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3D Geometry

JEE Mathematics · 279 questions · Page 4 of 28 · Click an option or "Show Solution" to reveal answer

Q31
If x = a, y = b, z = c is a solution of the system of linear equations x + 8y + 7z = 0 9x + 2y + 3z = 0 x + y + z = 0 such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals :
A −- 1
B 0
C 1
D 2
Correct Answer
Option C
Solution

Given, x + 8y + 7z = 0 9x + 2y + 3z = 0 x + y + z = 0 Solving those equations, we get x = λ\lambda, y = 6λ\lambda, z = -7λ\lambda This point lies on the plane x + 2y + z = 6 ∴\therefore λ\lambda + 2(6λ\lambda) + (-7λ\lambda) = 0 ⇒\Rightarrow λ\lambda = 1 ∴\therefore x = 1, y = 6, z = -7 ∴\therefore a = 1, b = 6, c = -7 So, 2a + b + c = 2(1) + 6 + (-7) = 1

Q32
The line of intersection of the planes r→.(3i^−j^+k^)=1  \overrightarrow r .\left( {3\widehat i - \widehat j + \widehat k} \right) = 1\,\, and r→.(i^+4j^−2k^)=2,\overrightarrow r .\left( {\widehat i + 4\widehat j - 2\widehat k} \right) = 2, is :
A x−47−2=y7=z−5713{{x - {4 \over 7}} \over { - 2}} = {y \over 7} = {{z - {5 \over 7}} \over {13}}
B x−472=y−7=z+5713{{x - {4 \over 7}} \over 2} = {y \over { - 7}} = {{z + {5 \over 7}} \over {13}}
C x−6132=y−513−7=z−13{{x - {6 \over {13}}} \over 2} = {{y - {5 \over {13}}} \over { - 7}} = {z \over { - 13}}
D x−6132=y−5137=z−13{{x - {6 \over {13}}} \over 2} = {{y - {5 \over {13}}} \over 7} = {z \over { - 13}}
Correct Answer
Option C
Solution
n→=n1→×n2→\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}}

⇒\Rightarrow

∣i^j^k^3−1114−2∣=i^(−2)−j^(−7)+k^(13)\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 3 & { - 1} & 1 \\ 1 & 4 & { - 2} \end{array} \right| = \widehat i\left( { - 2} \right) - \widehat j\left( { - 7} \right) + \widehat k\left( {13} \right)
n→=−2i^+7j^+13k^\overrightarrow n = - 2\widehat i + 7\widehat j + 13\widehat k

Now,

3x−y+z=13x - y + z = 1
x+4y−2z=2x + 4y - 2z = 2

but z == 0 & solving the given x == 6/13 & y = 5/13 ∴\therefore required equation of a line is

x−6/132=y−5/13−7=z−13{{x - 6/13} \over 2} = {{y - 5/13} \over { - 7}} = {z \over { - 13}}
Q33
The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, x + y + z = 7 is :
A 23\sqrt {{2 \over 3}}
B 23{2 \over {\sqrt 3 }}
C 23{2 \over 3}
D 13{1 \over 3}
Correct Answer
Option A
Solution

PQ is the projection of line segment AB on the plane x + y + z = 7 P and Q are called foot of perpendicular on the plane x + y + z = 7 Let P = (x, y, z) then

x−51=y+11=z−41=−(5−1+4)12+12+12{{x - 5} \over 1} = {{y + 1} \over 1} = {{z - 4} \over 1} = {{ - \left( {5 - 1 + 4} \right)} \over {{1^2} + {1^2} + {1^2}}}
⇒    x−5=y+1=z−4=−83\Rightarrow \,\,\,\,x - 5 = y + 1 = z - 4 = - {8 \over 3}
∴   \therefore\,\,\,

x =

73{7 \over 3}

, y = −-

113,{{11} \over 3},

z =

43{4 \over 3}
∴   \therefore\,\,\,

Point P =

(73,−113,43)\left( {{7 \over 3}, - {{11} \over 3},{4 \over 3}} \right)

