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All questions of Chapter 11 - Three Dimensional Geometry for JEE Exam
Find the equation of the set of points which are equidistant from the points (1, 2 , 3) and (3, 2, -1)
a) x + 2z = 0
b) y + 2z = 0
c) x – 2z = 0
d) x – 2y = 0
Correct answer is option 'C'. Can you explain this answer?
|
Aryan Khanna answered |
Pt. A(1, 2 , 3)
Pt. B(3, 2, -1)
Let P(x,y,z)
So, AP = BP
((x-1)2 + (y-2)2 + (z-3)2)1/2 = ((x-3)2 + (y-2)2 + (z+1)2)1/2
(x-1)2 + (y-2)2 + (z-3)2) = (x-3)2 + (y-2)2 + (z+1)2
x2 +1 -2x + y2 + 4 - 4y + z2 + 9 – 6z = x2 +9 -6x + y2 + 4 - 4y + z2 + 1 + 2z
4x – 8z = 0
x – 2z = 0
Pt. B(3, 2, -1)
Let P(x,y,z)
So, AP = BP
((x-1)2 + (y-2)2 + (z-3)2)1/2 = ((x-3)2 + (y-2)2 + (z+1)2)1/2
(x-1)2 + (y-2)2 + (z-3)2) = (x-3)2 + (y-2)2 + (z+1)2
x2 +1 -2x + y2 + 4 - 4y + z2 + 9 – 6z = x2 +9 -6x + y2 + 4 - 4y + z2 + 1 + 2z
4x – 8z = 0
x – 2z = 0
The signs of the X,Y and Z coordinates of a point that lies in the octant OXYZ’ is- a)(- , + , +)
- b)(-, – , +)
- c)( + , – , -)
- d)( + , + , -)
Correct answer is option 'D'. Can you explain this answer?
The signs of the X,Y and Z coordinates of a point that lies in the octant OXYZ’ is
a)
(- , + , +)
b)
(-, – , +)
c)
( + , – , -)
d)
( + , + , -)
|
Vikas Kapoor answered |
X,Y,Z imply positive X,Y,Z axis & X’,Y’,Z’ imply negative X,Y,Z axis.
So, OXYZ’ will have a point of signs (+, +, -).
So, OXYZ’ will have a point of signs (+, +, -).
The direction cosines of the line joining the points (2, -1, 8) and (-4, -3, 5) are:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The direction cosines of the line joining the points (2, -1, 8) and (-4, -3, 5) are:
a)
b)
c)
d)
|
Geetika Shah answered |
Pt. A(2, -1, 8)
Pt. B(-4, -3, 5)
Direction Ratio DR of AB : ( -4-2 , -3+1 , 5-8 )
: (-6,-2,-3)
Direction cosine of AB : ( -6/(62+22+32)1/2 , -2/(62+22+32)1/2 , -3/(62+22+32)1/2)
: ( -6/7, -2/7, -3/7)
Pt. B(-4, -3, 5)
Direction Ratio DR of AB : ( -4-2 , -3+1 , 5-8 )
: (-6,-2,-3)
Direction cosine of AB : ( -6/(62+22+32)1/2 , -2/(62+22+32)1/2 , -3/(62+22+32)1/2)
: ( -6/7, -2/7, -3/7)
The equation of the plane passing through the line of intersection of the planes x-2y+3z+8=0 and 2x-7y+4z-3=0 and the point (3, 1, -2) is:- a)6x-15y+12z+29=0
- b)6x-15y+16z+29=0
- c)6x-15y+12z+32=0
- d)2x-5y+4z+9=0
Correct answer is option 'B'. Can you explain this answer?
The equation of the plane passing through the line of intersection of the planes x-2y+3z+8=0 and 2x-7y+4z-3=0 and the point (3, 1, -2) is:
a)
6x-15y+12z+29=0
b)
6x-15y+16z+29=0
c)
6x-15y+12z+32=0
d)
2x-5y+4z+9=0
|
Tejas Verma answered |
(x - 2y + 3z + 8) + μ(2x - 7y + 4z - 3) = 0
i.e, (1 + 2μ)x - (2 + 7μ)y + (3 + 4μ)z + (8 - 3μ) = 0......(1)
the required plane is passing through (3, 1, -2)
so, 3(1 + 2μ) - (1)(2 + 7μ) + (-2)(3 + 4μ)+ (8 - 3μ) = 0
3 + 6μ - 2 - 7μ -6 -8μ + 8 - 3μ = 0
by solving, μ = 1/4
putting μ in equation (1)
we get the required equation of plane as :- 6x - 15y + 16z + 29 = 0
i.e, (1 + 2μ)x - (2 + 7μ)y + (3 + 4μ)z + (8 - 3μ) = 0......(1)
the required plane is passing through (3, 1, -2)
so, 3(1 + 2μ) - (1)(2 + 7μ) + (-2)(3 + 4μ)+ (8 - 3μ) = 0
3 + 6μ - 2 - 7μ -6 -8μ + 8 - 3μ = 0
by solving, μ = 1/4
putting μ in equation (1)
we get the required equation of plane as :- 6x - 15y + 16z + 29 = 0
The Cartesian form of the equation of the plane 
is:- a)2x+3y-z=10
- b)2x+3y-z= √14
- c)2x+3y-z+√14=0
- d)2x+3y-z+10=0
Correct answer is option 'A'. Can you explain this answer?
