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3D Geometry

JEE Mathematics · 279 questions · Page 7 of 28 · Click an option or "Show Solution" to reveal answer

Q61
A plane P meets the coordinate axes at A, B and C respectively. The centroid of Δ\Delta ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is :
A x−11=y−11=z−22{{x - 1} \over 1} = {{y - 1} \over 1} = {{z - 2} \over 2}
B x−12=y−11=z−21{{x - 1} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}
C x−12=y−12=z−21{{x - 1} \over 2} = {{y - 1} \over 2} = {{z - 2} \over 1}
D x−11=y−12=z−22{{x - 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2}
Correct Answer
Option C
Solution

Let, Equation of plane is

xa+yb+zc{x \over a} + {y \over b} + {z \over c}

= 1 A = (

aa

, 0, 0) B = (0, b, 0), C = (0, 0, c) ∴\therefore Centroid =

(a3,b3,c3)\left( {{a \over 3},{b \over 3},{c \over 3}} \right)

= (1, 1, 2) ⇒\Rightarrow

aa

= 3, b = 3, c = 6 ∴\therefore Plane :

x3+y3+z6{x \over 3} + {y \over 3} + {z \over 6}

= 1 ⇒\Rightarrow 2x + 2y + z = 6 The equation of the line through this centroid (1, 1, 2) and perpendicular to the plane 2x + 2y + z = 6 is :

x−12=y−12=z−21{{x - 1} \over 2} = {{y - 1} \over 2} = {{z - 2} \over 1}
Q62
The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, –2, 3) and (1, 1, 0) lies on the plane :
A x – 2y + z = 1
B x + 2y – z = 1
C x – y – 2y = 1
D 2x + y – z = 1
Correct Answer
Option D
Solution

Equation of AB,

x−10=y+23=z−3−3=λ{{x - 1} \over 0} = {{y + 2} \over 3} = {{z - 3} \over { - 3}} = \lambda

∴\therefore Coordinates of any point on the line (M) =

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

As

PM→⊥AB→\overrightarrow {PM} \bot \overrightarrow {AB}

∴\therefore

PM→.AB→=0\overrightarrow {PM} .\overrightarrow {AB} = 0

⇒\Rightarrow

(−3).0+(3λ−4)(3)+(−3λ)(−3)\left( { - 3} \right).0 + \left( {3\lambda - 4} \right)\left( 3 \right) + \left( { - 3\lambda } \right)\left( { - 3} \right)

= 0 ⇒\Rightarrow λ\lambda =

23{2 \over 3}

∴\therefore M = (1, 0, 1) By checking each options we can see M lies on 2x + y – z = 1.

Q63
Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :
A (4, 3, 2)
B (6, 5, - 2)
C (3, 4, -2)
D (6, 5, 2)
Correct Answer
Option B
Solution

Plane passing through (2, 1, 0), (4, 1, 1) and (5, 0, 1) is

∣x−2y−1z−04−21−11−05−20−11−0∣\left| \begin{array}{lll}{x - 2} & {y - 1} & {z - 0} \\ {4 - 2} & {1 - 1} & {1 - 0} \\ {5 - 2} & {0 - 1} & {1 - 0} \end{array} \right|

= 0 ⇒\Rightarrow x + y – 2z = 3 ∴\therefore Image of R(2, 1, 6) in this plane is

x−21=y−11=z−6−2=−2(2+1−12−3)1+1+4{{x - 2} \over 1} = {{y - 1} \over 1} = {{z - 6} \over { - 2}} = - 2{{\left( {2 + 1 - 12 - 3} \right)} \over {1 + 1 + 4}}

∴\therefore (x, y, z) = (6, 5, –2)

Q64
The shortest distance between the lines x−33=y−8−1=z−31{{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1} and x+3−3=y+72=z−64{{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4} is :
A 3
B 7230{7 \over 2}\sqrt {30}
C 3303\sqrt {30}
D 2302\sqrt {30}
Correct Answer
Option C
Solution
a→\overrightarrow a

= < 3, 8, 3 >

b→\overrightarrow b

= < – 3, – 7, 6 >

p→\overrightarrow p

= < 3, – 1, 1 >

q→\overrightarrow q

= < –3, 2, 4 >

p→×q→=∣i^j^k^3−11−324∣\overrightarrow p \times \overrightarrow q = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 3 & { - 1} & 1 \\ { - 3} & 2 & 4 \end{array} \right|

= < -6, -15, 3 > Shortest distance =

∣(b→−a→).(p→×q→)∣p→×q→∣∣\left| {{{\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|

