Vector Practice Test 2 ( Addition & Resolution of Vector )
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1 / 29
If u·v = 0 (and neither u nor v is the zero vector), then u and v are:
Explanation:
A dot product of zero indicates the vectors are orthogonal (at right angles).
2 / 29
The vector from P(1, 2, 3) to Q(4, 0, 5) is:
Subtract coordinates: (4–1)i + (0–2)j + (5–3)k = 3i – 2j + 2k.
3 / 29
A unit vector u satisfies u×i = j. Then u is:
4 / 29
If u and v are two perpendicular unit vectors, then |u + v| equals:
Since u·v=0 and |u|=|v|=1, |u+v| = √(u·u + 2u·v + v·v) = √(1+0+1) = √2.
5 / 29
Which property holds for any unit vector u?
By definition, a unit vector has magnitude 1, so its dot product with itself is |u|² = 1.
6 / 29
The unit vector parallel to a = 3 i – 4 j + 12 k is:
|a|=√(3²+ (–4)²+12² ) = √(9+16+144) =√169=13. So the unit vector is a/|a| = (3/13) i – (4/13) j + (12/13) k.
7 / 29
Which of the following is a unit vector?
The magnitude of i+j+k is √(1²+1²+1²)=√3. Dividing by √3 gives a vector of length 1:
8 / 29
A force of 25 N acts at 180° to the x-axis. What is the magnitude of its x-component?
The x-component is Fₓ = F·cos θ = 25·cos 180° = 25·(–1) = –25 N, whose magnitude is 25 N.
9 / 29
A force of 8 N acts at 45° to the x-axis. What is the magnitude of its x-component?
Fₓ = F·cos θ = 8·cos 45° = 8·(√2/2) = 4√2 N.
10 / 29
A force of 12 N acts at 30° to the x-axis. What is the magnitude of its y-component?
Fᵧ = F·sin θ = 12·sin 30° = 12·0.5 = 6 N.
11 / 29
A force of 15 N acts at 90° to the x-axis. What is the magnitude of its x-component?
Fₓ = F·cos θ = 15·cos 90° = 15·0 = 0 N.
12 / 29
A force of 20 N acts at 60° to the x-axis. What is the magnitude of its x-component?
The x-component is Fx = F·cos θ = 20·cos 60° = 20·0.5 = 10 N.
13 / 29
A force of 10 N is acting along the x-axis. Which of the following is the magnitude of its y-component?
A force along the x-axis makes θ=0° with the x-axis, soFy = F·sin θ = 10·sin 0° = 10·0 = 0 N.
14 / 29
A force entirely along x has no y-component ⇒ 0 N.
15 / 29
For any nonzero scalar k, the vector kA is ______ to A.
Scalar multiples of A always lie along the same line as A, so kA is parallel to A.
16 / 29
The magnitude of the vector 3A is
|3A| = |3|·|A| = 3·|A|
17 / 29
If a vector is multiplied by a scalar k then the resultant may be
• If k=0, kA = 0 (the zero vector).• If k>0, kA points in the same direction as A.• If k<0, kA points opposite to A (–kA).
18 / 29
Two vectors are said to be collinear if the angle between them is
Collinear vectors lie along the same line, so they point in exactly the same (0°) or exactly opposite (180°) direction
19 / 29
The magnitude of the cross product A × B is given by
|A × B| = A B sin θ, with θ the angle between A and B.
20 / 29
The vector having zero magnitude and arbitrary direction is called
A “zero vector” has zero length and its direction is undefined; it is also called the “null vector.”
21 / 29
The cross product A × B of two parallel vectors is:
|A × B| = AB sin θ, and sin 0° = 0, so the result is the zero vector.
22 / 29
The dot product A · B of two perpendicular vectors is:
A · B = AB cos θ, and cos 90° = 0, so the result is 0.
23 / 29
A unit vector is defined as a vector whose magnitude is:
A unit vector has been scaled so its length is exactly 1.
24 / 29
Adding any vector A to the zero vector 0 gives:
A + 0 = A by definition of the zero vector.
25 / 29
If two perpendicular vectors of magnitudes A and B are added, the magnitude of the resultant R is:
By Pythagoras’ theorem (since the angle is 90°), |R| = √(A² + B²).
26 / 29
Vector addition is commutative. That means:
“Commutative” means you can swap the order of the vectors without changing the resultant: A + B = B + A.
27 / 29
Addition of vectors obey the law of _____
Graphical vector addition is done by placing the tail of one vector at the head of the other (head-to-tail). This is neither a scalar “simple addition” nor a product operation.
28 / 29
Two forces each of 10 N act at angles of 0° and 90° to the x-axis. What is the magnitude of their resultant?
F₁ at 0° has components (10, 0), F₂ at 90° has (0, 10).Resultant R = (10 + 0, 0 + 10) = (10, 10),|R| = √(10² + 10²) = √200 ≈ 14.14 N
29 / 29
A child pulls a box with a force of 200N at an angle of 60∘ above the horizontal. Then the horizontal and vertical components of the force will be:
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