The Cartesian form of the equation of the plane is:
is:a)
2x+3y-z=10
b)
2x+3y-z= √14
c)
2x+3y-z+√14=0
d)
2x+3y-z+10=0
|
Deepak Kapoor answered |
Let r = xi + yj + zk
(xi + yj + zk) . (2i + 3j - k) = 10
2x + 3y - k = 10
If a line makes angles 45°,150°, 135°, with x, y and z-axes respectively, find its direction cosines.- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If a line makes angles 45°,150°, 135°, with x, y and z-axes respectively, find its direction cosines.
a)
b)
c)
d)
|
|
Leelu Bhai answered |
Direction cosines of this line are cos 45, cos150
The angle between the lines x = 2y = – 3z and – 4x = 6y = – z is:- a)0°
- b)cos-1(1/√3)
- c)90°
- d)180°
Correct answer is option 'C'. Can you explain this answer?
The angle between the lines x = 2y = – 3z and – 4x = 6y = – z is:
a)
0°
b)
cos-1(1/√3)
c)
90°
d)
180°
|
Sushil Kumar answered |
x = 2y = -3z -4x = 6y = -z
x/1 = y/(½) = z(-⅓) x/(-¼) = y/(⅙) = z/(-1)
Cosθ = [(a1a2 + b1b2 + c1c2)/(a1 + b1 + c1)½ * (a2 + b2 + c2)½]
Cosθ ={[(1*(-¼)) + (½)(⅙) + (-⅓)(-1)]/[(1)2 + (½)2 + (-⅓)2]1/2 * [(-¼)2 + (⅙)2 + (-1)2]1/2}
= {[(-¼ + 1/12 - ⅓)]/[2 + 1 - ⅔]1/2 * [ -½ + ⅓ -½]½}
Cosθ = 0
θ = 90deg
x/1 = y/(½) = z(-⅓) x/(-¼) = y/(⅙) = z/(-1)
Cosθ = [(a1a2 + b1b2 + c1c2)/(a1 + b1 + c1)½ * (a2 + b2 + c2)½]
Cosθ ={[(1*(-¼)) + (½)(⅙) + (-⅓)(-1)]/[(1)2 + (½)2 + (-⅓)2]1/2 * [(-¼)2 + (⅙)2 + (-1)2]1/2}
= {[(-¼ + 1/12 - ⅓)]/[2 + 1 - ⅔]1/2 * [ -½ + ⅓ -½]½}
Cosθ = 0
θ = 90deg
If plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes, then the area of the triangle ABC equal to- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
If plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes, then the area of the triangle ABC equal to
a)
b)
c)
d)
|
Neha Joshi answered |
=AC = −ai^ + ck^
AB = −ai^ + bj^
Area of △ABC= ½|AB × AC∣
|AB × AC∣ =
−(bc)i^− (ac)j^ − (ab)k^
∣AB × AC∣ = (b2c2 + a2c2 + a2b2)1/2
Area = 1/2(a2b2 + b2c2 + c2a2)1/2
AB = −ai^ + bj^
Area of △ABC= ½|AB × AC∣
|AB × AC∣ =
−(bc)i^− (ac)j^ − (ab)k^
∣AB × AC∣ = (b2c2 + a2c2 + a2b2)1/2
Area = 1/2(a2b2 + b2c2 + c2a2)1/2
and 
is:
If the sum of the squares of the distances of a point from the three coordinate axes be 36, then its distance from the origin is- a)6
- b)3√2
- c)2√3
- d)6√2
Correct answer is option 'B'. Can you explain this answer?
If the sum of the squares of the distances of a point from the three coordinate axes be 36, then its distance from the origin is
a)
6
b)
3√2
c)
2√3
d)
6√2
|
Yash Patel answered |
Let (x,y,z) be the point.
Given sum of the squares of distance from point to the axes is 36.
⇒(x2+y2)+(y2+z2)+(z2+x2)=36
⇒2(x2+y2+z2)=36⇒x2+y2+z2=18
So the distance of the point from the origin is =3(2)1/2
Given sum of the squares of distance from point to the axes is 36.
⇒(x2+y2)+(y2+z2)+(z2+x2)=36
⇒2(x2+y2+z2)=36⇒x2+y2+z2=18
So the distance of the point from the origin is =3(2)1/2
If a line has the direction ratios -4, 18, -12 then what are its direction cosines?- a)-2, 9, -6
- b)-4, 18, -12
- c)2/11, 9/11, 6/11
- d)-2/11, 9/11, -6/11
Correct answer is option 'D'. Can you explain this answer?
