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3D Geometry

JEE Mathematics · 279 questions · Page 10 of 28 · Click an option or "Show Solution" to reveal answer

Q91
Let the plane passing through the point (−-1, 0, −-2) and perpendicular to each of the planes 2x + y −- z = 2 and x −- y −- z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to :
A 3
B 8
C 5
D 4
Correct Answer
Option D
Solution

Normal of required plane

(2i^+j^−k^)×(i^−j^−k^)\left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right)
=−2i^+j^−3k^= - 2\widehat i + \widehat j - 3\widehat k

Equation of plane

−2(x+1)+1(y−0)−3(z+2)=0- 2(x + 1) + 1(y - 0) - 3(z + 2) = 0
−2x+y−3z−8=0- 2x + y - 3z - 8 = 0
2x−y+3z+8=02x - y + 3z + 8 = 0
a+b+c=4a + b + c = 4
Q92
A plane P contains the line x+2y+3z+1=0=x−y−z−6x + 2y + 3z + 1 = 0 = x - y - z - 6, and is perpendicular to the plane −2x+y+z+8=0 - 2x + y + z + 8 = 0. Then which of the following points lies on P?
A (−-1, 1, 2)
B (0, 1, 1)
C (1, 0, 1)
D (2, −-1, 1)
Correct Answer
Option B
Solution

Equation of plane P can be assumed as P : x + 2y + 3z + 1 + λ\lambda (x −- y −- z −- 6) = 0 ⇒\Rightarrow P : (1 + λ\lambda)x + (2 −- λ\lambda)y + (3 −- λ\lambda)z + 1 −- 6λ\lambda = 0

⇒n→1=(1+λ)i^+(2−λ)j^+(3−λ)k^\Rightarrow {\overrightarrow n _1} = (1 + \lambda )\widehat i + (2 - \lambda )\widehat j + (3 - \lambda )\widehat k

∴\therefore

n→1 . n→2=0{\overrightarrow n _1}\,.\,{\overrightarrow n _2} = 0

⇒\Rightarrow 2(1 + λ\lambda) −- (2 −- λ\lambda) −- (3 −- λ\lambda) = 0 ⇒\Rightarrow 2 + 2λ\lambda −- 2 + λ\lambda −- 3 + λ\lambda = 0 ⇒\Rightarrow λ\lambda =

34{3 \over 4}

⇒\Rightarrow

P:7x4+54y+9z4−144=0P:{{7x} \over 4} + {5 \over 4}y + {{9z} \over 4} - {{14} \over 4} = 0

⇒\Rightarrow 7x + 5y + 9z = 14 (0, 1, 1) lies on P.

Q93
Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes r→ . (i^+j^+4k^)=16\overrightarrow r \,.\,\left( {\widehat i + \widehat j + 4\widehat k} \right) = 16 and r→ . (−i^+j^+k^)=6\overrightarrow r \,.\,\left( { - \widehat i + \widehat j + \widehat k} \right) = 6. Then which of the following points does NOT lie on P?
A (3, 3, 2)
B (6, −-6, 2)
C (4, 2, 2)
D (−-8, 8, 6)
Correct Answer
Option C
Solution
(x+y+4z−16)+λ(−x+y+z−6)=0(x + y + 4z - 16) + \lambda ( - x + y + z - 6) = 0

Passes through (1, 2, 3)

−1+λ(−2)⇒λ=−12- 1 + \lambda ( - 2) \Rightarrow \lambda = - {1 \over 2}
2(x+y+4z−16)−(−x+y+z−6)=02(x + y + 4z - 16) - ( - x + y + z - 6) = 0
3x+y+7z−26=03x + y + 7z - 26 = 0
Q94
The distance of the point (1, −-2, 3) from the plane x −- y + z = 5 measured parallel to a line, whose direction ratios are 2, 3, −-6 is :
A 3
B 5
C 2
D 1
Correct Answer
Option D
Solution
(1+2λ)+2−3λ+3−6λ=5(1 + 2\lambda ) + 2 - 3\lambda + 3 - 6\lambda = 5
⇒6−7λ=5⇒λ=17\Rightarrow 6 - 7\lambda = 5 \Rightarrow \lambda = {1 \over 7}

so,

P=(97,−117,157)P = \left( {{9 \over 7}, - {{11} \over 7},{{15} \over 7}} \right)
AP=(1−97)2+(−2+117)2+(3−157)2AP = \sqrt {{{\left( {1 - {9 \over 7}} \right)}^2} + {{\left( { - 2 + {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}}
AP=(449)+949+3649=1AP = \sqrt {\left( {{4 \over {49}}} \right) + {9 \over {49}} + {{36} \over {49}}} = 1
Q95
Equation of a plane at a distance 221\sqrt {{2 \over {21}}} from the origin, which contains the line of intersection of the planes x −- y −- z −- 1 = 0 and 2x + y −- 3z + 4 = 0, is :
A 3x−y−5z+2=03x - y - 5z + 2 = 0
B 3x−4z+3=03x - 4z + 3 = 0
C −x+2y+2z−3=0 - x + 2y + 2z - 3 = 0
D 4x−y−5z+2=04x - y - 5z + 2 = 0
Correct Answer
Option D
Solution

