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All questions of Vector Algebra and Three Dimensional Geometry for JEE Exam
Let α, β, γ be distinct real numbers. The points with position vectors 
- a)are collinear
- b)form an equilateral triangle
- c)form a scalene triangle
- d)form a right angled triangle
Correct answer is option 'B'. Can you explain this answer?
Let α, β, γ be distinct real numbers. The points with position vectors
a)
are collinear
b)
form an equilateral triangle
c)
form a scalene triangle
d)
form a right angled triangle
|
Anaya Patel answered |
Let the given position vectors be of point A, B and C respectively, then
⇒ ΔABC is an equilateral Δ.
If the vectors 
and 
are mutually orthogonal, then (λ, μ)=- a)(2, –3)
- b)(–2, 3)
- c)(3, –2)
- d)(–3, 2)
Correct answer is option 'D'. Can you explain this answer?
If the vectors and are mutually orthogonal, then (λ, μ)=
and are mutually orthogonal, then (λ, μ)=a)
(2, –3)
b)
(–2, 3)
c)
(3, –2)
d)
(–3, 2)
|
|
Anaya Patel answered |
Since 
are mutually orthogonal
On solving (i) and (ii), we get λ = -3, μ = 2
are mutually orthogonal On solving (i) and (ii), we get λ = -3, μ = 2
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is- a)3/2
- b)5/2
- c)7/2
- d)9/2
Correct answer is option 'C'. Can you explain this answer?
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
a)
3/2
b)
5/2
c)
7/2
d)
9/2
|
Lavanya Menon answered |
2x + y + 2z – 8 = 0 …(Plane 1)
…(Plane 2)Distance between Plane 1 and 2
in (1), we get
lie in a plane, then c is
arecoplanar if
A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1) B(2, 1, 3) and C(–1, 1, 2). Then the angle between the faces OAB and ABC will be- a)90°
- b)
- c)
- d)30°
Correct answer is option 'B'. Can you explain this answer?
A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1) B(2, 1, 3) and C(–1, 1, 2). Then the angle between the faces OAB and ABC will be
a)
90°
b)
c)
d)
30°
|
|
Lavanya Menon answered |
Vector perpendicular to the face OAB
Vector perpendicular to the face ABC
Angle between the faces = angle between their normals
If (2, 3, 5) is one end of a diameter of the sphere x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0, then the cooordinates of the other end of the diameter are- a)(4, 3, 5)
- b)(4, 3, – 3)
- c)(4, 9, – 3)
- d)(4, –3, 3).
Correct answer is option 'C'. Can you explain this answer?
If (2, 3, 5) is one end of a diameter of the sphere x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0, then the cooordinates of the other end of the diameter are
a)
(4, 3, 5)
b)
(4, 3, – 3)
c)
(4, 9, – 3)
d)
(4, –3, 3).
|
|
Ram Mohith answered |
The center for the givens sphere is (3,6,1) and one end of diameter is (2,3,5). So, we can find the other end using mid-point formula. Let the other end be (x,y,z) then,
3 = (2 + x)/2 => x = 4
6 = (3 + y)/2 => y = 9
1 = (5 + z)/2 => z = -3
Therefore, the other end of the diameter is (4,9,-3)
Two system of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a' , b' , c' from the origin then- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Two system of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a' , b' , c' from the origin then
a)
b)
c)
d)
|
Chirag Verma answered |
Eq. of planes be 
(⊥ r distance on plane from origin is same.)
If the vectors 
are the sides of a triangle ABC, then the length of the median through A is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If the vectors are the sides of a triangle ABC, then the length of the median through A is
are the sides of a triangle ABC, then the length of the median through A isa)
b)
c)
d)
|
|
Chirag Verma answered |
∵ M is mid point of BC
Length of median AM
If the line, 
lies in the plane, lx + my – z = 9,then l2 + m2 is equal to :- a)5
- b)2
- c)26
- d)18
Correct answer is option 'B'. Can you explain this answer?
