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All questions of Chapter 10 - Vectors for JEE Exam
The area of triangle whose adjacent sides are
is :- a)√70/2 sq. units
- b)9√2 /2 sq. units
- c)3√3 /2 sq. units
- d)2√3 /2 sq. units
Correct answer is option 'A'. Can you explain this answer?
The area of triangle whose adjacent sides are is :
is :a)
√70/2 sq. units
b)
9√2 /2 sq. units
c)
3√3 /2 sq. units
d)
2√3 /2 sq. units
|
Suresh Iyer answered |
Area of triangle = ½(a * b)
a = (1, 0, -2) b = (2, 3, 1)
= i(0 + 6) + j(-4 - 1) + k(3 - 0)
= 6i - 5j + 3k
|a * b| = (36 + 25 + 9)½
|a * b| = (70)½
Area of triangle = ½(a * b)
= [(70)½]/2
a = (1, 0, -2) b = (2, 3, 1)
= i(0 + 6) + j(-4 - 1) + k(3 - 0)
= 6i - 5j + 3k
|a * b| = (36 + 25 + 9)½
|a * b| = (70)½
Area of triangle = ½(a * b)
= [(70)½]/2
A vector of magnitude 14 units, which is parallel to the vector
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A vector of magnitude 14 units, which is parallel to the vector
a)
b)
c)
d)
|
EduRev JEE answered |
Given vector = i + 2j - 3k
Magnitude = √12 + 22 + (-3)2 = √14
Unit vector in direction of resultant = (i + 2j - 3k) / √14
Vector of magnitude 14 unit in direction of resultant,
⇒ 14[ (i + 2j - 3k) / √14 ]
⇒ √14(i + 2j - 3k)
Magnitude = √12 + 22 + (-3)2 = √14
Unit vector in direction of resultant = (i + 2j - 3k) / √14
Vector of magnitude 14 unit in direction of resultant,
⇒ 14[ (i + 2j - 3k) / √14 ]
⇒ √14(i + 2j - 3k)
The value of 
is equal to the box product - a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The value of is equal to the box product
is equal to the box product a)
b)
c)
d)
|
Geetika Shah answered |
Matrix {(1,2,-1) (1,-1,0) (1,-1,-1)} [a b c]
= {1(1) -2(-1) -1(0)} [a b c]
= 3[a b c]
= {1(1) -2(-1) -1(0)} [a b c]
= 3[a b c]
If a, b, c and d are the position vectors of the points A, B, C and D such that a + c = b + d, then ABCD is a- a)Trapezium
- b)Rectangle
- c)Square
- d)Parallelogram
Correct answer is option 'D'. Can you explain this answer?
If a, b, c and d are the position vectors of the points A, B, C and D such that a + c = b + d, then ABCD is a
a)
Trapezium
b)
Rectangle
c)
Square
d)
Parallelogram
|
Tejas Verma answered |
Given:
If we divide both sides with 2 we get,
If we divide both sides with 2 we get,
⇒ Mid pt. of AC = Mid. pt. of BD
∴ ABCD is a parallelogram.
∴ ABCD is a parallelogram.
If 
and 
, then the value of scalars x and y are:- a)x = 1 and y = -2
- b)x = -2 and y = 1
- c)x = 2 and y = -1
- d)x = 2 and y = 1
Correct answer is option 'C'. Can you explain this answer?
If and , then the value of scalars x and y are:
and , then the value of scalars x and y are:a)
x = 1 and y = -2
b)
x = -2 and y = 1
c)
x = 2 and y = -1
d)
x = 2 and y = 1
|
Sushil Kumar answered |
Given, a = i + 2j
b = -2i + j
c = 4i +3j
Also, c = xa +yb
Now putting the values in above equation,
4i + 3j = x(i + 2j) + y(-2i +j)
⇒ xi + 2xj - 2yi + yj
⇒ (x-2y)i + (2x+y)j
We get,
x - 2y = 4
2x + y = 3
After solving,
x = 2
y = -1
b = -2i + j
c = 4i +3j
Also, c = xa +yb
Now putting the values in above equation,
4i + 3j = x(i + 2j) + y(-2i +j)
⇒ xi + 2xj - 2yi + yj
⇒ (x-2y)i + (2x+y)j
We get,
x - 2y = 4
2x + y = 3
After solving,
x = 2
y = -1
If a and b are the position vectors of two points A and B and C is a point on AB produced such that AC = 3AB, then position vector of C will be- a)3b – 2a
- b)3a – b
- c)3a – 2b
- d)3b – a
Correct answer is option 'A'. Can you explain this answer?
