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3D Geometry

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

Q271
The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point :
A (–3, 0, -1)
B (3, 2, 1)
C (3, 3, -1)
D (–3, 1, 1)
Correct Answer
Option B
Solution

The equation of plane (x + y + z −- 1) + λ\lambda (2x + 3y −- z + 4) = 0 ⇒\Rightarrow (1 + 2λ\lambda)x + (1 + 3λ\lambda)y + (1 −- λ\lambda)z + 4λ\lambda −- 1 = 0 As plane is parallel to y axis so the normal vector of plane and dot product of

j^\widehat j

is zero. ∴\therefore 1 + 3λ\lambda = 0 ⇒\Rightarrow λ\lambda = −-

13{1 \over 3}

∴\therefore So the equation of the plane is x(1 −-

23{2 \over 3}

) + (1 −-

33{3 \over 3}

) y + (1 +

13{1 \over 3}

) −-

43{4 \over 3}

−- 1 = 0 ⇒\Rightarrow x (

13{1 \over 3}

) + z(

43{4 \over 3}

) −-

73{7 \over 3}

= 0 ⇒\Rightarrow x + 4z −- 7 = 0 By checking each options you can see only point (3, 2, 1) lies on the plane.

Q272
The equation of the plane containing the straight line x2=y3=z4{x \over 2} = {y \over 3} = {z \over 4} and perpendicular to the plane containing the straight lines x3=y4=z2{x \over 3} = {y \over 4} = {z \over 2} and x4=y2=z3{x \over 4} = {y \over 2} = {z \over 3} is :
A x −- 2y + z = 0
B 3x + 2y −- 3z = 0
C x + 2y −- 2z = 0
D 5x + 2y −- 4z = 0
Correct Answer
Option A
Solution

Vector ⊥\bot to given plane =

∣i^j^k^342423∣\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{array} \right|

=

i^(12−4)−j^(9−8)+k^(6−16)\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)

=

8i^−j^−10k^ 8\widehat i - \widehat j - 10\widehat k\,

. . . . (1) Vector parallel to given line =

2i^+3j^+4k^    2\widehat i + 3\widehat j + 4\widehat k\,\,\,\,

. . . (2) Vector

⊥   \bot \,\,\,

to both (1) & (2) vector =

∣i^j^k^8−1−10234∣\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 8 & { - 1} & { - 10} \\ 2 & 3 & 4 \end{array} \right|

=

i^(−4+30)−j^(32+20)+k^(24+2)\widehat i\left( { - 4 + 30} \right) - \widehat j\left( {32 + 20} \right) + \widehat k\left( {24 + 2} \right)

=

26i^−52j^+26k^26\widehat i - 52\widehat j + 26\widehat k

Dr's of normal of required plane is (26, −-52, 26) ⇒\Rightarrow (1, −-2, 1) Equation of plane whose Dr's of Normal is (1, −-2, 1) and passes through origin 1.(x −- 0) −- 2(y −- 0) + 1.(z −- 0) = 0 x −- 2y + z = 0

Q273
Two lines x−31=y+13=z−6−1{{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}} and x+57=y−2−6=z−34{{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4} intersect at the point R. The reflection of R in the xy-plane has coordinates :
A (2, 4, 7)
B (2, −- 4, −-7)
C (2, −- 4, 7)
D (−- 2, 4, 7)
Correct Answer
Option B
Solution

Point on L1 (λ\lambda + 3, 3λ\lambda −- 1, −-λ\lambda + 6) Point on L2 (7μ\mu −- 5, −-6μ\mu + 2, 4μ\mu + 3 ⇒\Rightarrow λ\lambda + 3 = 7μ\mu −- 5 . . . . (i) 3λ\lambda −- 1 = −-6μ\mu + 2 . . . .(ii) ⇒\Rightarrow λ\lambda = −-1, μ\mu = 1 point R(2, −- 4, 7) Reflection is (2, −-4, −- 7)

Q274
A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point (α\alpha , –3, 5), then α\alpha is equal to:
A -10
B 10
C 5
D -5
Correct Answer
Option C
Solution

As normal is perpendicular to both the lines so normal vector to the plane is

n→=(i^−2j^+2k^)×(2i^+3j^−k^)\overrightarrow n = \left( {\widehat i - 2\widehat j + 2\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)
n→=∣i^j^k^1−2223−1∣\overrightarrow n = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & { - 2} & 2 \\ 2 & 3 & { - 1} \end{array} \right|
n→=(2−6)i^−(−1−4)j^+(3+4)k^\overrightarrow n = \left( {2 - 6} \right)\widehat i - \left( { - 1 - 4} \right)\widehat j + \left( {3 + 4} \right)\widehat k
n→=−4i^+5j^+7k^\overrightarrow n = - 4\widehat i + 5\widehat j + 7\widehat k

Now equation of plane passing through (3,1,1) is ⇒\Rightarrow –4(x – 3) + 5(y – 1) + 7(z – 1) = 0 ⇒\Rightarrow –4x + 12 + 5y – 5 + 7z – 7 = 0 ⇒\Rightarrow –4x + 5y + 7z = 0 ...(

1) Plane is also passing through (α\alpha, –3, 5) so this point satisfies the equation of plane so put in equation (1) –4α\alpha + 5 × (–3) + 7 × (5) = 0 ⇒\Rightarrow –4α\alpha – 15 + 35 = 0

