Explanation:

If the dot product is zero the angle will be 90° and if the dot product is 1 the angle will be 0°

Explanation:
From dot product formula, A · B = |A||B| cosθ.
Given A · B equals |A||B|, so cosθ must be 1.
Cosθ is 1 when θ = 0°, so the vectors are in the same direction.

Explanation:

cosθ < 0 in this range ⇒ A·B < 0

Explanation:
Sign of the dot product depends on cosθ.
A · B > 0 means cosθ is positive.
Cosθ is positive for angles less than 90°, so the angle is acute.

Explanation:

Start from the given equality:
A · B = A · C

Move all terms to one side:
A · B – A · C = 0

Factor out A using the distributive property of the dot product:
A · (B – C) = 0

A dot product is zero precisely when the two vectors are perpendicular.
Therefore, A is perpendicular to the vector B – C (equivalently, B – C ⟂ A).

Explanation:

A·B = |A||B|cosθ.
If A·B = 0 and both A, B ≠ 0,
then
cosθ = 0
⇒ θ = 90°

Explanation:

A·B = |A||B|cosθ
= 1×1×cosθ = 1
⇒ cosθ = 1
⇒ θ = 0°

Explanation:
The dot product is defined using magnitudes of vectors and the angle between them.
If θ is the angle between A and B, then A · B = |A||B| cosθ.
It is called a scalar product because its result is a single number, not a vector.

Explanation:

A·B = (2)(-1) + (3)(4)
= -2 + 12 = 10

Explanation:

Scalar product is given by A·B = |A||B|cosθ.
If vectors are perpendicular, θ = 90°, and cos(90°) = 0,
⇒ A·B = 0

Explanation:

a · (b + λc)
= a · b + λ(a · c)
= (–2) + λ(2)= 0
⇒ λ = 1

Explanation:

This is false. Scalar product is
commutative
⇒ A·B = B·A.

Explanation:
Commutative property means A · B = B · A, which is true for dot product.
Distributive property means
A · (B + C) = A · B + A · C,
which also holds.
So scalar product is both commutative and distributive over vector addition.

Explanation:

A · B = |A||B|cosθ
⇒ 30 = 5×12×cosθ
⇒ cosθ = 0.5
⇒ θ = 60°

Explanation:
For equal vectors A and B with magnitude P, dot product is A · B = P² cosθ.
Given A · B = P², so P² cosθ = P².
Dividing by P² (non-zero) gives cosθ = 1 and θ = 0°.

Explanation:

A·A = |A||A|cos0°
= |A|²

Explanation:

A·B = |A||B|cosθ = 0
implies
cosθ = 0,
so θ = 90°

Explanation:

Scalar projection = |A||cosθ|.
Max when cosθ = 1
⇒ θ = 0°

Explanation:

cosθ < 0 for obtuse angles ⇒ A·B = |A||B|cosθ becomes negative.

Explanation:
Apply dot product formula A · B = |A||B| cosθ.
Here 8 = 4 × 4 × cosθ.
So 8 = 16 cosθ and
cosθ = 8 / 16
= 1 / 2.
Angle whose cosine is 1/2 is 60°,
so θ = 60°.

Explanation:
In A · A, the angle between the two A vectors is zero.
So A · A = |A||A| cos0°.
Cos0° is 1,
so A · A = |A|².

Explanation:

A · B = (x)(2) + (3)(y)
= 0 ⇒ 2x + 3y = 0

Explanation:

Explanation:
Dot product for equal vectors is P² cosθ.
Given P² cosθ = 0 and P ≠ 0, so cosθ must be zero.
Cosθ is zero at θ = 90°, hence they are perpendicular.

Explanation:
Dot product A · B = |A||B| cosθ.
Negative dot product means cosθ is negative.
Cosθ is negative when θ lies between 90° and 180°, so the angle is obtuse.

Explanation:

• Acute angle (0° – 90°)
→dot product > 0 (positive)
____________
• Right angle (90°)
→dot product = 0
_____________
• Obtuse angle (90° – 180°)
→dot product < 0 (negative)

Explanation:

A = 2i + 3j
B = i + j

We need to find the component of A parallel to B (i.e., projection of A on B).

Formula:
Projection of A on B = [(A · B) / |B|²] × B

Step 1: Find dot product A · B
A · B = (2)(1) + (3)(1) = 5

Step 2: Find |B|²
|B|² = 1² + 1² = 2

Step 3: Apply formula
Projection = (5 / 2)(i + j)

Explanation:

Dot product results in a scalar quantity, not a vector.
A·B = |A||B|cosθ is a number, not a direction-based quantity.

Explanation:

(A + B)² = A · A + B · B + 2A · B

Explanation:
Dot product of a vector with itself is equal to square of its magnitude.
So A · A = |A|².
Here |A| = 10,
so A · A = 10²
= 100.

Explanation:

All listed properties are standard for scalar products.

Explanation:

A·B = |A||B|cos(120°)
= 10×10×cos(120°)
= 100×(-0.5)
= -50

Explanation:

Explanation:
For perpendicular vectors, the angle between them is 90°.
Cos90° is equal to zero.
So A · B = |A||B| cos90° = 0 for non-zero perpendicular vectors.

Explanation:
Dot product in two dimensions is found by multiplying corresponding components and adding them.
So A · B = A × C for i-components plus B × D for j-components.
Thus A · B
= AC + BD.

Explanation:
Dot product formula is A · B = |A||B| cosθ.
This involves magnitude of A, magnitude of B and cos of the angle between them.
So the dot product depends on both magnitudes and the angle between the vectors.