If a and b are the position vectors of two points A and B and C is a point on AB produced such that AC = 3AB, then position vector of C will be
a)
3b – 2a
b)
3a – b
c)
3a – 2b
d)
3b – a
|
Snehal Choudhary answered |
AC = 3AB
(c - a)= 3(b - a). (c - a)
= 3b -3a
c= 3b--2a
Unit vectors along the axes OX, OY and OZ are denoted by - a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Unit vectors along the axes OX, OY and OZ are denoted by
a)
b)
c)
d)
|
Krishna Iyer answered |
represents the unit vectors along the co ordinate axis i.e. OX ,OY and OZ respectively.The unit vector in the direction of 
, where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The unit vector in the direction of , where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:
, where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:a)
b)
c)
d)
|
|
Sushil Kumar answered |
Given, Point A (2,-3,7)
Point B (1,3,-4)
Let vector in the direction of AB be C.
∴ C = B - A
⇒ (1,3,-4) - (2,-3,7)
⇒ ( 1-2 , 3+3 , -4-7 )
⇒ (-1,6,-11)
⇒ -1i + 6j -11k
Magnitude of vector C
|C| = √(-1)2 + 62 + (-11)2
⇒ √1+36+121
⇒ √158
Magnitude of vector C
|C| = √(-1)2 + 62 + (-11)2
⇒ √1+36+121
⇒ √158
Unit vector = (Vector)/(Magnitude of vector)
Unit vector C = (C vector)/(Magnitude of C vector) = (-1i + 6j -11k)/√158
Unit vector C = (C vector)/(Magnitude of C vector) = (-1i + 6j -11k)/√158
The points with position vectors 
are collinear vectors, Value of a =- a)-20
- b)20
- c)-40
- d)40
Correct answer is option 'C'. Can you explain this answer?
The points with position vectors are collinear vectors, Value of a =
are collinear vectors, Value of a =a)
-20
b)
20
c)
-40
d)
40
|
Om Desai answered |
Position vector A = 60i+3j
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.
The value of 
is:- a)0
- b)3
- c)1/3
- d)1
Correct answer is option 'B'. Can you explain this answer?
The value of is:
is:a)
0
b)
3
c)
1/3
d)
1
|
Sahil Soni answered |
Cross multiply in maths take place in cycle like i》j》k i×j=k j×k=i k×i=j but j×i=-k k×j=-i i×k=-j and dotmultiply takes place as i.i=1j.j=1K.K=1BUTI.J=0J.K=0k.i=0so the correct answer is b
can be equal to
What is the additive identity of a vector?- a)zero vector
- b)Negative of the vector
- c)unit vector
- d)The vector itself
Correct answer is option 'A'. Can you explain this answer?
What is the additive identity of a vector?
a)
zero vector
b)
Negative of the vector
c)
unit vector
d)
The vector itself
|
|
Nandini Iyer answered |
In the Additive Identity of vectors, the additive identity is zero vector 0.
For any vector V additive identity is defined as,
0 + V = V and V + 0 = V
For any vector V additive identity is defined as,
0 + V = V and V + 0 = V
If 
is equal to - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If is equal to
is equal to a)
b)
c)
d)
|
Md Jisan answered |
(u × v) = a square - b square , now mod of (u × v) = √( a to the power 4 + b to the power 4 + 2 (ab) square cosx ) . So ab cosx will give (a.b); now the determination of the value is much easy
For any two vectors a and b, we always have
- a)|a – b| ≥ |a| – |b|
- b)|a + b| ≤ |a| + |b|
- c)|a + b| ≤ |a| – |b|
- d)|a – b| = |a + b|
Correct answer is option 'B'. Can you explain this answer?