If a line has the direction ratios -4, 18, -12 then what are its direction cosines?
a)
-2, 9, -6
b)
-4, 18, -12
c)
2/11, 9/11, 6/11
d)
-2/11, 9/11, -6/11
|
|
Vikas Kapoor answered |
DR of the line : (-4, 18 -12)
DC of the line : (-4/k, 18/k, -12/k)
where k = ((42) + (182) + (12)2)1/2
= (16 + 324 + 144)1/2
= (484)1/2
= 22
So, DC : (-4/22, 18/22, -12/22)
: (-2/11 , 9/11 , -6/11)
DC of the line : (-4/k, 18/k, -12/k)
where k = ((42) + (182) + (12)2)1/2
= (16 + 324 + 144)1/2
= (484)1/2
= 22
So, DC : (-4/22, 18/22, -12/22)
: (-2/11 , 9/11 , -6/11)
The equation of the plane passing through the intersection of the planes 
and 
and the point (1, 2, 1) is:- a)18x+6y+14z-23=0
- b)18x+7y+14z-46=0
- c)9x+3y+7z-23=0
- d)18x+7y+14z-38=0
Correct answer is option 'B'. Can you explain this answer?
The equation of the plane passing through the intersection of the planes and and the point (1, 2, 1) is:
and and the point (1, 2, 1) is:a)
18x+6y+14z-23=0
b)
18x+7y+14z-46=0
c)
9x+3y+7z-23=0
d)
18x+7y+14z-38=0
|
|
Aryan Khanna answered |
n1 = 2i + j + k
n2 = 2i + 3j - 4k
p1 = 4, p2 = -6
r.(n1 + λn2) = p1 + λp2
=> r . [2i + j + k + λ(2i + 3j - 4k)] = 4 - 6λ
=> r . [ i(2 + 2λ) + j(1 + 3λ) + k(1 - 4k)] = 4 - 6λ
Taking r = xi + yj + zk
(2 + 2λ)x + (1 + 3λ)y + (1 - 4k)z = 4 - 6λ
(2x + y + - z - 4) + λ(2x + 3y - 4k + 6) = 0
Given points are (1,2,1)
(2 + 2 - 1 - 4) + λ(2 + 6 - 4 + 6) = 0
-1 + λ(10) = 0
λ = 1/10
Substitute λ = 1/10, we get
18x + 7y + 14z - 46=0
n2 = 2i + 3j - 4k
p1 = 4, p2 = -6
r.(n1 + λn2) = p1 + λp2
=> r . [2i + j + k + λ(2i + 3j - 4k)] = 4 - 6λ
=> r . [ i(2 + 2λ) + j(1 + 3λ) + k(1 - 4k)] = 4 - 6λ
Taking r = xi + yj + zk
(2 + 2λ)x + (1 + 3λ)y + (1 - 4k)z = 4 - 6λ
(2x + y + - z - 4) + λ(2x + 3y - 4k + 6) = 0
Given points are (1,2,1)
(2 + 2 - 1 - 4) + λ(2 + 6 - 4 + 6) = 0
-1 + λ(10) = 0
λ = 1/10
Substitute λ = 1/10, we get
18x + 7y + 14z - 46=0
The distance of the point (2, 3, – 5) from the plane x + 2y – 2z = 9 is:- a)2 units
- b)3/2 units
- c)3 units
- d)10/3 units
Correct answer is option 'C'. Can you explain this answer?
The distance of the point (2, 3, – 5) from the plane x + 2y – 2z = 9 is:
a)
2 units
b)
3/2 units
c)
3 units
d)
10/3 units
|
Nikita Singh answered |
Length of perpendicular from (2,3,-5) to the plane x + 2y − 2z − 9 = 0.
= |(2 + 2×3 −2×(−5) − 9)|√12 + 22 + (−2)2
= |2 + 6 + 10 − 9|/√9
= 9/3
= 3 units.
= |(2 + 2×3 −2×(−5) − 9)|√12 + 22 + (−2)2
= |2 + 6 + 10 − 9|/√9
= 9/3
= 3 units.
The length of the perpendicular from the origin to the plane 3x + 2y – 6z = 21 is:- a)3
- b)14
- c)21
- d)7
Correct answer is option 'A'. Can you explain this answer?
The length of the perpendicular from the origin to the plane 3x + 2y – 6z = 21 is:
a)
3
b)
14
c)
21
d)
7
|
|
Leelu Bhai answered |
Given equation of plane is : 3x + 2y - 6z - 21= 0
the length of perpendicular from a given point
(x' , y', z') on a plane ax + by + cz + d = 0 is given as :-
d = modulus of [{ax' + by' + cz' + d}/{√(a² + b² + c)²}]
so, d = modulus of [{(3*0) + (2*0) + (-6*0) + (-21)}/{√(3² + 2² + (-6)²)}]
d= modulus of (-21/√49) = (-21/7) = 3 units
hence option A is correct....
the length of perpendicular from a given point
(x' , y', z') on a plane ax + by + cz + d = 0 is given as :-
d = modulus of [{ax' + by' + cz' + d}/{√(a² + b² + c)²}]
so, d = modulus of [{(3*0) + (2*0) + (-6*0) + (-21)}/{√(3² + 2² + (-6)²)}]
d= modulus of (-21/√49) = (-21/7) = 3 units
hence option A is correct....
