\,\,m = - {1 \over {\sqrt 3 }},n = {1 \over {\sqrt 3 }}

Line

L,L,

makes an angle α\alpha with

+ve+ve
xx

-axis ∴\therefore

l=cos⁡ α    ⇒   cos⁡α  =13l = \cos \,\alpha \,\,\,\, \Rightarrow \,\,\,\cos \alpha \,\, = {1 \over {\sqrt 3 }}
Q255
The d.r.d.r. of normal to the plane through (1,0,0),(0,1,0)(1, 0, 0), (0, 1, 0) which makes an angle π/4\pi /4 with plane x+y=3x+y=3 are :
A 1,2,11,\sqrt 2 ,1
B 1,1,21,1,\sqrt 2
C 1,1,21, 1, 2
D 2,1,1\sqrt 2 ,1,1
Correct Answer
Option B
Solution

Equation of plane through

(1,0,0)\left( {1,0,0} \right)

is

a(x−1)+by+cz=0          ...(i)a\left( {x - 1} \right) + by + cz = 0\,\,\,\,\,\,\,\,\,\,...\left( i \right)
(i)(i)

passes through

(0,1,0).\left( {0,1,0} \right).
−a+b=0⇒b=a;- a + b = 0 \Rightarrow b = a;

Also,

cos⁡45∘\cos {45^ \circ }
=a+a2(2a2+c2)⇒2a=2a2+c2= {{a + a} \over {\sqrt {2\left( {2{a^2} + {c^2}} \right)} }} \Rightarrow 2a = \sqrt {2{a^2} + {c^2}}
⇒2a2=c2\Rightarrow 2{a^2} = {c^2}
⇒c=2a.\Rightarrow c = \sqrt 2 a.

So

d.rd.r

of normal area

a,a,
a2aa\sqrt {2a}

i.e.

1,1,
1,2.1,\sqrt 2 .
Q256
Distance between two parallel planes 2x+y+2z=82x+y+2z=8 and 4x+2y+4z+5=04x+2y+4z+5=0 is :
A 32{3 \over 2}
B 52{5 \over 2}
C 72{7 \over 2}
D 92{9 \over 2}
Correct Answer
Option C
Solution
2x+y+2z−8=0            ...(Plane  1)2x + y + 2z - 8 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,1} \right)
2x+y+2z+52=0           ...(Plane  2)2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,2} \right)

Distance between Plane

11

and

22
=∣−8−5222+12+22∣= \left| {{{ - 8 - {5 \over 2}} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|
=∣−216∣=72= \left| {{{ - 21} \over 6}} \right| = {7 \over 2}
Q257
The distance between the line r→=2i^−2j^+3k^+λ(i−j+4k),\overrightarrow r = 2\widehat i - 2\widehat j + 3\widehat k + \lambda \left( {i - j + 4k} \right), and the plane r→.(i^+5j^+k^)=5\overrightarrow r .\left( {\widehat i + 5\widehat j + \widehat k} \right) = 5 is
A 109{{10} \over 9}
B 1033{{10} \over {3\sqrt 3 }}
C 310{{3} \over 10}
D 103{{10} \over 3}
Correct Answer
Option B
Solution

A point on lines is

(2,−2,3)\left( {2, - 2,3} \right)

its perpendicular distance from the plane

x+5y+z−5=0x + 5y + z - 5 = 0

is

=∣2−10+3−51+25+1∣=1033= \left| {{{2 - 10 + 3 - 5} \over {\sqrt {1 + 25 + 1} }}} \right| = {{10} \over {3\sqrt 3 }}
Q258
The intersection of the spheres x2+y2+z2+7x−2y−z=13{x^2} + {y^2} + {z^2} + 7x - 2y - z = 13 and x2+y2+z2−3x+3y+4z=8{x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane
A 2x−y−z=12x-y-z=1
B x−2y−z=1x-2y-z=1
C x−y−2z=1x-y-2z=1
D x−y−z=1x-y-z=1
Correct Answer
Option A
Solution

The equation of spheres are

S1:x2+y2+z2+7x−2y−z−13=0{S_1}:{x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0

and

S2:x2+y2+z2−3x+3y+4z−8=0{S_2}:{x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0

Their plane of intersection is

S1−S2=0{S_1} - {S_2} = 0
⇒10x−5y−5z−5=0\Rightarrow 10x - 5y - 5z - 5 = 0
⇒2x−y−z=1\Rightarrow 2x - y - z = 1
Q259
If 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, then kk can have :
A any value
B exactly one value
C exactly two values
D exactly three values
Correct Answer
Option C
Solution

Given lines will be coplanar If

    ∣−11111−kk21∣=0\,\,\,\,\left| \begin{array}{lll}{ - 1} & 1 & 1 \\ 1 & 1 & { - k} \\ k & 2 & 1 \end{array} \right| = 0
⇒−1(1+2k)−(1+k2)+1(2−k)=0\Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0
⇒k=0,−3\Rightarrow k = 0, - 3
Q260
The shortest distance between the lines x2=y2=z1{x \over 2} = {y \over 2} = {z \over 1} and x+2−1=y−48=z−54{{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4} lies in the interval :
A [0, 1)
B [1, 2)
C (2, 3]
D (3, 4]
Correct Answer
Option C
Solution

Shortest distance between the lines

x−x1a1=y−y1b1=z−z1c1{{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}}

and

x−x2a2=y−y2b2=z−z2c2{{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}}

is

∣∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2∣\left| {{{\left| \begin{array}{lll}{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \\ {{a_1}} & {{b_1}} & {{c_1}} \\ {{a_2}} & {{b_2}} & {{c_2}} \end{array} \right|} \over {\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}} \right|

∴\therefore Shortest distance between two given lines are,

∣∣−245221−184∣(8−8)2+(−1−8)2+(16+2)2∣\left| {{{\left| \begin{array}{lll}{ - 2} & 4 & 5 \\ 2 & 2 & 1 \\ { - 1} & 8 & 4 \end{array} \right|} \over {\sqrt {{{\left( {8 - 8} \right)}^2} + {{\left( { - 1 - 8} \right)}^2} + {{\left( {16 + 2} \right)}^2}} }}} \right|

=

∣−36+90405∣\left| {{{ - 36 + 90} \over {\sqrt {405} }}} \right|

=

5420.1{{54} \over {20.1}}

= 2.68

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