For any two vectors a and b, we always have
a)
|a – b| ≥ |a| – |b|
b)
|a + b| ≤ |a| + |b|
c)
|a + b| ≤ |a| – |b|
d)
|a – b| = |a + b|
|
|
Nandini Iyer answered |
|a + b|2 = |a|2 + |b|2 + 2|a||b|.cosθ
|a|2 + |b|2 = |a|2 + |b|2 + 2|a| + |b| ∵ −1 ⩽ cosθ ⩽ 1
⇒ 2|a||b|.cosθ ⩽ 2|a||b|
So, |a + b|2 ⩽ (|a| + |b|
⇒ |a + b| ≤ |a| + |b|
This is also known as Triangle Inequality of vectors.
|a|2 + |b|2 = |a|2 + |b|2 + 2|a| + |b| ∵ −1 ⩽ cosθ ⩽ 1
⇒ 2|a||b|.cosθ ⩽ 2|a||b|
So, |a + b|2 ⩽ (|a| + |b|
)2
⇒ |a + b| ≤ |a| + |b|
This is also known as Triangle Inequality of vectors.
are unit vectors along X-axis, Y-axis and Z-axis respectively, then
ABCD is a parallelogram. If coordinates of A,B,C are (2,3), (1,4) and (0, -2). Coordinates of D =- a)(-1,3)
- b)(1,-3)
- c)(1,3)
- d)(-1,-3)
Correct answer is option 'B'. Can you explain this answer?
ABCD is a parallelogram. If coordinates of A,B,C are (2,3), (1,4) and (0, -2). Coordinates of D =
a)
(-1,3)
b)
(1,-3)
c)
(1,3)
d)
(-1,-3)
|
|
Naina Bansal answered |
Here's a generic way to solve problems like this.
on the vector
is:
and 
be three non-zero vectors such that 
is a unit vector perpendicular to both 
. if the angle between 
is π/6, then 
is equal to 
The angles α, β, γ made by the vector
with the positive directions of X, Y and Z-axes respectively, then the direction cosines of the vector are: - a)cos α, cos β, cos γ
- b)sin α, sin β, sin γ
- c)180° - α, 180° - β, 180° γ
- d)tan α, tan β, tan γ
Correct answer is option 'A'. Can you explain this answer?
The angles α, β, γ made by the vector with the positive directions of X, Y and Z-axes respectively, then the direction cosines of the vector are:
with the positive directions of X, Y and Z-axes respectively, then the direction cosines of the vector are: a)
cos α, cos β, cos γ
b)
sin α, sin β, sin γ
c)
180° - α, 180° - β, 180° γ
d)
tan α, tan β, tan γ
|
|
Sarita Yadav answered |
Direction Cosines for angle α, β, γ are:
cos α, cos β, cos γ
cos α, cos β, cos γ
Let 
be vectors of length 3,4,5 respectively. Let 
be perpendicular to 
,
and 
. then 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Let be vectors of length 3,4,5 respectively. Let be perpendicular to , and . then
be vectors of length 3,4,5 respectively. Let be perpendicular to , and . then a)
b)
c)
d)
|
Vivek Patel answered |
|A| = 3, |B| = 4, |C| = 5
Since A.(B + C) = B.(C+A) = C(A+B) = 0...........(1)
|A+B+C|2 = |A| + |B|2 + |C|2 + 2(A.B + B+C + C.A)
= 9+16+25+0
from eq(1) {A.B + B+C + C.A = 0}
therefore, |A+B+C|2 = 50
=> |A+B+C| = 5(2)1/2
Since A.(B + C) = B.(C+A) = C(A+B) = 0...........(1)
|A+B+C|2 = |A| + |B|2 + |C|2 + 2(A.B + B+C + C.A)
= 9+16+25+0
from eq(1) {A.B + B+C + C.A = 0}
therefore, |A+B+C|2 = 50
=> |A+B+C| = 5(2)1/2
If 
are two vectors, such that 
, then 
= ……- a)3
- b)√7
- c)√5
- d)√3
Correct answer is option 'D'. Can you explain this answer?