Let Q = (x1, y1, z1) then

a1−41=y1+11=z1−31=−(4−1+3)12+12+12{{{a_1} - 4} \over 1} = {{{y_1} + 1} \over 1} = {{{z_1} - 3} \over 1} = {{ - \left( {4 - 1 + 3} \right)} \over {{1^2} + {1^2} + {1^2}}}
⇒    \Rightarrow \,\,\,\,

x1 −- 4 = y1 +1= z1 −- 3 = −- 2

∴   \therefore\,\,\,

x = 2, y = −- 3, z = 1

∴   \therefore\,\,\,

Point Q = (2, −- 3, 1) Now length of PQ is

(73−2)2+(−113+3)2+(43−1)2\sqrt {{{\left( {{7 \over 3} - 2} \right)}^2} + {{\left( { - {{11} \over 3} + 3} \right)}^2} + {{\left( {{4 \over 3} - 1} \right)}^2}}
=19+49+19= \sqrt {{1 \over 9} + {4 \over 9} + {1 \over 9}}
=69= \sqrt {{6 \over 9}}
=23= \sqrt {{2 \over 3}}
Q34
A variable plane passes through a fixed point (3,2,1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz -plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is :
A x3+y2+z1=1{x \over 3} + {y \over 2} + {z \over 1} = 1
B x + y + z = 6
C 1x+1y+1z=116{1 \over x} + {1 \over y} + {1 \over z} = {{11} \over 6}
D 3x+2y+1z=1{3 \over x} + {2 \over y} + {1 \over z} = 1
Correct Answer
Option D
Solution

If a, b, c are the intercepts of the variable plane on the x,y,z axes respectively, then the equation of the plane is

xa+yb+zc=1{x \over a} + {y \over b} + {z \over c} = 1

And the point of intersection of the planes parallel to the xy, yz and zx planes is

(a,b,c)\left( {a,b,c} \right)

. As the point (3, 2, 1) lies on the variables plane, so

3a+2b+1c=1{3 \over a} + {2 \over b} + {1 \over c} = 1

Therefore, the required locus is

3x+2y+1z=1{3 \over x} + {2 \over y} + {1 \over z} = 1
Q35
If L1 is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L2 is the line of intersection of the planes x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L2, is :
A 12{1 \over {\sqrt 2 }}
B 142{1 \over {4\sqrt 2 }}
C 132{1 \over {3\sqrt 2 }}
D 122{1 \over {2\sqrt 2 }}
Correct Answer
Option C
Solution

L1 is the line of intersection of plane 1 and plane 2. L2 is the line of intersection of plane

33

and plane 4.

Line L1 and L2 are present on plane 5.

Now we have to find the distance of origin from the plane 5.

The plane 5 is passing through the Line L1, and L1 is the line of intersection of first two planes.

∴\therefore Equation of plane 5,

(2x−2y+3z−2)(2x-2y+3z-2)

+

λ(x−y+z+1)0\lambda \left( {x - y + z + 1} \right)0

Now find a point on L2 which is line of intersection of x + 2y −- z −- 3 = 0

    \,\,\,\,

.....(1) and 3x −- y + 2z −- 1 = 0

    \,\,\,\,

.....(2) Put x = 0 at both (1) and (2), 2y −- z = 3

    .....\,\,\,\,.....

(3) −-y + 2z = 1

    ....\,\,\,\,....

(4) by solving (3) and (4) we get

y=73, z=53y = {7 \over 3},\,z = {5 \over 3}

So, the point on line L2 is

=(0,73,53)= \left( {0,{7 \over 3},{5 \over 3}} \right)

This point is also present on the plane

5.5.

So, putting this point on equation of plane

5,5,
−145+5−2+λ(−73+53)=0- {{14} \over 5} + 5 - 2 + \lambda \left( { - {7 \over 3} + {5 \over 3}} \right) = 0
⇒−53+λ3=0\Rightarrow - {5 \over 3} + {\lambda \over 3} = 0
⇒λ=5\Rightarrow \lambda = 5

∴\therefore equation of plane 5 is 2x −- 2y + 3z −- 2 + λ\lambda(x −- y + z + 1) = 0 ⇒\Rightarrow 2x −- 2y + 3z −- 2 + 5 (x −- y + z + 1) = 0 ⇒\Rightarrow 7x −- 7y + 8z + 3 = 0 Distance of this plane 5 from (0, 0, 0)

