Vector Algebra Test

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Vector Algebra Test - 2

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Question 1
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For what values of x and y, the vectorsare equal?

A
x = 3, y = 3/2

B
x = 3, y = 6

C
x = 3/2, y = 3

D
x = 6, y = 3

Solution

Two vectors are equal if their corresponding components are equal.
Hence, 2x = 3
or x = 3/2
and 2x = y
or y = 2 * 3/2
y = 3
Question 2
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Two or more vectors having the same initial point are called

A
co-terminus vectors.

B
zero vectors

C
co-initial vectors

D
unit vectors

Solution

Two or more vectors having the same initial point are called coinitial vectors.
Question 3
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If ,the vectors a and b are ______ .

A
coinitial

B
free vector

C
zero vector

D
collinear

Solution

Both of these vectors are pointing the same direction but they have different magnitude. Hence, they are collinear vectors .
Question 4
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If the magnitude of the position vector is 7, the value of x is:

A
±1

B
±5

C
±3

D
±2

Solution

|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
Question 5
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If and are the position vectors of the points A, B, C and D such that then ABCD is:

A
Trapezium

B
Rectangle

C
Square

D
Parallelogram

Solution

Given:

If we divide both sides with 2 we get,
⇒ Mid pt. of AC = Mid. pt. of BD
∴ABCD is a parallelogram.
Question 6
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If , then

A

B

C

D

Solution

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.
Question 7
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What is the additive identity of a vector?

A
zero vector

B
Negative of the vector

C
unit vector

D
The vector itself

Solution

In the Additive Identity of vectors, the additive identity is zero vector 0.
For any vectorV additive identity is defined as,
0 + V = V and V + 0 = V
Question 8
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The angles α, β, γmade 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

Solution

Direction Cosines for angle α, β, γare:
cos α, cos β, cos γ
Question 9
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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

Solution

Position vector of A = OA = a
Position vector of B = OB = b
Now, AC = OC - OA
Also, AB = OB - OA
Given, AC = 3AB
⇒OC - OA = 3(OB - OA)
⇒OC = 3b - 2a
Question 10
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Vector of magnitude 1 is called.

A
Constant vector

B
Zero vector

C
Unit vector

D
None of these

Solution

A vector whose magnitude (i.e., length) is equal to 1 is called a unit vector . There are exactly two unit vectors in any given direction and one is the negative of the other.
Question 11
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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

Solution

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
Question 12
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The direction of zero vector.

A
Is towards the origin

B
Is towards a fixed point

C
Does not exist

D
Is indeterminate

Solution

Zero vector is the unit vector having zero length, hence the direction is undefined .
Question 13
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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

Solution

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
Unit vector = (Vector)/(Magnitude of vector)
Unit vector C = (C vector)/(Magnitude of C vector) = (-1i + 6j -11k)/√158
Question 14
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If a be magnitude of vector then

A
a >0

B
a ≤0

C
a ≥0

D
a <0

Solution

Since a is the magnitude of the vector, it is always positive and it can be 0 in case of zero vectors.
So, a ≥0
Question 15
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A vector of magnitude 14 units, which is parallel to the vector

A

B

C

D

Solution

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)
Question 16
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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|

Solution

|a + b|² = |a|² + |b|² + 2|a||b|.cos θ
|a|² + |b|² = |a|² + |b|² + 2|a| + |b| ∵−1 ⩽cos θ⩽1
⇒2|a||b|.cos θ ⩽ 2|a||b|
So, |a + b|² ⩽ (|a| + |b| )²
⇒ |a + b| ≤|a| + |b|
This is also known as Triangle Inequality of vectors.
Question 17
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If l, m, n are the direction cosines of a position vector then which of the following is true?

A
l² + m² - n² = 0

B
lmn = 1

C
l² + m² + n² = 1

D
l² m² + n² = 1

Solution

Consider is the position vector of a point M(x,y,z) and α, β, γare the angles, made by the vector with the positive directions of x, y and z respectively. The cosines of the angles, cos ⁡α, cos ⁡β, cos ⁡γare the direction cosines of the vector denoted by l, m, n, then
cos² ⁡α+ cos² ⁡β+ cos² ⁡γ=1 i.e.l² + m² + n² = 1.
Question 18
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A vector whose initial and terminal points coincide, is called

A
zero vector

B
unit vector

C
equal vectors

D
coterminus vectors

Solution

A vector whose initial and terminal points coincide has no particular direction and 0 magnitude. Therefore, it is called zero vector .
Question 19
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A point from a vector starts is called______and where it ends is called its______.

A
terminal point, end point.

B
initial point, terminal point

C
Origin, end point

D
initial point, end point

Solution

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point , where it begins, and a terminal point , where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point.
Question 20
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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

Solution

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