If are two vectors, such that , then = ……
are two vectors, such that , then = ……a)
3
b)
√7
c)
√5
d)
√3
|
|
Neha Sharma answered |
|a - b|2 = |a|2 + |b|2 - 2|a||b|
|a - b|2 = (3)2 + (2)2 - 2(5)
|a - b|2 = 9 + 4 - 10
|a - b|2 = 3
|a - b| = (3)½.
|a - b|2 = (3)2 + (2)2 - 2(5)
|a - b|2 = 9 + 4 - 10
|a - b|2 = 3
|a - b| = (3)½.
If the magnitude of the position vector
is 7, the value of x is:- a)±1
- b)±5
- c)±3
- d)±2
Correct answer is option 'C'. Can you explain this answer?
If the magnitude of the position vector is 7, the value of x is:
is 7, the value of x is:a)
±1
b)
±5
c)
±3
d)
±2
|
Shiksha Academy answered |
|a| = (x2 + 22 + (2x)2)1/2
7 = (x2 + 22 + (2x)2)1/2
⇒ 49 = x2 + 22 + 4x2
⇒ 49 = 4 + 5x2
⇒ 5x2 = 45
⇒ x2 = 9
x = ±3
are the sides of a triangle ABC. The length of the median through A is 
If u and v are unit vectors and θ is the acute angle between them, then 2u × 3v is a unit vector for- a)Exactly two values of θ
- b)More than two values of θ
- c)No value of θ
- d)Exactly one value of θ
Correct answer is option 'D'. Can you explain this answer?
If u and v are unit vectors and θ is the acute angle between them, then 2u × 3v is a unit vector for
a)
Exactly two values of θ
b)
More than two values of θ
c)
No value of θ
d)
Exactly one value of θ
|
Bhargavi Sen answered |
If u and v are unit vectors, it means that their magnitudes are equal to 1. Thus, ||u|| = ||v|| = 1.
Given that ||u + v|| = 2, we want to prove that ||u - v|| = 2.
Using the triangle inequality, we have:
||u + v|| ≤ ||u|| + ||v||
Since ||u|| = ||v|| = 1, the inequality becomes:
||u + v|| ≤ 1 + 1
||u + v|| ≤ 2
Since we are given that ||u + v|| = 2, the inequality becomes:
2 ≤ 2
This inequality is true, so the triangle inequality holds.
Now, let's consider ||u - v||:
||u - v|| = ||u + (-v)||
Since -v is the additive inverse of v, its magnitude is the same as v's magnitude, so ||-v|| = ||v|| = 1.
Using the triangle inequality again, we have:
||u + (-v)|| ≤ ||u|| + ||-v||
Since ||u|| = 1 and ||-v|| = 1, the inequality becomes:
||u + (-v)|| ≤ 1 + 1
||u + (-v)|| ≤ 2
However, since we are given that ||u + v|| = 2, we can replace u + (-v) with u + v in the inequality:
||u + v|| ≤ 2
Therefore, ||u - v|| ≤ 2.
But we also know that ||u - v|| ≥ 0, since magnitudes are always non-negative.
Since ||u - v|| ≤ 2 and ||u - v|| ≥ 0, the only possible value for ||u - v|| is 2.
Therefore, ||u - v|| = 2.
In conclusion, if u and v are unit vectors and ||u + v|| = 2, then ||u - v|| = 2.
Given that ||u + v|| = 2, we want to prove that ||u - v|| = 2.
Using the triangle inequality, we have:
||u + v|| ≤ ||u|| + ||v||
Since ||u|| = ||v|| = 1, the inequality becomes:
||u + v|| ≤ 1 + 1
||u + v|| ≤ 2
Since we are given that ||u + v|| = 2, the inequality becomes:
2 ≤ 2
This inequality is true, so the triangle inequality holds.
Now, let's consider ||u - v||:
||u - v|| = ||u + (-v)||
Since -v is the additive inverse of v, its magnitude is the same as v's magnitude, so ||-v|| = ||v|| = 1.
Using the triangle inequality again, we have:
||u + (-v)|| ≤ ||u|| + ||-v||
Since ||u|| = 1 and ||-v|| = 1, the inequality becomes:
||u + (-v)|| ≤ 1 + 1
||u + (-v)|| ≤ 2
However, since we are given that ||u + v|| = 2, we can replace u + (-v) with u + v in the inequality:
||u + v|| ≤ 2
Therefore, ||u - v|| ≤ 2.