=349+49+64= {3 \over {\sqrt {49 + 49 + 64} }}
=3162= {3 \over {\sqrt {162} }}
=392= {3 \over {9\sqrt 2 }}
=132= {1 \over {3\sqrt 2 }}
Q36
The magnitude of the projection of the vector 2i∧+3j∧+k∧\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge on the vector perpendicular to the plane containing the vectors i∧+j∧+k∧\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge and i∧+2j∧+3k∧\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge , is :
A 32{{\sqrt 3 } \over 2}
B 6\sqrt 6
C 32\sqrt {3 \over 2}
D 36\sqrt 6
Correct Answer
Option C
Solution

Let vector

p→\overrightarrow p

is perpendicular to the both vectors

i∧+j∧+k∧\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge

and

i∧+2j∧+3k∧\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge

. ∴\therefore

p→\overrightarrow p

= (

i∧+j∧+k∧\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge

) ×\times (

i∧+2j∧+3k∧\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge

) =

∣i^j^k^111123∣\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array} \right|

=

i^−2j^+k^\widehat i - 2\widehat j + \widehat k

Now a vector

a→\overrightarrow a

=

2i∧+3j∧+k∧\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge

is given and we have to findout projection of vector

a→\overrightarrow a

on

p→\overrightarrow p

. ∴\therefore Projection of vector

a→\overrightarrow a

on

p→\overrightarrow p

=

∣a→∣cos⁡θ\left| {\overrightarrow a } \right|\cos \theta

=

∣a→∣×a→.p→∣a→∣∣p→∣\left| {\overrightarrow a } \right| \times {{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow p } \right|}}

=

a→.p→∣p→∣{{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}

=

(2i^+3j^+k^).(i^−2j^+k^)1+4+1{{\left( {2\widehat i + 3\widehat j + \widehat k} \right).\left( {\widehat i - 2\widehat j + \widehat k} \right)} \over {\sqrt {1 + 4 + 1} }}

=

2−6+16{{2 - 6 + 1} \over {\sqrt 6 }}

=

−36{{ - 3} \over {\sqrt 6 }}

Magnitude of projection of vector

a→\overrightarrow a

on

p→\overrightarrow p

=

∣−36∣\left| {{{ - 3} \over {\sqrt 6 }}} \right|

=

36{{3 \over {\sqrt 6 }}}

=

32{{\sqrt 3 } \over {\sqrt 2 }}
Q37
The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y+ 4z = 5 which is perpendicular to the plane x – y + z = 0 is :
A r→×(i∧−k∧)−2=0\mathop r\limits^ \to \times \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) - 2 = 0
B r→.(i∧+k∧)+2=0\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge + \mathop k\limits^ \wedge } \right) + 2 = 0
C r→.(i∧−k∧)+2=0\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0
D r→×(i∧−k∧)+2=0\mathop r\limits^ \to \times \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0
Correct Answer
Option C
Solution

Given, P1 : x + y + z = 1 P1 : 2x + 3y + 4z = 5 Equation of the plane passing through the line of intersection of the plane P1 and P2 is : P1 + λ\lambdaP2 = 0 ⇒\Rightarrow (x + y + z –1) + λ\lambda(2x + 3y + 4z – 5) = 0 ⇒\Rightarrow x(1 + 2λ\lambda) + y(1 + 3λ\lambda) + z(1 + 4λ\lambda) - 5λ\lambda - 1 = 0 .....(

1) Direction Ratio (D.R) of this plane = (1 + 2λ\lambda, 1 + 3λ\lambda, 1 + 4λ\lambda) Plane (1) is perpendicular to x - y + z = 0, whose D.R = (1, -1, 1) As they are perpendicular so dot product of D.R = 0 ∴\therefore (1) (1 + 2λ\lambda) + (–1) (1 + 3λ\lambda) + (1) (1 + 4λ\lambda) = 0 ⇒\Rightarrow 1 + 2λ\lambda –1 – 3λ\lambda + 1 + 4λ\lambda = 0 ⇒\Rightarrow λ\lambda =