The angle of the line through Z-axis is:
The Cartesian equation of plane passing through (1, 1, 1) and containing the x-axis is- a)z – x = 0
- b)y – z = 0
- c)x – y = 0
- d)2x – y – z = 0
Correct answer is option 'B'. Can you explain this answer?
The Cartesian equation of plane passing through (1, 1, 1) and containing the x-axis is
a)
z – x = 0
b)
y – z = 0
c)
x – y = 0
d)
2x – y – z = 0
|
|
Naina Sharma answered |
Eq. of x-axis is
Equation of plane passing through (1, 1, 1) and containing
Equation of plane passing through (1, 1, 1) and containing
In the following case, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0- a)The planes are perpendicular
- b)The planes are parallel
- c)The planes are at 45∘
- d)The planes are at 55∘
Correct answer is option 'B'. Can you explain this answer?
In the following case, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
a)
The planes are perpendicular
b)
The planes are parallel
c)
The planes are at 45∘
d)
The planes are at 55∘
|
Pranavi Iyer answered |
Find the direction cosines of the x axis.- a)1, 0, 0
- b)0, 0, 0
- c)0, 1, 0
- d)0, 0, 1
Correct answer is option 'A'. Can you explain this answer?
Find the direction cosines of the x axis.
a)
1, 0, 0
b)
0, 0, 0
c)
0, 1, 0
d)
0, 0, 1
|
Infinity Academy answered |
To find Direction Cosines of X-axis.
Take any two points on X-axis : A(a,0,0) & B(b,0,0)
DR of AB : (b-a,0,0)
DC of AB : ((b-a)/(((b-a)2 + 0 + 0)1/2), 0, 0)
: ((b-a)/(b-a) , 0 , 0)
: (1,0,0)
Take any two points on X-axis : A(a,0,0) & B(b,0,0)
DR of AB : (b-a,0,0)
DC of AB : ((b-a)/(((b-a)2 + 0 + 0)1/2), 0, 0)
: ((b-a)/(b-a) , 0 , 0)
: (1,0,0)
The equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5) is:- a)– x – 5y + 6z + 18 = 0
- b)x – 5y + 6z + 18 = 0
- c)x + 5y – 6z + 18 = 0
- d)– x – 5y – 6z + 18 = 0
Correct answer is 'C'. Can you explain this answer?
The equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5) is:
a)
– x – 5y + 6z + 18 = 0
b)
x – 5y + 6z + 18 = 0
c)
x + 5y – 6z + 18 = 0
d)
– x – 5y – 6z + 18 = 0
|
|
Leelu Bhai answered |
find its caryesian equation.
A point moves so that the sum of the squares of its distances from the six faces of a cube given by x = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is- a)x2 + y2 + z2 = 1
- b)x2 + y2 + z2 = 2
- c)x + y + z = 1
- d)x + y + z = 2
Correct answer is option 'B'. Can you explain this answer?
A point moves so that the sum of the squares of its distances from the six faces of a cube given by x = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is
a)
x2 + y2 + z2 = 1
b)
x2 + y2 + z2 = 2
c)
x + y + z = 1
d)
x + y + z = 2
|
|
Ram Mohith answered |
Let the point be (x,y,z)
Distance of this point from x = 1 is |x -1| and from x = +1 is |x + 1|. Similarly you can find the distance from the other faces. The sum of squares of distances will be,
(x - 1)^2 + (x + 1)^2 + (y - 1)^2 + (y + 1)^2 + (z - 1)^2 + (z + 1)^2 = 10
2(x^2 + y^2 + z^2 ) + 6 = 10
x^2 + y^2 + z^2 = 2
Distance of this point from x = 1 is |x -1| and from x = +1 is |x + 1|. Similarly you can find the distance from the other faces. The sum of squares of distances will be,
(x - 1)^2 + (x + 1)^2 + (y - 1)^2 + (y + 1)^2 + (z - 1)^2 + (z + 1)^2 = 10
2(x^2 + y^2 + z^2 ) + 6 = 10
x^2 + y^2 + z^2 = 2
The equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5) is:- a)– x – 5y + 6z + 18 = 0
- b)x – 5y + 6z + 18 = 0
- c)x + 5y – 6z + 18 = 0
- d)– x – 5y – 6z + 18 = 0
Correct answer is option 'C'. Can you explain this answer?
The equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5) is:
a)
– x – 5y + 6z + 18 = 0
b)
x – 5y + 6z + 18 = 0
c)
x + 5y – 6z + 18 = 0
d)
– x – 5y – 6z + 18 = 0
|
Preeti Iyer answered |
The equation of the plane passing through the point (3, – 3, 1) is:
a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points
(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.
Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratios of the normal to the plane is – 1, – 5, 6.
So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0
i.e., x + 5y – 6z + 18 = 0.
a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points
(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.
Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratios of the normal to the plane is – 1, – 5, 6.
So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0
i.e., x + 5y – 6z + 18 = 0.