If the line, lies in the plane, lx + my – z = 9,then l2 + m2 is equal to :
lies in the plane, lx + my – z = 9,then l2 + m2 is equal to :a)
5
b)
2
c)
26
d)
18
|
Rohit Jain answered |
Line lies in the plane ⇒ (3, –2, –4) lie in the plane
⇒ 3ℓ – 2m + 4 = 9 or 3ℓ – 2m = 5 ..... (1)
Also, ℓ, m,–1 are dr's of line perpendicular to plane and 2, –1, 3 are dr's of line lying in the plane
⇒ 2ℓ – m – 3 = 0 or 2ℓ – m = 3 .....(2)
Solving (1) and (2) we get ℓ = 1 and m = –1
⇒ ℓ2 + m2 = 2
⇒ 3ℓ – 2m + 4 = 9 or 3ℓ – 2m = 5 ..... (1)
Also, ℓ, m,–1 are dr's of line perpendicular to plane and 2, –1, 3 are dr's of line lying in the plane
⇒ 2ℓ – m – 3 = 0 or 2ℓ – m = 3 .....(2)
Solving (1) and (2) we get ℓ = 1 and m = –1
⇒ ℓ2 + m2 = 2
are coplanar, is
be three non - coplanar vectors and 
are vectors defined by the relations 
then the value of the expression 
is equal to
are non coplanar
Let 
be three vectors. A vector in the plane of 
whose projection on 
- a)
- b)
- c)
- d)
Correct answer is option 'A,C'. Can you explain this answer?
Let be three vectors. A vector in the plane of whose projection on
be three vectors. A vector in the plane of whose projection on a)
b)
c)
d)
|
|
Chirag Verma answered |
We have
Given magnitude of projection of
∴ The required vector is either,
are coplanar. Then a can take value(s)
are 3 vectors, such that 
, then 
is equal to- a)1
- b)0
- c)–7
- d)7
Correct answer is option 'C'. Can you explain this answer?
are 3 vectors, such that , then is equal toa)
1
b)
0
c)
–7
d)
7
|
Vikash Yadav answered |
Square a+b+c=0
(a+b+c)^2 = a^2 + b^2 + c^2 + 2 a.b + 2 b.c + 2 c.a
0 = 1 + 4 + 9 + 2(a.b + b.c + c.a)
(a.b + b.c + c.a)= -14/2 = -7
lies in the plane of the vectors 
and bisects the angle between 
Then which one of the following gives possible values of α and β?
If a line makes an angle of π /4 with the positive directions of each of x- axis and y- axis, then the angle that the line makes with the positive direction of the z-axis is- a)π /4
- b)π /2
- c)π /6
- d)π /3
Correct answer is option 'B'. Can you explain this answer?
If a line makes an angle of π /4 with the positive directions of each of x- axis and y- axis, then the angle that the line makes with the positive direction of the z-axis is
a)
π /4
b)
π /2
c)
π /6
d)
π /3
|
Tushar Choudhury answered |
Let the angle of line makes with the positive direction of z-axis is a direction cosines of line with the +ve directions of x-axis, y-axis, and z-axis is l, m, n respectively.
Hence, angle with positive direction of the
and vectrs (1, a,a2),(1, b, b2) and (1, c,c2) are non- coplanar, then the product abc equals
The angle between the lines whose direction cosines satisfy the equations l + m +n= 0 and l2 = m2 + n2 is- a)π/6
- b)π/2
- c)π/3
- d)π/4
Correct answer is option 'C'. Can you explain this answer?
The angle between the lines whose direction cosines satisfy the equations l + m +n= 0 and l2 = m2 + n2 is
a)
π/6
b)
π/2
c)
π/3
d)
π/4
|
Nabanita Nambiar answered |
Given l + m + n = 0 and l2 = m2 + n2
Now, (–m –n)2 = m2 + n2
⇒ mn = 0 ⇒ m = 0 or n = 0
If m = 0 then l = –n
Now, (–m –n)2 = m2 + n2
⇒ mn = 0 ⇒ m = 0 or n = 0
If m = 0 then l = –n
be three vector s. A vector
the plane of 
whose projection on 
, is given by 
The value of k such that 
lies in the plane 2x – 4y + z = 7, is- a)7
- b)–7
- c)no real value
- d)4
Correct answer is option 'A'. Can you explain this answer?
The value of k such that lies in the plane 2x – 4y + z = 7, is
lies in the plane 2x – 4y + z = 7, isa)
7
b)
–7
c)
no real value
d)
4
|
|
Gaurav Kumar answered |
As the line 
lies in th e plan e 2x - 4 y +z= 7, the point (4, 2, k) through which line passes must also lie on the given plane and hence 2 × 4 – 4 × 2 + k = 7 ⇒ k = 7
lies in th e plan e 2x - 4 y +z= 7, the point (4, 2, k) through which line passes must also lie on the given plane and hence 2 × 4 – 4 × 2 + k = 7 ⇒ k = 7
is a unit vector such that 
is equal to
perpendicular 
The two lines x = ay+b , z = cy+d ; and x = a' y+b' , z = c' y+d' are perpendicular to each other if- a)aa '+ cc ' =-1
- b)aa '+ cc '=1
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The two lines x = ay+b , z = cy+d ; and x = a' y+b' , z = c' y+d' are perpendicular to each other if
a)
aa '+ cc ' =-1
b)
aa '+ cc '=1
c)
d)
|
|
Harsh Singhal answered |
First find direction of two line by taking cross product of their eq.of plane & then their dot product become zero
Let the line 
lie in the plane x + 3y – αz + β = 0. Then (α, β) equals- a)(–6, 7)
- b)(5, –15)
- c)(–5, 5)
- d)(6, –17)
Correct answer is option 'A'. Can you explain this answer?