But we also know that ||u - v|| ≥ 0, since magnitudes are always non-negative.
Since ||u - v|| ≤ 2 and ||u - v|| ≥ 0, the only possible value for ||u - v|| is 2.
Therefore, ||u - v|| = 2.
In conclusion, if u and v are unit vectors and ||u + v|| = 2, then ||u - v|| = 2.
then
Coinitial Vectors are- a)Two or more vectors having the same final point
- b)Two or more pseudo vectors having the same initial point
- c)Two or more force vectors having the same initial point
- d)Two or more vectors having the same initial point
Correct answer is option 'D'. Can you explain this answer?
Coinitial Vectors are
a)
Two or more vectors having the same final point
b)
Two or more pseudo vectors having the same initial point
c)
Two or more force vectors having the same initial point
d)
Two or more vectors having the same initial point
|
Bhavana Dey answered |
Two vectors whose initial point is same are called co- initial vectors
A vector whose initial and terminal points coincide, is called- a)zero vector
- b)unit vector
- c)equal vectors
- d)coterminus vectors
Correct answer is option 'A'. Can you explain this answer?
A vector whose initial and terminal points coincide, is called
a)
zero vector
b)
unit vector
c)
equal vectors
d)
coterminus vectors
|
|
Nandini Patel answered |
Zero Vector Or Null Vector – A vector whose initial and terminal points coincide is known as zero vector.Unit Vector
A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the points of intersection are given by- a)(2a, 3a, 3a), (2a, a, a)
- b)(3a, 2a, 3a), (a, a, a)
- c)(3a, 2a, 3a), (a, a, 2a)
- d)(3a, 3a, 3a), (a, a, a)
Correct answer is option 'B'. Can you explain this answer?
A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the points of intersection are given by
a)
(2a, 3a, 3a), (2a, a, a)
b)
(3a, 2a, 3a), (a, a, a)
c)
(3a, 2a, 3a), (a, a, 2a)
d)
(3a, 3a, 3a), (a, a, a)
|
Nilotpal Goyal answered |
To find the coordinates of the points of intersection, we need to solve the given equations simultaneously.
Given information:
Direction cosines of the line = 2, 1, 2
Equations of the lines:
1. x = y
2. a = z
3. x = 2y = 2z
Let's solve these equations one by one:
1. x = y
Substitute x = y in equation 3.
y = 2y = 2z
Divide the equation by y to get:
1 = 2 = 2z/y
Simplifying, we get:
z = y/2
Substitute the value of z in equation 2:
a = y/2
So, the coordinates of the point of intersection of the line with equation 1 and equation 2 are (y, y/2, a).
2. a = z
Substitute a = z in equation 3.
x = 2y = 2a
Divide the equation by 2 to get:
x/2 = y/2 = a
So, the coordinates of the point of intersection of the line with equation 1 and equation 2 are (x/2, y/2, y/2).
3. x = 2y = 2z
Substitute x = 2y = 2z in equation 1.
2y = y
y = 0
Substitute y = 0 in equation 2.
a = 0
So, the coordinates of the point of intersection of the line with equation 1 and equation 2 are (0, 0, 0).
From the above calculations, we can see that the coordinates of the points of intersection are given by (y, y/2, a), (x/2, y/2, y/2), and (0, 0, 0).
Comparing these coordinates with the given options, we can see that option B, (3a, 2a, 3a), (a, a, a), matches with the calculated coordinates.
Hence, the correct answer is option B.
Given information:
Direction cosines of the line = 2, 1, 2
Equations of the lines:
1. x = y
2. a = z
3. x = 2y = 2z
Let's solve these equations one by one:
1. x = y
Substitute x = y in equation 3.
y = 2y = 2z
Divide the equation by y to get:
1 = 2 = 2z/y
Simplifying, we get:
z = y/2
Substitute the value of z in equation 2:
a = y/2
So, the coordinates of the point of intersection of the line with equation 1 and equation 2 are (y, y/2, a).
2. a = z
Substitute a = z in equation 3.
x = 2y = 2a
Divide the equation by 2 to get:
x/2 = y/2 = a
So, the coordinates of the point of intersection of the line with equation 1 and equation 2 are (x/2, y/2, y/2).