A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular drawn from origin to this plane is- a)x2 + y2 + z2 – x – 2y – 3z = 0
- b)x2 + 2y2 + 3z2 – x – 2y – 3z = 0
- c)x2 + 4y2 + 9z2 + x + 2y + 3 = 0
- d)x2 + y2 + z2 + x + 2y + 3z = 0
Correct answer is option 'A'. Can you explain this answer?
A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular drawn from origin to this plane is
a)
x2 + y2 + z2 – x – 2y – 3z = 0
b)
x2 + 2y2 + 3z2 – x – 2y – 3z = 0
c)
x2 + 4y2 + 9z2 + x + 2y + 3 = 0
d)
x2 + y2 + z2 + x + 2y + 3z = 0
|
|
Neha Joshi answered |
α(x − α) + β(y − β) + γ(z − γ) = 0
α(1 − α) + β (2 − β) + γ ( 3 − γ) = 0
α + 2β + 3γ = α2 + β2 + γ2
α2 + β2 + γ2 − α − 2β − 3γ = 0
x2 + y2 + z2 − x − 2y− 3z = 0
α(1 − α) + β (2 − β) + γ ( 3 − γ) = 0
α + 2β + 3γ = α2 + β2 + γ2
α2 + β2 + γ2 − α − 2β − 3γ = 0
x2 + y2 + z2 − x − 2y− 3z = 0
The equation of the plane, which is at a distance of 5 unit from the origin and has 
as a normal vector, is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The equation of the plane, which is at a distance of 5 unit from the origin and has as a normal vector, is:
as a normal vector, is:a)
b)
c)
d)
|
|
Preeti Iyer answered |
x = 3i - 2j - 6k
|x| = ((3)2 + (2)2 + (6)2)
|x| = (49)½
|x| = 7
x = x/|x|
= (3i - 2j - 6k)/7
The required equation of plane is r.x = d
⇒ r.(3i - 2j - 6k)/7 = 5
⇒ r.(3i - 2j - 6k) = 35
|x| = ((3)2 + (2)2 + (6)2)
|x| = (49)½
|x| = 7
x = x/|x|
= (3i - 2j - 6k)/7
The required equation of plane is r.x = d
⇒ r.(3i - 2j - 6k)/7 = 5
⇒ r.(3i - 2j - 6k) = 35
Direction: Read the following text and answer the following questions on the basis of the same:A mobile tower stands at the top of a hill. Consider the surface on which the tower stands as a plane having points A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1) on it. The mobile tower is tied with 3 cables from the points A, B and C such that it stands vertically on the ground. The top of the tower is at the point (2, 3, 1) as shown in the figure.
Q. The equation of the plane passing through the points A, B and C is- a)3x – 2y + 4z = –11
- b)3x + 2y + 4z = 11
- c)3x – 2y – 4z = 11
- d)–3x + 2y + 4z = –11
Correct answer is option 'B'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
A mobile tower stands at the top of a hill. Consider the surface on which the tower stands as a plane having points A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1) on it. The mobile tower is tied with 3 cables from the points A, B and C such that it stands vertically on the ground. The top of the tower is at the point (2, 3, 1) as shown in the figure.
Q. The equation of the plane passing through the points A, B and C is
a)
3x – 2y + 4z = –11
b)
3x + 2y + 4z = 11
c)
3x – 2y – 4z = 11
d)
–3x + 2y + 4z = –11
|
Krishna Iyer answered |
A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1)
(x - 1)(1 + 2) -y (-2, 0) + (z - 2) (4 - 0)
3x - 3 + 2y + 4z - 8 = 0
3x + 2y + 4z = 11
If the distance of the point P(1, -2, 1) from the plane x + 2y - 2z = a, where a > 0 is 5, then foot of perpendicular from P to the plane is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
If the distance of the point P(1, -2, 1) from the plane x + 2y - 2z = a, where a > 0 is 5, then foot of perpendicular from P to the plane is
a)
b)
c)
d)
|
Lavanya Menon answered |
Distance of pt P from plane = 5
Gen. pt. on line is (λ + 1, 2λ - 2, -2λ + 1)
This pt line on plane
(λ + 1 + 2 2λ - 2) - 2 (-2λ + 1) = 10
Gen. pt. on line is (λ + 1, 2λ - 2, -2λ + 1)
This pt line on plane
(λ + 1 + 2 2λ - 2) - 2 (-2λ + 1) = 10
If the lines 
and 
are concurrent then- a)h = –2,k = –6
- b)h = 1/2, k = 2
- c)h = 6, k = 2
- d)h = 2, k = 1/2
Correct answer is option 'D'. Can you explain this answer?
If the lines and are concurrent then
and are concurrent thena)
h = –2,k = –6
b)
h = 1/2, k = 2
c)
h = 6, k = 2
d)
h = 2, k = 1/2
|
|
Sneha Singh answered |
Their determinant equal to zero
and 
and at greatest distance from the point (0, 0, 0) is
and 
is:
The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 is- a)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0
- b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0
- c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0
- d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0
Correct answer is option 'A'. Can you explain this answer?