Required equation of plane

P1+λP2=0{P_1} + \lambda {P_2} = 0
(x−y−z−1)+λ(2x+y−3z+4)=0(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0

Given that its dist. From origin is

221{2 \over {\sqrt {21} }}

Thus,

∣4λ−1∣(2λ+1)2+(λ−1)2+(−3λ−1)2=221{{|4\lambda - 1|} \over {\sqrt {{{(2\lambda + 1)}^2} + {{(\lambda - 1)}^2} + {{( - 3\lambda - 1)}^2}} }} = {{\sqrt 2 } \over {\sqrt {21} }}
⇒21(4λ−1)2=2(14λ2+8λ+3)\Rightarrow 21{(4\lambda - 1)^2} = 2(14{\lambda ^2} + 8\lambda + 3)
⇒336λ2−168λ+21=28λ2+16λ+6\Rightarrow 336{\lambda ^2} - 168\lambda + 21 = 28{\lambda ^2} + 16\lambda + 6
⇒308λ2−184λ+15=0\Rightarrow 308{\lambda ^2} - 184\lambda + 15 = 0
⇒308λ2−154λ−30λ+15=0\Rightarrow 308{\lambda ^2} - 154\lambda - 30\lambda + 15 = 0
⇒(2λ−1)(154λ−15)=0\Rightarrow (2\lambda - 1)(154\lambda - 15) = 0
⇒λ=12\Rightarrow \lambda = {1 \over 2}

or

15154{{15} \over {154}}

for

λ=12\lambda = {1 \over 2}

reqd. plane is

4x−y−5z+2=04x - y - 5z + 2 = 0
Q96
The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m −- n = 0 and mn + nl + lm = 0, is :
A π2{\pi \over 2}
B π−cos⁡−1(49)\pi - {\cos ^{ - 1}}\left( {{4 \over 9}} \right)
C cos⁡−1(89){\cos ^{ - 1}}\left( {{8 \over 9}} \right)
D π3{\pi \over 3}
Correct Answer
Option A
Solution

n = 2 (l + m) lm + n(l + m) = 0 lm + 2(l + m)2 = 0 2l2 + 2m2 + 5ml = 0

2(lm)2+2+5(lm)=02{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0

2t2 + 5t + 2 = 0 (t + 2)(2t + 1) = 0

⇒t=−2;−12\Rightarrow t = - 2; - {1 \over 2}

(i)

lm=−2{l \over m} = - 2
nm=−2{n \over m} = - 2

(−-2m, m, −-2m) (−-2, 1, −-2) (ii)

lm=−12{l \over m} = - {1 \over 2}

n = −-2l (l, −-2l, −-2l) (1, −-2, −-2)

cos⁡θ=−2−2+499=0⇒0=π2\cos \theta = {{ - 2 - 2 + 4} \over {\sqrt 9 \sqrt 9 }} = 0 \Rightarrow 0 = {\pi \over 2}
Q97
The equation of the plane passing through the line of intersection of the planes r→.(i^+j^+k^)=1\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1 and r→.(2i^+3j^−k^)+4=0\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0 and parallel to the x-axis is :
A r→.(j^−3k^)+6=0\overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0
B r→.(i^+3k^)+6=0\overrightarrow r .\left( {\widehat i + 3\widehat k} \right) + 6 = 0
C r→.(i^−3k^)+6=0\overrightarrow r .\left( {\widehat i - 3\widehat k} \right) + 6 = 0
D r→.(j^−3k^)−6=0\overrightarrow r .\left( {\widehat j - 3\widehat k} \right) - 6 = 0
Correct Answer
Option A
Solution

Equation of planes are

r→.(i^+j^+k^)−1=0⇒x+y+z−1=0\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) - 1 = 0 \Rightarrow x + y + z - 1 = 0

and

r→.(2i^+3j^−k^)+4=0⇒2x+3y−z+4=0\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0 \Rightarrow 2x + 3y - z + 4 = 0

equation of planes through line of intersection of these planes is :-

(x+y+z−1)+λ(2x+3y−z+4)=0(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0
⇒(1+2λ)x+(1+3λ)y+(1−λ)z−1+4λ=0\Rightarrow (1 + 2\lambda )x + (1 + 3\lambda )y + (1 - \lambda )z - 1 + 4\lambda = 0

But this plane is parallel to x-axis whose direction are (1, 0, 0) ∴\therefore

(1+2λ)1+(1+3λ)0+(1−λ)0=0(1 + 2\lambda )1 + (1 + 3\lambda )0 + (1 - \lambda )0 = 0
λ=−12\lambda = - {1 \over 2}

∴\therefore Required plane is

0x+(1−32)y+(1+12)z−1+4(−12)=00x + \left( {1 - {3 \over 2}} \right)y + \left( {1 + {1 \over 2}} \right)z - 1 + 4\left( {{{ - 1} \over 2}} \right) = 0
⇒−y2+32z−3=0\Rightarrow {{ - y} \over 2} + {3 \over 2}z - 3 = 0
⇒y−3z+6=0\Rightarrow y - 3z + 6 = 0
⇒r→.(j^−3k^)+6=0\Rightarrow \overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0

Ans.