3. x = 2y = 2z
Substitute x = 2y = 2z in equation 1.
2y = y
y = 0
Substitute y = 0 in equation 2.
a = 0
So, the coordinates of the point of intersection of the line with equation 1 and equation 2 are (0, 0, 0).
From the above calculations, we can see that the coordinates of the points of intersection are given by (y, y/2, a), (x/2, y/2, y/2), and (0, 0, 0).
Comparing these coordinates with the given options, we can see that option B, (3a, 2a, 3a), (a, a, a), matches with the calculated coordinates.
Hence, the correct answer is option B.
If a unit vector 
makes angles 
and an acute angle θ with 
, then the components of are- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
If a unit vector makes angles and an acute angle θ with , then the components of are
makes angles and an acute angle θ with , then the components of area)
b)
c)
d)
|
|
Riya Banerjee answered |
Let 
It is given that left| 
, then ,
It is given that left| , then , Putting these values in (1) , we get :
The vector joining the points A(2, – 3, 1) and B(1, – 2, – 5) directed from B to A is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The vector joining the points A(2, – 3, 1) and B(1, – 2, – 5) directed from B to A is:
a)
b)
c)
d)
|
Siddhant Kumar answered |
The vector goes from B to A means that its initial coordinates is at B and final at A.
so the vector BA will be [ (2-1)i + (-3+2) j + (1+5)k ] = i-j+6k.
so the vector BA will be [ (2-1)i + (-3+2) j + (1+5)k ] = i-j+6k.
The distance between the point (2, 3, 1) and (–1, 2, – 3) is:
- a)√26 units
- b)5 units
- c)√20 units
- d)4 units
Correct answer is option 'A'. Can you explain this answer?
The distance between the point (2, 3, 1) and (–1, 2, – 3) is:
a)
√26 units
b)
5 units
c)
√20 units
d)
4 units
|
|
Giriraj Gupta answered |
Use the distance formula
distance=root of [ (x1-x2)^2+(y1-y2)^2+(z1-z2)^2]
distance=root of [ (x1-x2)^2+(y1-y2)^2+(z1-z2)^2]
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
a)
b)
c)
d)
|
Lavanya Menon answered |
D1 = a+b
D2 = a-b
D1 = 3i + 0j + 0k
D2 = i + 2j + 2k
|D1| = 3
D1.D2 = |D1| . |D2| . cos θ
3 + 0 + 0 = (3) . (3) cos θ
3 = 9 cosθ
cos-1 = (⅓)
D2 = a-b
D1 = 3i + 0j + 0k
D2 = i + 2j + 2k
|D1| = 3
D1.D2 = |D1| . |D2| . cos θ
3 + 0 + 0 = (3) . (3) cos θ
3 = 9 cosθ
cos-1 = (⅓)
and 
respectively are the vertices of a right angled triangle with C = π/2 are
If 
, then- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
If , then
, thena)
b)
c)
d)
|
|
Lavanya Menon answered |
a = 2i + 3j - 6k
|a| = √4+9+36 = √49 = 7
b = 6i - 2j + 3k
|b| = √36+4+9 = √49 = 7
|a| = |b|
Hence, option A is correct.
|a| = √4+9+36 = √49 = 7
b = 6i - 2j + 3k
|b| = √36+4+9 = √49 = 7
|a| = |b|
Hence, option A is correct.
The position vectors of the end points of diameter of a circle are 
and 
, then the position vector of the centre of the circle is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The position vectors of the end points of diameter of a circle are and , then the position vector of the centre of the circle is:
and , then the position vector of the centre of the circle is:a)
b)
c)
d)
|
|
Om Desai answered |
{(1+5)î +(1-3)j + (1-1)k} / 2
= {6i - 2j + 0k}/2
= 3i - j
= {6i - 2j + 0k}/2
= 3i - j
4-points whose position vector 
are coplanar and 
then the least value of 
- a)1
- b)1/14
- c)2
- d)1/10
Correct answer is option 'B'. Can you explain this answer?