The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 is
a)
x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0
b)
x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0
c)
x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0
d)
x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0
|
Kritika Sarkar answered |
To find the equation of the plane through the intersection of two given planes, we need to first determine the line of intersection and then find a normal vector for the plane.
1. Finding the line of intersection:
The two given planes are:
Plane 1: x + y + z = 1
Plane 2: 2x + 3y + z + 4 = 0
To find the line of intersection, we can set the equations of the two planes equal to each other:
x + y + z = 1
2x + 3y + z + 4 = 0
By subtracting the second equation from the first equation, we can eliminate z:
x + y + z - (2x + 3y + z + 4) = 1 - 0
-x - 2y - 4 = -1
Simplifying the equation, we get:
x + 2y = 3
This equation represents the line of intersection of the two planes.
2. Finding a normal vector for the plane:
Since the line of intersection lies on both planes, the normal vector of the required plane should be perpendicular to this line. Therefore, we can choose the direction ratios of the line, -1 and 2, as coefficients of the normal vector.
The normal vector of the required plane is given by the cross product of the direction ratios of the line of intersection:
n = (1, 2, 0) x (-1, 2, 0) = (0, 0, -4)
3. Writing the equation of the plane:
Now we have a point on the plane, (1, 0, 0), and a normal vector, (0, 0, -4). We can use the point-normal form of the equation of a plane to write the equation of the required plane.
The equation of the plane is given by:
0(x - 1) + 0(y - 0) + (-4)(z - 0) = 0
Simplifying the equation, we get:
-4z + 4 = 0
z = 1
Therefore, the equation of the plane through the intersection of the given planes is:
x + 2y - 4z + 4 = 0
Comparing this equation with the options provided, we see that option A is the correct answer:
x(1 2k) + y(1 3k) + z(1 k) + (-1 4k) = 0
1. Finding the line of intersection:
The two given planes are:
Plane 1: x + y + z = 1
Plane 2: 2x + 3y + z + 4 = 0
To find the line of intersection, we can set the equations of the two planes equal to each other:
x + y + z = 1
2x + 3y + z + 4 = 0
By subtracting the second equation from the first equation, we can eliminate z:
x + y + z - (2x + 3y + z + 4) = 1 - 0
-x - 2y - 4 = -1
Simplifying the equation, we get:
x + 2y = 3
This equation represents the line of intersection of the two planes.
2. Finding a normal vector for the plane:
Since the line of intersection lies on both planes, the normal vector of the required plane should be perpendicular to this line. Therefore, we can choose the direction ratios of the line, -1 and 2, as coefficients of the normal vector.
The normal vector of the required plane is given by the cross product of the direction ratios of the line of intersection:
n = (1, 2, 0) x (-1, 2, 0) = (0, 0, -4)
3. Writing the equation of the plane:
Now we have a point on the plane, (1, 0, 0), and a normal vector, (0, 0, -4). We can use the point-normal form of the equation of a plane to write the equation of the required plane.
The equation of the plane is given by:
0(x - 1) + 0(y - 0) + (-4)(z - 0) = 0
Simplifying the equation, we get:
-4z + 4 = 0
z = 1
Therefore, the equation of the plane through the intersection of the given planes is:
x + 2y - 4z + 4 = 0
Comparing this equation with the options provided, we see that option A is the correct answer:
x(1 2k) + y(1 3k) + z(1 k) + (-1 4k) = 0
For which value of a lines 
and 
are perpendicular?
- a)11/70
- b)5
- c)1
- d)70/11
Correct answer is option 'B'. Can you explain this answer?
For which value of a lines and are perpendicular?
and are perpendicular?a)
11/70
b)
5
c)
1
d)
70/11
|
|
Tejas Verma answered |
(x-1)/(-3) = (y-2)/(2p/7) = (z-3)/2
(x-1)(-3p/7) = (y - 5)/1 = (z - 6)/(-5)
The direction ratio of the line are -3, 2p/7, -2 and (-3p)/7, 1, -5
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0
Therefore, (-3)(-3p/7) + (2p/7)(1) + 2(-5) = 0
(9p/7) + (2p/7) = 10
11p = 70
p = 70/1
(x-1)(-3p/7) = (y - 5)/1 = (z - 6)/(-5)
The direction ratio of the line are -3, 2p/7, -2 and (-3p)/7, 1, -5
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0
Therefore, (-3)(-3p/7) + (2p/7)(1) + 2(-5) = 0
(9p/7) + (2p/7) = 10
11p = 70
p = 70/1
A variable plane passes through a fixed point (a, b, c) and meets the coordinate axes in A, B, C. Locus of the point common to the planes through A, B, C and parallel to coordinate plane, is- a)
- b)
- c)ax + by + cz = 1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
A variable plane passes through a fixed point (a, b, c) and meets the coordinate axes in A, B, C. Locus of the point common to the planes through A, B, C and parallel to coordinate plane, is
a)
b)
c)
ax + by + cz = 1
d)
None of these
|
|
Silent Nisarg answered |
Option a is correct
Direction: Read the following text and answer the following questions on the basis of the same:A cricket match is organized between two Clubs A and B for which a team from each club is chosen. Remaining players of Club A and Club B are respectively sitting on the plane represented by the equation 
to cheer the team of their own clubs.