=

∣(−6,−15,3).(−6,−15,3)36+225+9∣\left| {{{\left( { - 6, - 15,3} \right).\left( { - 6, - 15,3} \right)} \over {\sqrt {36 + 225 + 9} }}} \right|

=

∣36+225+936+225+9∣\left| {{{36 + 225 + 9} \over {\sqrt {36 + 225 + 9} }}} \right|

=

270\sqrt {270}

=

3303\sqrt {30}
Q65
The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point :
A (0, 6, –2)
B (–2, 0, 1)
C (0, –6, 2)
D (2, 0 –1)
Correct Answer
Option B
Solution

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

1) As point (1, 2, 1) also passes through the plane, so it satisfy the equation, a(1 −- 2) + b(2 −- 1) + c(1 −- 2) = 0 ⇒\Rightarrow −-a + b −- c = 0 ....(

2) Given line 2x = 3y and z = 1, So, symmetric form of the line

x3=y2=z−10{x \over 3} = {y \over 2} = {{z - 1} \over 0}

∴\therefore Direction ratio of this line is (3, 2, 0) and Direction ration of plane = (a, b, c) As plane is parallel to the line so the normal of the plane is perpendicular to the line.

∴\therefore Dot product of direction ratio = 0 3a + 2b + 0(c) = 0 .....(

3) Equation of plane,

∣x−2y−1z−2−11−1320∣=0\left| \begin{array}{lll}{x - 2} & {y - 1} & {z - 2} \\ { - 1} & 1 & { - 1} \\ 3 & 2 & 0 \end{array} \right| = 0
⇒3(1−y+2−z)−2(−x+2+z−2)=0\Rightarrow 3(1 - y + 2 - z) - 2( - x + 2 + z - 2) = 0
⇒9−3y−3z+2x−2z=0\Rightarrow 9 - 3y - 3z + 2x - 2z = 0
⇒2x−3y−5z+9=0\Rightarrow 2x - 3y - 5z + 9 = 0

By checking all options you can see (−-2, 0, 1) satisfy the equation.

Q66
Let α\alpha be the angle between the lines whose direction cosines satisfy the equations l + m −- n = 0 and l2 + m2 −- n2 = 0. Then the value of sin4α\alpha + cos4α\alpha is :
A 38{{3 \over 8}}
B 34{{3 \over 4}}
C 12{{1 \over 2}}
D 58{{5 \over 8}}
Correct Answer
Option D
Solution
l2+m2+n2=1{l^2} + {m^2} + {n^2} = 1

∴\therefore

2n2=12{n^2} = 1

(∵\because l2 + m2 −- n2 = 0)

⇒n=±12\Rightarrow n = \pm {1 \over {\sqrt 2 }}

∴\therefore

l2+m2=12{l^2} + {m^2} = {1 \over 2}

&

l+m=12l + m = {1 \over {\sqrt 2 }}
⇒12−2lm=12\Rightarrow {1 \over 2} - 2lm = {1 \over 2}
⇒lm=0\Rightarrow lm = 0

or

m=0m = 0

∴\therefore

l=0,m=12l = 0,m = {1 \over {\sqrt 2 }}

or

l=12l = {1 \over {\sqrt 2 }}
<0,12,12>< 0,{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }} >

or

<12,0,12>< {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }} >

∴\therefore

cos⁡α=0+0+12=12\cos \alpha = 0 + 0 + {1 \over 2} = {1 \over 2}

∴\therefore

sin⁡4α+cos⁡4α=1−12sin⁡2(2α)=1−12.34=58{\sin ^4}\alpha + {\cos ^4}\alpha = 1 - {1 \over 2}{\sin ^2}(2\alpha ) = 1 - {1 \over 2}.{3 \over 4} = {5 \over 8}
Q67
The mirror image of the point (1, 2, 3) in a plane is (−73,−43,−13)\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right). Which of the following points lies on this plane ?
A (1, –1, 1)
B (–1, –1, –1)
C (–1, –1, 1)
D (1, 1, 1)
Correct Answer
Option A
Solution

Let A(1, 2, 3), B

(−73,−43,−13)\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)

∴\therefore Midpoint of AB = M =

(−73+12,−43+22,−13+32)\left( {{{{{ - 7} \over 3} + 1} \over 2},{{{{ - 4} \over 3} + 2} \over 2},{{{{ - 1} \over 3} + 3} \over 2}} \right)

=

(−23,13,43)\left( {{{ - 2} \over 3},{1 \over 3},{4 \over 3}} \right)

DR of AM =

(1+23,2−13,3−43)\left( {1 + {2 \over 3},2 - {1 \over 3},3 - {4 \over 3}} \right)

=

(53,53,53)\left( {{5 \over 3},{5 \over 3},{5 \over 3}} \right)

= (1, 1, 1) Equation of plane

a(x+23)+b(y−13)+c(z−43)a\left( {x + {2 \over 3}} \right) + b\left( {y - {1 \over 3}} \right) + c\left( {z - {4 \over 3}} \right)

= 0 ⇒\Rightarrow

1(x+23)+1(y−13)+1(z−43)1\left( {x + {2 \over 3}} \right) + 1\left( {y - {1 \over 3}} \right) + 1\left( {z - {4 \over 3}} \right)

= 0 ⇒\Rightarrow x + y + z = 1 ∴\therefore (1, –1, 1) lies on the plane.