Explanation:

A·B = (3)(4) + (4)(–3)
= 12 – 12 = 0
⇒ angle= 90°

Explanation:
Work is defined as the dot product of force and displacement.
So W = F · s.
Using the scalar product definition, W = |F||s| cosθ
= F s cosθ.

Explanation:

Explanation:

A · B = |A||B|cosθ
= 0 ⇒ cosθ
= 0
⇒ θ = 90°

Explanation:

(A + B) · (A – B)
= A · A – B · B
= |A|² – |B|²

Explanation:
Dot product A · B = |A||B| cosθ.
Given A · B = −|A||B|, so cosθ = −1.
Cosθ is −1 at θ = 180°, so the vectors are opposite in direction.

Explanation:

For any vector A = a î + b ĵ + c k̂, the dot product with itself is
 A · A = a² + b² + c².
The MCQs states A · A = 49,
so a² + b² + c² = 49.

Explanation:

A·B = |A||B|cosθ
= 5×10×cos(60°)
= 50×0.5 = 25

Explanation:

A · B = a·b + b·(–a)
= ab – ab
= 0

Explanation:
Work is equal to the dot product of force and displacement.
Torque and angular momentum are vector quantities written using cross products.
Momentum is also a vector, not a scalar.

Explanation:
Component of A along a direction is the projection of A on that direction.
When n̂ is a unit vector along that direction, projection is A · n̂.
This value is a scalar giving how much of A lies along n̂.

Explanation:
Component parallel to n̂ has the same direction as n̂.
Its magnitude is given by the projection A · n̂.
So the parallel component vector is (A · n̂) n̂.

Explanation:
For non-zero vectors A and B, A · B = |A||B| cosθ.
Given A · B = 0, so cosθ must be zero.
Cosθ is zero when θ = 90°, meaning the vectors are perpendicular.

Explanation:
Dot product in three dimensions is obtained by multiplying corresponding components and adding them.
So A · B = Ax Bx for x-components plus Ay By for y-components plus Az Bz for z-components.
Thus A · B = Ax Bx + Ay By + Az Bz.

Explanation:
Dot product of force and displacement gives work done.
Unit of force is newton and unit of displacement is metre.
So unit of work is newton metre, which is called joule.

Explanation:

Magnitude of A
|A| = √(2² + 3²) = √13.

Angle θ with the y-axis:

Adjacent side = y-component = 3

Opposite side = x-component = 2

Therefore tan θ = (opposite)/(adjacent) = 2 ⁄ 3.

Hence θ = tan⁻¹ ( 2 ⁄ 3 ).

Explanation:

A · B = (2)(–1) + (–3)(4) + (1)(2)
= –2 – 12 + 2
= –12

Explanation:
Vector A lies along the x-axis and vector B lies along the y-axis.
These directions are perpendicular to each other.
So A · B = |A||B| cos90° = 0.

Explanation:
Dot product is A · B = Ax Bx + Ay By.
Here Ax = 2, Ay = −3,
Bx = 4 and By = 1.
So A · B = 2 × 4 + (−3) × 1
= 8 − 3
= 5.

Explanation:

Step 1: Find A + B

Add components of A and B:

• i-component: 2 + 1 = 3

• j-component: 1 + 2 = 3

• k-component: -1 + 3 = 2

So,
A + B = 3i + 3j + 2k

First, calculate the dot product:

(A + B) · C = (3)(6) + (3)(-2) + (2)(-6)
= 18 − 6 − 12 = 0

as the dot product is zero angle will be 90°

Explanation:

A·B = |A||B|cosθ.
If A·B = |A||B|
⇒ cosθ = 1
⇒ θ = 0°

Explanation:

Let the two force vectors be F₁ and F₂.
Given: (F₁ + F₂) is perpendicular to (F₁ − F₂)
⇒ (F₁ + F₂) · (F₁ − F₂) = 0

Now expand the dot product:
F₁ · F₁ − F₁ · F₂ + F₂ · F₁ − F₂ · F₂ = 0
⇒ |F₁|² − |F₂|² = 0
⇒ |F₁| = |F₂|

Explanation:

Vector components
A = a î + b ĵ
• x-component = a
• y-component = b

Explanation:
Use A · B = |A||B| cosθ.
Here 15 = 5 × 3 × cosθ.
So 15 = 15 cosθ and cosθ = 1.
Angle whose cosine is 1 is 0°, so vectors are in same direction.

Explanation:
Dot product A · B = |A||B| cosθ.
Cosθ can be positive, negative or zero depending on θ.
Therefore the scalar product can be positive, negative or zero.

Explanation:
Use dot product formula
A · B = |A||B| cosθ.
Here A · B = 4 × 5 × cos60°.
Cos60° is 1/2, so A · B
= 20 × 1/2
= 10.

Explanation:
A · B = 0 indicates that vectors are perpendicular or one is zero.
A × B ≠ 0 means their cross product has non-zero magnitude, so neither is zero and they are not parallel.
Together these conditions show that A and B are perpendicular non-zero vectors.

Explanation:
Dot product in component form is A · B = Ax Bx + Ay By.
Here Ax = 3, Ay = 4,
Bx = 1 and By = 2.
So A · B = 3 × 1 + 4 × 2
= 3 + 8
= 11.

Explanation:
Dot product formula is
A · B = |A||B| cosθ.
Given A · B = 0 and both vectors are non-zero, so cosθ must be zero.
Cosθ is zero at θ = 90°, hence the vectors are perpendicular.