Q. The intercept form of the equation of the plane on which players of Club B are seated is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
A cricket match is organized between two Clubs A and B for which a team from each club is chosen. Remaining players of Club A and Club B are respectively sitting on the plane represented by the equation to cheer the team of their own clubs.
to cheer the team of their own clubs.Q. The intercept form of the equation of the plane on which players of Club B are seated is
a)
b)
c)
d)
|
Hansa Sharma answered |
Direction: Read the following text and answer the following questions on the basis of the same:The equation of motion of a missile are x = 3t, y = –4t, z = t, where the time ‘t’ is given in seconds, and the distance is measured in kilometres.
Q. At what distance will the rocket be from the starting point (0, 0, 0) in 5 seconds?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
The equation of motion of a missile are x = 3t, y = –4t, z = t, where the time ‘t’ is given in seconds, and the distance is measured in kilometres.
Q. At what distance will the rocket be from the starting point (0, 0, 0) in 5 seconds?
a)
b)
c)
d)
|
Cstoppers Instructors answered |
Here,
t = 5 seconds
x = 3t = 3 x 5 = 15
y = -4t = -4 x 5 = -20
z = t = 5
(x, y, z) (15, -20, 5)
Distance from starting point(0, 0, 0)
Direction: Read the following text and answer the following questions on the basis of the same:Consider the plane π1 : 2x – 3y + 4z + 9 = 0 and the point P(1, –2, 3). π1 is a plane parallel to π1 and containing the point P.
Q. Distance between π1 and π2 is _______ units.- a)5
- b)
- c)
- d)2√3
Correct answer is option 'B'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
Consider the plane π1 : 2x – 3y + 4z + 9 = 0 and the point P(1, –2, 3). π1 is a plane parallel to π1 and containing the point P.
Q. Distance between π1 and π2 is _______ units.
a)
5
b)
c)
d)
2√3
|
|
Tanuja Kapoor answered |
Distance between π1 and π2
The non zero value of ‘a’ for which the lines 2x – y + 3z + 4 = 0 = ax + y – z + 2 and x – 3y + z = 0 = x + 2y + z + 1 are co-planar is- a)–2
- b)4
- c)6
- d)0
Correct answer is option 'A'. Can you explain this answer?
The non zero value of ‘a’ for which the lines 2x – y + 3z + 4 = 0 = ax + y – z + 2 and x – 3y + z = 0 = x + 2y + z + 1 are co-planar is
a)
–2
b)
4
c)
6
d)
0
|
Parth Yadav answered |
2x - y +3z + 4 = 0
x - 3y + z = 0
x + 2y + z + 1 = 0
x = 12/5
y = -1/5
z= -3
this point also satisfied by
ax - y + z + 2 = 0
a(12 / 5) - (-1/5) + (-3) = 0
⇒ 12a/5 + 1/5 -3 = 0
a= -2
x - 3y + z = 0
x + 2y + z + 1 = 0
x = 12/5
y = -1/5
z= -3
this point also satisfied by
ax - y + z + 2 = 0
a(12 / 5) - (-1/5) + (-3) = 0
⇒ 12a/5 + 1/5 -3 = 0
a= -2
Find the direction cosines of a line which makes equal angles with all three the coordinate axes.- a)± 1/√2, ± 1/√2, ± 1/√2
- b)1/√3, 1/√3, 1/√3
- c)1/√2, 1/√2, 1/√2
- d)± 1/√3, ± 1/√3, ± 1/√3
Correct answer is option 'D'. Can you explain this answer?
Find the direction cosines of a line which makes equal angles with all three the coordinate axes.
a)
± 1/√2, ± 1/√2, ± 1/√2
b)
1/√3, 1/√3, 1/√3
c)
1/√2, 1/√2, 1/√2
d)
± 1/√3, ± 1/√3, ± 1/√3
|
|
Sakshi Shah answered |
Understanding Direction Cosines
Direction cosines are the cosines of the angles that a line makes with the coordinate axes. For a line to make equal angles with all three axes, the direction cosines must be equal in magnitude.
Equal Angles with Axes
If a line makes equal angles with the x, y, and z axes, then we can denote the direction cosines as:
- l = cos(α)
- m = cos(β)
- n = cos(γ)
Where α, β, and γ are the angles with the x, y, and z axes, respectively.
Since the angles are equal, we have:
- l = m = n
Magnitude Condition
The sum of the squares of the direction cosines must equal 1:
- l² + m² + n² = 1
Substituting l = m = n gives:
- 3l² = 1
- l² = 1/3
- l = ±1/√3
Thus, the direction cosines can be expressed as:
- l = ±1/√3
- m = ±1/√3
- n = ±1/√3
Conclusion
The direction cosines that make equal angles with the coordinate axes are:
- ±1/√3, ±1/√3, ±1/√3
This corresponds to option 'D':
- Direction cosines: ±1/√3, ±1/√3, ±1/√3
Therefore, the correct answer is indeed option 'D'.