Q68
The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes 3x + y - 2z = 5 and 2x - 5y - z = 7, is :
A 6x - 5y + 2z + 10 =0
B 3x - 10y - 2z + 11 = 0
C 6x - 5y - 2z - 2 = 0
D 11x + y + 17z + 38 = 0
Correct Answer
Option D
Solution

Given, equation of planes are 3x + y - 2z = 5 2x - 5y - z = 7 and point ( 1, 2, 3).

Normal vector of required plane = n =

∣i^j^k^31−22−5−1∣\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 3 & 1 & { - 2} \\ 2 & { - 5} & { - 1} \end{array} \right|

=

i^{\widehat i}

(-1 - 10) -

j^{\widehat j}

( -3 + 4) +

k^{\widehat k}

( -15 - 2) = -11

i^{\widehat i}

-

j^{\widehat j}

- 17

k^{\widehat k}

Now, the equation of plane passing through (1, 2, -3) having normal vector -11

i^{\widehat i}

-

j^{\widehat j}

- 17

k^{\widehat k}

is -[11(x - 1) + (y - 2) + 17(z + 3)] = 0 ⇒\Rightarrow 11x + y + 17z + 38 = 0

Q69
The distance of the point (1, 1, 9) from the point of intersection of the line x−31=y−42=z−52{{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2} and the plane x + y + z = 17 is :
A 19219\sqrt 2
B 2192\sqrt {19}
C 38
D 38\sqrt {38}
Correct Answer
Option D
Solution

Given, P(1, 1, 9). Equation of plane x + y + z = 17 Equation of line ⇒\Rightarrow

x−31=y−42=z−52{{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}

⇒\Rightarrow

x−31=y−42=z−52=λ{{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2} = \lambda

(let) ⇒\Rightarrow x = λ\lambda + 3; y = 2λ\lambda + 4; z = 2λ\lambda + 5 ∴\therefore The point we have is (λ\lambda + 3, 2λ\lambda + 4, 2λ\lambda + 5).

∵\because This point lies on the plane x + y + z = 17.

∴\therefore λ\lambda + 3 + 2λ\lambda + 4 + 2λ\lambda + 5 = 17 ⇒\Rightarrow λ\lambda = 1 ∴\therefore The coordinate of point is (4, 6, 7) ∴\therefore Required distance between (1, 1, 9) and (4, 6, 7) is

=(4−1)2+(6−1)2+(7−9)2= \sqrt {{{(4 - 1)}^2} + {{(6 - 1)}^2} + {{(7 - 9)}^2}}
=9+25+4=38= \sqrt {9 + 25 + 4} = \sqrt {38}
Q70
Let a, b∈ \in R. If the mirror image of the point P(a, 6, 9) with respect to the line x−37=y−25=z−1−9{{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}} is (20, b, −-a−-9), then | a + b |, is equal to :
A 88
B 90
C 86
D 84
Correct Answer
Option A
Solution

Given, P(a, 6, 9) Equation of line

x−37=y−25=z−1−9{{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}

Image of point P with respect to line is point Q(20, b, −-a −-9) Mid-point of P and Q =

(a+202,6+b2,−a2)\left( {{{a + 20} \over 2},{{6 + b} \over 2},{{ - a} \over 2}} \right)

This point lies on line ∴\therefore

a+202−37=6+b2−25=−a2−1−9{{{{a + 20} \over 2} - 3} \over 7} = {{{{6 + b} \over 2} - 2} \over 5} = {{{{ - a} \over 2} - 1} \over { - 9}}
⇒a+1414=b+210=a+218\Rightarrow {{a + 14} \over {14}} = {{b + 2} \over {10}} = {{a + 2} \over {18}}
⇒a+1414=a+218\Rightarrow {{a + 14} \over {14}} = {{a + 2} \over {18}}

and

b+210=a+218{{b + 2} \over {10}} = {{a + 2} \over {18}}

Solving, we get a = −- 56, b = −- 32 ∴\therefore

∣a+b∣=∣−56−32∣=88\left| {a + b} \right| = \left| { - 56 - 32} \right| = 88
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