Vector Practice Test 4 ( Scalar Product )
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The angle which the vector A = 2i + 3j makes with the y-axis, where i and j are unit vectors along x- and y-axis, respectively, is:
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 ).
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The vector sum of two forces is perpendicular to their vector difference. In that case, the forces:
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|
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A and B are two vectors given by A = 2i + 3j and B = i + j. The component of A parallel to B is:
A = 2i + 3jB = 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 · BA · B = (2)(1) + (3)(1) = 5
Step 2: Find |B|²|B|² = 1² + 1² = 2
Step 3: Apply formulaProjection = (5 / 2)(i + j)
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The magnitude of the resultant of two vectors of magnitude 3 units and 4 units is 1 unit. What is the value of their dot product?
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If A = 2i + j − k, B = i + 2j + 3k, and C = 6i − 2j − 6k, then the angle between (A + B) and C will be:
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°
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The angle between the two vectors (–2i + 3j + k) and (i + 2j – 4k) is:
If the dot product is zero the angle will be 90° and if the dot product is 1 the angle will be 0°
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A particle moves from position null to (11i + 11j + 15k) due to a uniform force of (4i + j + 3k) N. If the displacement is in m, then the work done will be:
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If A = aî + bĵ + ck̂ and A · A = 49, then which of the following is true?
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.
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. If |A| = 5, |B| = 12, and A · B = 30, then the angle between A and B is:
A · B = |A||B|cosθ ⇒ 30 = 5×12×cosθ ⇒ cosθ = 0.5 ⇒ θ = 60°
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If the scalar projection of vector A on vector B is maximum, then angle between them is:
Scalar projection = |A||cosθ|. Max when cosθ = 1 ⇒ θ = 0°
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Let vector A = aî + bĵ. The angle between A and the x-axis is 60°. Then the value of b/a is:
Vector componentsA = a î + b ĵ• x-component = a• y-component = b
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Let A = 2î – 3ĵ + k̂, B = –î + 4ĵ + 2k̂. Find A · B.
A · B = (2)(–1) + (–3)(4) + (1)(2) = –2 – 12 + 2 = –12
13 / 35
If vector A = xî + 3ĵ and vector B = 2î + yĵ, and A · B = 0, then the relation between x and y is:
A · B = (x)(2) + (3)(y) = 0 ⇒ 2x + 3y = 0
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If angle between two vectors is acute, then their dot product is:
• Acute angle (0° – 90°) →dot product > 0 (positive) ____________• Right angle (90°) →dot product = 0 _____________• Obtuse angle (90° – 180°) →dot product < 0 (negative)
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If A · B = 0 and |A| = 3, |B| = 4, the angle between A and B is:
A · B = |A||B|cosθ = 0 ⇒ cosθ = 0 ⇒ θ = 90°
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For any vectors A and B, (A + B) · (A + B) equals:
(A + B)² = A · A + B · B + 2A · B
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Let a = 2î + ĵ – k̂, b = î – 3ĵ + k̂, and c = î + ĵ + k̂. If a · (b + λc) = 0, then λ = ?
a · (b + λc) = a · b + λ(a · c) = (–2) + λ(2)= 0 ⇒ λ = 1
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The value of (A + B) · (A – B) is:
(A + B) · (A – B) = A · A – B · B = |A|² – |B|²
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If A · B = A · C and A ≠ 0, then
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).
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If A = aî + bĵ and B = bî – aĵ, then A · B is:
A · B = a·b + b·(–a) = ab – ab = 0
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If the dot product of two vectors is negative, the angle between them lies between:
cosθ < 0 in this range ⇒ A·B < 0
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If A = 3i + 4j and B = 4i – 3j, find the angle between them using scalar product.
A·B = (3)(4) + (4)(–3) = 12 – 12 = 0 ⇒ angle= 90°
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Two vectors A and B satisfy A·B = |A||B|. The angle between them is:
A·B = |A||B|cosθ. If A·B = |A||B| ⇒ cosθ = 1 ⇒ θ = 0°
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If the angle between two vectors is obtuse (>90°), then their dot product is:
cosθ < 0 for obtuse angles ⇒ A·B = |A||B|cosθ becomes negative.
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The scalar product of two non-zero vectors is zero. Which of the following must be true?
A·B = |A||B|cosθ. If A·B = 0 and both A, B ≠ 0, then cosθ = 0 ⇒ θ = 90°
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What is the scalar product of two vectors inclined at 120°, both having magnitude 10?
A·B = |A||B|cos(120°) = 10×10×cos(120°) = 100×(-0.5) = -50
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Which of the following statements is incorrect about scalar product?
This is false. Scalar product is commutative ⇒ A·B = B·A.
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If |A| = 5, |B| = 10, and angle between them is 60°, then A·B = ?
A·B = |A||B|cosθ = 5×10×cos(60°) = 50×0.5 = 25
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The scalar product of vector A with itself is:
A·A = |A||A|cos0° = |A|²
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Which of the following is true about scalar product?
All listed properties are standard for scalar products.
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The scalar product of two unit vectors is 1. The angle between them is:
A·B = |A||B|cosθ = 1×1×cosθ = 1 ⇒ cosθ = 1 ⇒ θ = 0°
32 / 35
If A·B = 0 and neither A nor B is zero, then:
A·B = |A||B|cosθ = 0 implies cosθ = 0, so θ = 90°
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What is the scalar product of vector A = 2i + 3j and B = -i + 4j?
A·B = (2)(-1) + (3)(4) = -2 + 12 = 10
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The scalar (dot) product of two vectors is a:
Dot product results in a scalar quantity, not a vector.A·B = |A||B|cosθ is a number, not a direction-based quantity.
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If two vectors are perpendicular, then their scalar product is:
Scalar product is given by A·B = |A||B|cosθ.If vectors are perpendicular, θ = 90°, and cos(90°) = 0,⇒ A·B = 0
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