Direction cosines are the cosines of the angles that a line makes with the coordinate axes. For a line to make equal angles with all three axes, the direction cosines must be equal in magnitude.
Equal Angles with Axes
If a line makes equal angles with the x, y, and z axes, then we can denote the direction cosines as:
- l = cos(α)
- m = cos(β)
- n = cos(γ)
Where α, β, and γ are the angles with the x, y, and z axes, respectively.
Since the angles are equal, we have:
- l = m = n
Magnitude Condition
The sum of the squares of the direction cosines must equal 1:
- l² + m² + n² = 1
Substituting l = m = n gives:
- 3l² = 1
- l² = 1/3
- l = ±1/√3
Thus, the direction cosines can be expressed as:
- l = ±1/√3
- m = ±1/√3
- n = ±1/√3
Conclusion
The direction cosines that make equal angles with the coordinate axes are:
- ±1/√3, ±1/√3, ±1/√3
This corresponds to option 'D':
- Direction cosines: ±1/√3, ±1/√3, ±1/√3
Therefore, the correct answer is indeed option 'D'.
The equation of the plane passing through the point (1, – 3, –2) and perpendicular to planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8, is- a)2x – 4y + 3z – 8 = 0
- b)2x – 4y – 3z + 8 = 0
- c)2x – 4y + 3z + 8 = 0
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The equation of the plane passing through the point (1, – 3, –2) and perpendicular to planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8, is
a)
2x – 4y + 3z – 8 = 0
b)
2x – 4y – 3z + 8 = 0
c)
2x – 4y + 3z + 8 = 0
d)
None of these
|
Nidhi Sen answered |
-2, 3) and perpendicular to the vector (2, 1, -4) can be found using the point-normal form of the equation of a plane:
(x - x1)(a) + (y - y1)(b) + (z - z1)(c) = 0
where (x1, y1, z1) is the given point and (a, b, c) is the normal vector to the plane.
Substituting the given values, we get:
(x - 1)(2) + (y + 2)(1) + (z - 3)(-4) = 0
Expanding and simplifying, we get:
2x + y - 4z - 15 = 0
Therefore, the equation of the plane is:
2x + y - 4z - 15 = 0
(x - x1)(a) + (y - y1)(b) + (z - z1)(c) = 0
where (x1, y1, z1) is the given point and (a, b, c) is the normal vector to the plane.
Substituting the given values, we get:
(x - 1)(2) + (y + 2)(1) + (z - 3)(-4) = 0
Expanding and simplifying, we get:
2x + y - 4z - 15 = 0
Therefore, the equation of the plane is:
2x + y - 4z - 15 = 0
The coplanar points A, B, C, D are (2 – x, 2, 2), (2, 2 – y, 2), (2, 2, 2 – z) and (1, 1, 1) respectively. Then- a)
- b)x + y + z = 1
- c)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The coplanar points A, B, C, D are (2 – x, 2, 2), (2, 2 – y, 2), (2, 2, 2 – z) and (1, 1, 1) respectively. Then
a)
b)
x + y + z = 1
c)
d)
None of these
|
|
Neha Joshi answered |
We have four coplanar points.
The three vectors connecting two of them at a time are thus coplanar.
⇒{(−x, y, 0) (-x, 0, z) (1−x, 1, 1)} = 0
⇒−x (−z) −y (−x−z+xz)+0=0
⇒xz + xy + yz = xyz
⇒1/y + 1/z + 1/x = 1
The three vectors connecting two of them at a time are thus coplanar.
⇒{(−x, y, 0) (-x, 0, z) (1−x, 1, 1)} = 0
⇒−x (−z) −y (−x−z+xz)+0=0
⇒xz + xy + yz = xyz
⇒1/y + 1/z + 1/x = 1
Direction: Read the following text and answer the following questions on the basis of the same:The Indian coast guard, while patrolling, saw a suspicious boat with people. They were nowhere looking like fishermen. The coast guard were closely observing the movement of the boat for an opportunity to seize the boat. They observed that the boat is moving along a planar surface. At an instant of time, the coordinates of the position of the coast guard helicopter and the boat is (1, 3, 5) and (2, 5, 3) respectively.
Q. At that given instant of time, the equation of line passing through the positions of the helicopter and boat is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
The Indian coast guard, while patrolling, saw a suspicious boat with people. They were nowhere looking like fishermen. The coast guard were closely observing the movement of the boat for an opportunity to seize the boat. They observed that the boat is moving along a planar surface. At an instant of time, the coordinates of the position of the coast guard helicopter and the boat is (1, 3, 5) and (2, 5, 3) respectively.
Q. At that given instant of time, the equation of line passing through the positions of the helicopter and boat is
a)
b)
c)
d)
|
Top Rankers answered |
Here, direction cosines are 1, 2, –2.
Equation of line passing through the positions of the helicopter and boat is
⇒ 
Chapter doubts & questions for Chapter 11 - Three Dimensional Geometry - Mathematics (Maths) Class 12 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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