Let the line lie in the plane x + 3y – αz + β = 0. Then (α, β) equals
lie in the plane x + 3y – αz + β = 0. Then (α, β) equalsa)
(–6, 7)
b)
(5, –15)
c)
(–5, 5)
d)
(6, –17)
|
|
Sonu Kumar answered |
Line lie in the plane and line passes through (2, 1,-2) then plane also passes through this point.
by satisfying find relation between alpha and beta.
and second direction ratio of line and plane are perpendicular then by dot products we get the value of alpha.
by satisfying find relation between alpha and beta.
and second direction ratio of line and plane are perpendicular then by dot products we get the value of alpha.
thus what will be the value of 
given that 
where 
are any three vectors such that 
are
are not perpendicular and 
are two vectors satisfying 
Then the vector 
Let 
be unit vectors such that 
Which one of the following is correct ?- a)
- b)
- c)
- d)
are muturally perpendicular
Correct answer is option 'B'. Can you explain this answer?
Let be unit vectors such that Which one of the following is correct ?
be unit vectors such that Which one of the following is correct ?a)
b)
c)
d)
are muturally perpendicular
|
Ankita Das answered |
These vectors are forming an equilateral triangle so option b will be correct
A line with positive direction cosines passes through the point P(2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals- a)1
- b)√2
- c)√3
- d)2
Correct answer is option 'C'. Can you explain this answer?
A line with positive direction cosines passes through the point P(2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals
a)
1
b)
√2
c)
√3
d)
2
|
Samarth Saha answered |
The line has +ve and equal direction cosines, these are 
or direction ratios are 1, 1, 1. Also thelines passes through P (2, – 1, 2).
∴ Equation of line is
or direction ratios are 1, 1, 1. Also thelines passes through P (2, – 1, 2).∴ Equation of line is
be a point on this line where it meets the plane 2 x + y + z = 9Then Q must satisfy the eqn of plane
∴ Q has coordintes (3, 0, 3)
are coplanar, then the plane (s) containing these two lines is (are)
Let 
If the vectors 
in the plane of 
then x equals- a)– 4
- b)– 2
- c)0
- d)1.
Correct answer is option 'B'. Can you explain this answer?
Let If the vectors in the plane of then x equals
If the vectors in the plane of then x equalsa)
– 4
b)
– 2
c)
0
d)
1.
|
|
Anand Kumar answered |
Since, vector a,b and c are coplanar.
Therefore, by condition of coplanarity,
vector c=m(vector a)+n(vector b)
substitute the given values of the individual vector.
further, relate coefficient of i,j and k of LHS to RHS
there is will have three relations between X,m and n . solve further for m,n and X.
Therefore, by condition of coplanarity,
vector c=m(vector a)+n(vector b)
substitute the given values of the individual vector.
further, relate coefficient of i,j and k of LHS to RHS
there is will have three relations between X,m and n . solve further for m,n and X.
If 
are non-coplanar vectors and l is a real number, then the vectors 
are non coplanar for- a)no value of λ
- b)all except on e value of λ
- c)all except two values of λ
- d)all values of λ
Correct answer is option 'C'. Can you explain this answer?
If are non-coplanar vectors and l is a real number, then the vectors are non coplanar for
are non-coplanar vectors and l is a real number, then the vectors are non coplanar fora)
no value of λ
b)
all except on e value of λ
c)
all except two values of λ
d)
all values of λ
|
Sahil Ghosh answered |
coplanar if
∵ Forces are noncoplanar for all λ, except 
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y +4z + 5 = 0 is- a)9/2
- b)5/2
- c)7/2
- d)3/2
Correct answer is option 'C'. Can you explain this answer?
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y +4z + 5 = 0 is
a)
9/2
b)
5/2
c)
7/2
d)
3/2
|
Kiran Sengupta answered |
To find the distance between two parallel planes, we need to find the perpendicular distance between them. The normal vectors of the planes will be parallel, so we can find the distance between the planes by finding the distance between a point on one plane and the other plane.