Q98
Let the equation of the plane, that passes through the point (1, 4, −-3) and contains the line of intersection of the planes 3x −- 2y + 4z −- 7 = 0 and x + 5y −- 2z + 9 = 0, be α\alphax + β\betay + γ\gammaz + 3 = 0, then α\alpha + β\beta + γ\gamma is equal to :
A −-23
B −-15
C 23
D 15
Correct Answer
Option A
Solution

3x −- 2y + 4z −- 7 + λ\lambda(x + 5y −- 2z + 9) = 0 (3 + λ\lambda)x + (5λ\lambda −- 2)y + (4 −- 2λ\lambda)z + 9λ\lambda −- 7 = 0 passing through (1, 4, −-3) ⇒\Rightarrow 3 + λ\lambda + 20λ\lambda −- 8 −- 12 + 6λ\lambda + 9λ\lambda −- 7 = 0 ⇒\Rightarrow λ\lambda =

23{2 \over 3}

⇒\Rightarrow equation of plane is −-11x −- 4y −- 8z + 3 = 0 ⇒\Rightarrow α\alpha + β\beta + γ\gamma = −-23

Q99
The distance of the point (−-1, 2, −-2) from the line of intersection of the planes 2x + 3y + 2z = 0 and x −- 2y + z = 0 is :
A 12{1 \over {\sqrt 2 }}
B 52{5 \over 2}
C 422{{\sqrt {42} } \over 2}
D 342{{\sqrt {34} } \over 2}
Correct Answer
Option D
Solution

P1 : 2x + 3y + 2z = 0 ⇒\Rightarrow

n→1=2i^+3j^+2k^{\overrightarrow n _1} = 2\widehat i + 3\widehat j + 2\widehat k

P2 : x −- 2y + z = 0 ⇒\Rightarrow

n→2=i^−2j^+k^{\overrightarrow n _2} = \widehat i - 2\widehat j + \widehat k

Direction vector of line L which is line of intersection of P1 & P2

r→=n→1×n→2=7i^−7k^\overrightarrow r = {\overrightarrow n _1} \times {\overrightarrow n _2} = 7\widehat i - 7\widehat k

DR's of L are (1, 0, −-1) ⇒\Rightarrow Equation of L :

x1=y0=z−1=λ{x \over 1} = {y \over 0} = {z \over { - 1}} = \lambda

DR's of

PQ→\overrightarrow {PQ}

= (λ\lambda + 1, −-2, 2 −- λ\lambda) ∵\because

PQ→⊥r→\overrightarrow {PQ} \bot \overrightarrow r
⇒(λ+1)(1)+(−2)(0)+(2−λ)(−1)=0\Rightarrow (\lambda + 1)(1) + ( - 2)(0) + (2 - \lambda )( - 1) = 0
⇒λ=12⇒Q(12,0,−12)\Rightarrow \lambda = {1 \over 2} \Rightarrow Q\left( {{1 \over 2},0,{{ - 1} \over 2}} \right)
⇒PQ=342\Rightarrow PQ = {{\sqrt {34} } \over 2}
Q100
Let the acute angle bisector of the two planes x −- 2y −- 2z + 1 = 0 and 2x −- 3y −- 6z + 1 = 0 be the plane P. Then which of the following points lies on P?
A (3,1,−12)\left( {3,1, - {1 \over 2}} \right)
B (−2,0,−12)\left( { - 2,0, - {1 \over 2}} \right)
C (0, 2, −-4)
D (4, 0, −-2)
Correct Answer
Option B
Solution
P1:x−2y−2z+1=0{P_1}:x - 2y - 2z + 1 = 0
P2:2x−3y−6z+1=0{P_2}:2x - 3y - 6z + 1 = 0
∣x−2y−2z+11+4+4∣=∣2x−3y−6z+122+32+62∣\left| {{{x - 2y - 2z + 1} \over {\sqrt {1 + 4 + 4} }}} \right| = \left| {{{2x - 3y - 6z + 1} \over {\sqrt {{2^2} + {3^2} + {6^2}} }}} \right|
x−2y−2z+13=±2x−3y−6z+17{{x - 2y - 2z + 1} \over 3} = \pm {{2x - 3y - 6z + 1} \over 7}

Since

a1a2+b1b2+c1c2=20>0{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 20 > 0

∴\therefore Negative sign will give acute bisector

7x−14y−14z+7=−[6x−9y−18z+3]7x - 14y - 14z + 7 = - [6x - 9y - 18z + 3]
⇒13x−23y−32z+10=0\Rightarrow 13x - 23y - 32z + 10 = 0
(−2,0,−12)\left( { - 2,0, - {1 \over 2}} \right)

satisfy it ∴\therefore Ans. (b)

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