−13- {1 \over 3}

Putting the value of λ\lambda in equation (1), we get ⇒\Rightarrow

x3−z3+23=0{x \over 3} - {z \over 3} + {2 \over 3} = 0

⇒\Rightarrow x - z + 2 = 0 Vector form of this plane,

r→.(i∧−k∧)+2=0\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0
Q38
If a point R(4, y, z) lies on the line segment joining the points P(2, –3, 4) and Q(8, 0, 10), then the distance of R from the origin is :
A 2142 \sqrt {14}
B 53 \sqrt {53}
C 2212 \sqrt {21}
D 6
Correct Answer
Option A
Solution

Equation of PQ is

x−28−2=y+30−(−3)=z−410−4{{x - 2} \over {8 - 2}} = {{y + 3} \over {0 - \left( { - 3} \right)}} = {{z - 4} \over {10 - 4}}

⇒\Rightarrow

x−26=y+33=z−46{{x - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}

Point R (4, y, z) lies on this ∴\therefore

4−26=y+33=z−46{{4 - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}

⇒\Rightarrow

13=y+33=z−46{1 \over 3} = {{y + 3} \over 3} = {{z - 4} \over 6}

y = -2 and y = 6 ∴\therefore R = (4, -2, 6) Distance of R(4, -2, 6) from the origin O(0, 0, 0) is RO =

42+(−2)2+62\sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {6^2}}

=

56=214\sqrt {56} = 2\sqrt {14}
Q39
A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle π4{\pi \over 4} with the plane y – z + 5 = 0, also passes through the point
A (2,1,4)\left( {\sqrt 2 ,1,4} \right)
B (−2,1,4)\left(- {\sqrt 2 ,1,4} \right)
C (−2,−1,−4)\left( -{\sqrt 2 ,-1,-4} \right)
D (2,−1,4)\left( {\sqrt 2 ,-1,4} \right)
Correct Answer
Option A
Solution

Let ax + by + cz = 1 be the equation of the plane it passed through point (0, –1, 0). ∴\therefore -b = 1 ⇒\Rightarrow b = -1 Also it passes through point (0, 0, 1) ∴\therefore c = 1 So the plane is ax - y + z = 1.

This plane an angle

π4{\pi \over 4}

with the plane y – z + 5 = 0. Normal to the plane ax - y + z = 1 is

a→{\overrightarrow a }

=

ai^−j^+k^a\widehat i - \widehat j + \widehat k

Normal to the plane y – z + 5 = 0 is

b→{\overrightarrow b }

=

j^−k^\widehat j - \widehat k

cos θ\theta =

∣a→.b→∣a→∣∣b→∣∣\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|

⇒\Rightarrow

12{1 \over {\sqrt 2 }}

=

∣a→.b→∣a→∣∣b→∣∣\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|

⇒\Rightarrow

∣0−1−1∣a2+1+112+12{{\left| {0 - 1 - 1} \right|} \over {\sqrt {{a^2} + 1 + 1} \sqrt {{1^2} + {1^2}} }}

=

12{1 \over {\sqrt 2 }}

⇒\Rightarrow

a2+2=4{a^2} + 2 = 4

⇒\Rightarrow a = ±\pm

2\sqrt 2

∴\therefore Equation of plane ±\pm

2\sqrt 2

x - y + z = 1 Now by checking each options you can see equation -

2\sqrt 2

x - y + z = 1 satisfy by the point

(2,1,4)\left( {\sqrt 2 ,1,4} \right)

.

Q40
The equation of a plane containing the line of intersection of the planes 2x – y – 4 = 0 and y + 2z – 4 = 0 and passing through the point (1, 1, 0) is :
A x – 3y – 2z = –2
B 2x – z = 2
C x – y – z = 0
D x + 3y + z = 4
Correct Answer
Option C
Solution

The equation of any plane passing through the intersection of the planes 2x – y – 4 = 0 and y + 2z – 4 = 0 is : (2x – y – 4) + λ\lambda(y + 2z – 4) = 0 ........(

1) As this plane passes through (1, 1, 0) then this point satisfy the equation (1).

∴\therefore (2 – 1 – 4) + λ\lambda(1 + 0 – 4) = 0 ⇒\Rightarrow λ\lambda = -1 Equation of required plane will be (2x – y – 4) – (y + 2z – 4) = 0 ⇒\Rightarrow 2x – 2y – 2z = 0 ⇒\Rightarrow x – y – z = 0

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