Given planes:
1) 2x + y + 2z = 8
2) 4x + 2y + 4z + 5 = 0
Step 1: Find the normal vectors of the planes
The normal vector of a plane is the coefficients of x, y, and z in the equation of the plane. So, the normal vectors for the given planes are:
1) Plane 1: (2, 1, 2)
2) Plane 2: (4, 2, 4)
Step 2: Find a point on one of the planes
To find a point on Plane 1, we can assume a value for one of the variables and solve for the other two variables. Let's assume x = 0. Substituting x = 0 in Plane 1 equation, we get:
y + 2z = 8
Assuming y = 0, we get:
2z = 8
z = 4
So, a point on Plane 1 is (0, 0, 4).
Step 3: Find the distance between the point and Plane 2
To find the distance between a point and a plane, we can use the formula:
Distance = |ax + by + cz + d| / √(a^2 + b^2 + c^2)
where (a, b, c) are the coefficients of x, y, and z in the equation of the plane, and d is the constant term.
Substituting the values for Plane 2, we get:
Distance = |4(0) + 2(0) + 4(4) + 5| / √(4^2 + 2^2 + 4^2)
= |16 + 5| / √(16 + 4 + 16)
= |21| / √(36)
= 21 / 6
= 7/2
Therefore, the distance between the two parallel planes is 7/2, which corresponds to option C.
Given planes:
1) 2x + y + 2z = 8
2) 4x + 2y + 4z + 5 = 0
Step 1: Find the normal vectors of the planes
The normal vector of a plane is the coefficients of x, y, and z in the equation of the plane. So, the normal vectors for the given planes are:
1) Plane 1: (2, 1, 2)
2) Plane 2: (4, 2, 4)
Step 2: Find a point on one of the planes
To find a point on Plane 1, we can assume a value for one of the variables and solve for the other two variables. Let's assume x = 0. Substituting x = 0 in Plane 1 equation, we get:
y + 2z = 8
Assuming y = 0, we get:
2z = 8
z = 4
So, a point on Plane 1 is (0, 0, 4).
Step 3: Find the distance between the point and Plane 2
To find the distance between a point and a plane, we can use the formula:
Distance = |ax + by + cz + d| / √(a^2 + b^2 + c^2)
where (a, b, c) are the coefficients of x, y, and z in the equation of the plane, and d is the constant term.
Substituting the values for Plane 2, we get:
Distance = |4(0) + 2(0) + 4(4) + 5| / √(4^2 + 2^2 + 4^2)
= |16 + 5| / √(16 + 4 + 16)
= |21| / √(36)
= 21 / 6
= 7/2
Therefore, the distance between the two parallel planes is 7/2, which corresponds to option C.
If 
are unit vectors, then 
does NOT exceed- a)4
- b)9
- c)8
- d)6
Correct answer is option 'B'. Can you explain this answer?
If are unit vectors, then does NOT exceed
are unit vectors, then does NOT exceeda)
4
b)
9
c)
8
d)
6
|
|
Ankita Das answered |
Ur ans will be 9 because when angle between a,b,c is 120 it will be maximize
Let 
Then the vector 
satisfying 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Let Then the vector satisfying
Then the vector satisfying a)
b)
c)
d)
|
Sanchita Chavan answered |
which is not possible
be such that 
If the projection 
is equal to that of 
and 
are perpendicular to each other then 
Let
be a unit vector in R3 and 
Given that there exists a vector 
in 
Which of the following statement(s) is (are) correct?- a)There is exactly one choice for such
- b)There are in finitely many choices for such
- c)If
in the xy-plane then |u1| = |u2| - d)If
in the xz-plane then 2 |u1| = |u3|
Correct answer is option 'B,C'. Can you explain this answer?
Let be a unit vector in R3 and Given that there exists a vector in Which of the following statement(s) is (are) correct?
be a unit vector in R3 and Given that there exists a vector in Which of the following statement(s) is (are) correct?a)
There is exactly one choice for such
b)
There are in finitely many choices for such
c)
If in the xy-plane then |u1| = |u2|
in the xy-plane then |u1| = |u2|d)
If in the xz-plane then 2 |u1| = |u3|
in the xz-plane then 2 |u1| = |u3|
|
Ameya Sengupta answered |
where α is the angle between
From (i) and (ii) cos α = 1 ⇒ a = 0°
is perpendicular to the plane containing 
is perpendicular to 
Clearly there can be infinite many choices for 
the value of 
is equal to
then the value 
- a)–3
- b)5
- c)3
- d)–5
Correct answer is option 'D'. Can you explain this answer?
then the value a)
–3
b)
5
c)
3
d)
–5
|
|
Vikash Yadav answered |
Use vector triple product
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