⇒α\Rightarrow\alpha

= 5

Q275
If the line, x−31=y+2−1=z+λ−2{{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}} lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, x−112=y9=z4{{x - 1} \over {12}} = {y \over 9} = {z \over 4} is :
A 2
B 1
C 0
D 3
Correct Answer
Option C
Solution

Point (3, −- 2, −- λ\lambda) on p line 2x −- 4y + 3z −- 2 == 0 == 6 + 8 −- 3λ\lambda −- 2 = 0 == 3λ\lambda == 12 λ\lambda == 4 Now,

x−31=y+2−1=z+4−2=k1{{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1}

. . .(i)

x−112=y9=z4=k2{{x - 1} \over {12}} = {y \over 9} = {z \over 4} = {k_2}

. . .(ii) Point on equation (i) P (k1 + 3, −- k1 −- 2, −- 2k1 −- 4) Point on equation (ii) Q(12k2 + 1, 9k2, 4k2) k1 + 3 == 12k2 + 1

∣−k1−2=9k2∣\left| { - {k_1} - 2 = 9{k_2}} \right|

−- 2k1 −- 4 == 4k2 k2 == 0 k1 == −- 2 p (1, 0, 0) lie on equation of a line 1 gives shortest distance == 0

Q276
The distance of the point (1, − 2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x − 2y + z + 12 = 0, is :
A 222\sqrt 2
B 2
C 2\sqrt 2
D 12{1 \over {\sqrt 2 }}
Correct Answer
Option A
Solution

Equation of plane passing through point (1, 2, 2) is, a(x −- 1) + b(y −- 2) + c(z −- 2) = 0 . . .(

1) This plane is perpendicular to the plane x −- y + 2z = 3 and 2x −- 2y + z + 12 = 0 When two planes, a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are perpendicular to each other then a1a2 + b1b2 + c1c2 = 0 So, we can say, a ×\times 1 + b ×\times (−- 1) + c ×\times 2 = 0 ⇒\Rightarrow a −- b + 2c = 0 . . . (2) and, a ×\times 2 + b(−-2) + c(1) = 0 ⇒\Rightarrow 2a −- 2b + c = 0 . . .(

3) Solving (2) and (3)

a−1+4{a \over { - 1 + 4}}

=

b4−1{b \over {4 - 1}}

=

c−2+2{c \over { - 2 + 2}}

= λ\lambda ⇒\Rightarrow a = 3λ\lambda, b = 3λ\lambda, c = 0 Putting the values of a, b, c in equation (1) we get, 3λ\lambda (x −- 1) + 3λ\lambda (y −- 2) = 0 ⇒\Rightarrow 3(x −- 1) + 3(y −- 2) = 0 ⇒\Rightarrow 3x + 3y −- 9 = 0 ⇒\Rightarrow x + y −- 3 = 0 ∴\therefore Distance of point (1, −-2, 4) from plane x + y −- 3 = 0 is, =

∣1−2−31+1∣\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|

=

42{4 \over {\sqrt 2 }}

= 2

2\sqrt 2
Q277
The plane passing through the point (4, –1, 2) and parallel to the lines x+23=y−2−1=z+12{{x + 2} \over 3} = {{y - 2} \over { - 1}} = {{z + 1} \over 2} and x−21=y−32=z−43{{x - 2} \over 1} = {{y - 3} \over 2} = {{z - 4} \over 3} also passes through the point -
A (1, 1, −- 1)
B (1, 1, 1)
C (−- 1, −- 1, −-1)
D (−- 1, −- 1, 1)
Correct Answer
Option B
Solution

Let

n→\overrightarrow n

be the normal vector to the plane passing through (4, −-1, 2) and parallel to the lines L1 & L2 then

n→\overrightarrow n

=

∣i^j^k^3−12123∣\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 3 & { - 1} & 2 \\ 1 & 2 & 3 \end{array} \right|

∴\therefore

n→\overrightarrow n

=

−7i^−7j^+7k^- 7\widehat i - 7\widehat j + 7\widehat k

∴\therefore Equation of plane is −- 1(x −- 4) −- 1(y + 1) + 1(z −- 2) = 0 ∴\therefore x + y −- z −- 1 = 0 Now check options

Q278
Let A be a point on the line r→=(1−3μ)i^+(μ−1)j^+(2+5μ)k^\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k and B(3, 2, 6) be a point in the space. Then the value of μ\mu for which the vector AB→\overrightarrow {AB} is parallel to the plane x −- 4y + 3z = 1 is -
A 18{1 \over 8}
B 12{1 \over 2}
C 14{1 \over 4}
D −- 14{1 \over 4}
Correct Answer
Option C
Solution

Let point A is

(1−3μ)i^+(μ−1)j^+(2+5μ)k^\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k

and point B is (3, 2, 6) then

AB→=(2+3μ)i^+(3−μ)j^+(4−5μ)k^\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k

which is parallel to the the plane x −- 4y + 3z = 1 ∴\therefore 2 + 3μ\mu −- 12 + 4μ\mu + 12 −- 15μ\mu = 0 8μ\mu = 2 μ\mu =

14{1 \over 4}
Q279
If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
A ab' + bc' + 1 = 0
B cc' + a + a' = 0
C bb' + cc' + 1 = 0
D aa' + c + c' = 0
Correct Answer
Option D
Solution

Equation of 1st line is

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

Dr's of 1st line = (

aa

, 1 , c) Equation of 2nd line is

x−b′a′=y−b′c′=z1{{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}

Dr's of 2nd line = (

a′a'

, c' , 1) Lines are perpendicular, so the dot product of the Dr's of two lines are zero. ∴\therefore

aaaa

' + c + c' = 0

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