Vectors Miscellaneous Exercise
Question:
What pulled the Princess out of her gloom?
Answer:

First, it is important to understand why the Princess was feeling gloomy.

Once the reason for the Princess’ gloom is identified, it can be determined what could have pulled her out of her gloom.

Possible solutions could include activities such as spending time with friends, engaging in hobbies, or going on an adventure.

Ultimately, the exact solution to the Princess’ gloom will depend on the individual and the specific circumstances.
Question:
Write down a unit vector in XY plane, making an angle of 30∘ with the positive direction of x−axis .
Answer:
A unit vector in XYplane making an angle of 30∘ with the positive direction of x−axis can be written as:
u = (cos 30°, sin 30°) = (√3/2, 1/2)
Question:
Find the scalar components and magnitude of the vector joining the points P(x1,y1,z1) and Q(x2,y2,z2).
Answer:
Step 1: Calculate the difference between the xcoordinates of the two points: x2  x1.
Step 2: Calculate the difference between the ycoordinates of the two points: y2  y1.
Step 3: Calculate the difference between the zcoordinates of the two points: z2  z1.
Step 4: Calculate the scalar components of the vector by taking the square root of the sum of the squares of the differences calculated in Steps 13:
√(x2  x1)2 + (y2  y1)2 + (z2  z1)2
Step 5: Calculate the magnitude of the vector by taking the square root of the sum of the squares of the scalar components:
√(x2  x1)2 + (y2  y1)2 + (z2  z1)2
Question:
Find the value of x for which x(i^+j^+k^) is a unit vector .
Answer:
Step 1: A unit vector is a vector with a magnitude of 1. Therefore, we need to find the magnitude of x(i^+j^+k^).
Step 2: The magnitude of a vector can be found using the formula: v = √(v₁² + v₂² + v₃²).
Step 3: Substitute i^+j^+k^ into the formula to get: x(i^+j^+k^) = √(x² + x² + x²) = √(3x²).
Step 4: Set the magnitude equal to 1 and solve for x.
x(i^+j^+k^) = 1
√(3x²) = 1
3x² = 1
x² = 1/3
x = √(1/3)
Question:
A girl walks 4 km towards west, then she walk 3 km in a direction 30o east of north and stops. Determine the girl’s displacement from her initial point of departure.
Answer:

The girl’s initial point of departure is the origin (0,0).

She walks 4 km towards west, so her new coordinates are (4,0).

She then walks 3 km in a direction 30o east of north. To find the new coordinates, we use the trigonometric ratio of sine and cosine.
sin 30o = 3/r cos 30o = 4/r
where r is the magnitude (or distance) of the girl’s displacement.
Therefore, r = 4√3
Using the trigonometric ratios, we can find the new x and ycoordinates of the girl’s position.
x = 4 + 4√3 cos 30o = 4 + 4√3 (4/r) = 4 + 16/√3
y = 0 + 4√3 sin 30o = 0 + 4√3 (3/r) = 12/√3
 The girl’s displacement from her initial point of departure is the magnitude of the vector joining the two points, which is given by the Pythagorean theorem.
displacement = √((x2  x1)2 + (y2  y1)2)
= √((4 + 16/√3)2 + (12/√3)2)
= √(256/3 + 144/3)
= √(400/3)
= 20√3 units
Question:
If a=b+c, then is it true that ∣a∣=∣b∣+∣c∣? Justify your answer.
Answer:
No, it is not true that a=b+c.
To justify this answer, we can look at an example. Let a=2, b=1, and c=1.
Then a=b+c, or a=2.
However, a=2=2, while b+c=1+1=2+1=3.
Since a does not equal b+c, it is not true that a=b+c.
Question:
If a=i^+j^+k^,b=2i^−j^+3k^ and c=i^−2j^+k^, find a unit vector parallel to the vector 2a−b+3c.
Answer:

Expand the vectors: a = i + j + k b = 2i  j + 3k c = i  2j + k

Substitute the vectors into the equation: 2a  b + 3c = 2(i + j + k)  (2i  j + 3k) + 3(i  2j + k)

Simplify the equation: 2a  b + 3c = 2i + 2j + 2k  2i + j  3k + 3i  6j + 3k

Combine like terms: 2a  b + 3c = 4i  5j + 6k

Divide the equation by its magnitude: 2a  b + 3c = (4i  5j + 6k) / √(42 + 52 + 62)

Simplify the equation: 2a  b + 3c = (4/√77)i  (5/√77)j + (6/√77)k

The unit vector parallel to the vector 2a  b + 3c is: (4/√77)i  (5/√77)j + (6/√77)k