4-points whose position vector are coplanar and then the least value of
are coplanar and then the least value of a)
1
b)
1/14
c)
2
d)
1/10
|
Ishan Choudhury answered |
As the vectors are complainer sum of coefficient =0
sinα + 2sinβ + 3sinγ = 1
this can be also LHS can also be called to be dot product of two vectors (1,2,3) and (sinα,sinβ,sinγ)
dot product of these two vectors is
sinα + 2sinβ + 3sinγ
aˉ.bˉ=|a||b|cosθ
(sinα+2sinβ+3sinγ)/1 = (sin2α+sin2β+sin2γ)×14cosθ
1/14cosθ = sin^2α+sin^2β+sin^2γ
∴ Minimum value is 1/14
sinα + 2sinβ + 3sinγ = 1
this can be also LHS can also be called to be dot product of two vectors (1,2,3) and (sinα,sinβ,sinγ)
dot product of these two vectors is
sinα + 2sinβ + 3sinγ
aˉ.bˉ=|a||b|cosθ
(sinα+2sinβ+3sinγ)/1 = (sin2α+sin2β+sin2γ)×14cosθ
1/14cosθ = sin^2α+sin^2β+sin^2γ
∴ Minimum value is 1/14
and lies in the plane of 
and 
and 
then 
is equal to 
If 
are position vectors of the points (- 1, 1) and (m, – 2). then for what value of m, the vectors are collinear. - a)1
- b)2
- c)-1
- d)-2
Correct answer is option 'B'. Can you explain this answer?
If are position vectors of the points (- 1, 1) and (m, – 2). then for what value of m, the vectors are collinear.
are position vectors of the points (- 1, 1) and (m, – 2). then for what value of m, the vectors are collinear. a)
1
b)
2
c)
-1
d)
-2
|
|
Neha Sharma answered |
Given a = (-1,1) and b = (m,-2)
Given that above two vectors are collinear, so they are parallel
⇒ -1/m = 1/-2
⇒ m = 2
Given that above two vectors are collinear, so they are parallel
⇒ -1/m = 1/-2
⇒ m = 2
is :
If the distance of the point P (1, –2, 1) from the plane x + 2y – 2z = α , where α > 0, is 5, then the foot of the 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 = α , where α > 0, is 5, then the foot of the perpendicular from P to the plane is
a)
b)
c)
d)
|
|
Tejas Verma answered |
As perpendicular distance of x + 2y – 2z = α from the point (1, –2, 1) is 5
∴ Plane becomes x + 2y – 2z – 10 = 0
It lies on x + 2y – 2z – 10 = 0
∴ Plane becomes x + 2y – 2z – 10 = 0
It lies on x + 2y – 2z – 10 = 0
For what values of x and y, the vectors
are equal?- a)
- b)x = 3, y = 6
- c)
- d)x = 6, y = 3
Correct answer is option 'C'. Can you explain this answer?
For what values of x and y, the vectorsare equal?
are equal?a)
b)
x = 3, y = 6
c)
d)
x = 6, y = 3
|
Satyam Sriraj answered |
For equal vector
2x=3
and 2x=y
now
solve it.
2x=3
and 2x=y
now
solve it.
, then 
........
Unit vector perpendicular to the plane of the triangle ABC with position vectors 
of the vertices
A, B, C is- a)
- b)
- c)
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Unit vector perpendicular to the plane of the triangle ABC with position vectors of the vertices
A, B, C is
of the verticesA, B, C is
a)
b)
c)
d)
None of these
|
|
Tejas Verma answered |
Δ = 1/2(a * b) = 1/2(b*c) = 1/2(c*a)
2Δ = a*b = b*c = c*a
unit vector = 1/2Δ[a*b + b*c + c*a]
2Δ = a*b = b*c = c*a
unit vector = 1/2Δ[a*b + b*c + c*a]
are any two vectors, then
The direction of zero vector.- a)Is towards the origin
- b)Is towards a fixed point
- c)Does not exist
- d)Is indeterminate
Correct answer is option 'D'. Can you explain this answer?
The direction of zero vector.
a)
Is towards the origin
b)
Is towards a fixed point
c)
Does not exist
d)
Is indeterminate
|
Supriya Senapati answered |
Zero vector is the unit vector having zero length ,hence the direction is undefined
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