Find all vectors v in 2 dimensions having ||v ||=5 where the i -component of v is 3i
vectors:
let the vector be <3,y>
then 3^2 + y^2 = 5^2
y = ±4
so our vectors having a magnitude of 5 of the form <3,y>
are <3,4> and <3,-4>
Well, if the i-component of v is 3i, then we can write the vector v as (3i, y).
We know that the magnitude of v, ||v||, is given by ||v|| = √(x^2 + y^2), where x is the i-component and y is the j-component of v.
In this case, ||v|| = √((3i)^2 + y^2) = 5.
Simplifying this equation, we get 9 + y^2 = 25.
Taking the square root of both sides, we have y^2 = 16.
So, y can be either 4 or -4.
Therefore, the vectors v in 2 dimensions having ||v|| = 5 and i-component of v as 3i are (3i, 4) and (3i, -4).
To find all vectors v in 2 dimensions with ||v|| = 5 and the i-component of v as 3i, we can use the Pythagorean theorem.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (||v||) is equal to the sum of the squares of the lengths of the other two sides (components of v).
Let's denote the j-component of v as 3j. Since ||v|| = 5, we have:
√((3i)^2 + (3j)^2) = 5
Simplifying the equation, we get:
√(9i^2 + 9j^2) = 5
Squaring both sides of the equation, we have:
9i^2 + 9j^2 = 25
Dividing both sides by 9, we get:
i^2 + j^2 = 25/9
This equation represents a circle centered at the origin with a radius of √(25/9) = 5/3.
Therefore, all vectors v in 2 dimensions with ||v|| = 5 and the i-component of v as 3i lie on the circle with a radius of 5/3.
In vector form, these vectors can be represented as:
v = (3i, √(25/9 - 9i^2))
To find the full set of vectors, we can substitute different values of i into the equation and calculate j using the equation above.
To find all vectors v in 2 dimensions with ||v|| = 5, where the i-component of v is 3i, we can use the vector's magnitude formula.
The magnitude or length of a vector v = (a, b) in 2 dimensions can be calculated using the formula:
||v|| = sqrt(a^2 + b^2)
In this case, we have the magnitude as 5, which means:
5 = sqrt(a^2 + b^2)
Squaring both sides of the equation, we get:
25 = a^2 + b^2
Since the i-component of v is 3i, we know that a = 3i. Substituting this into the equation, we have:
25 = (3i)^2 + b^2
25 = 9i^2 + b^2
25 = -9 + b^2
Solving for b^2, we get:
b^2 = 34
Taking the square root of both sides, we have two possible solutions for b:
b = sqrt(34) or b = -sqrt(34)
Therefore, we have two vectors:
v1 = (3i, sqrt(34))
v2 = (3i, -sqrt(34))
These are the two vectors in 2 dimensions with ||v|| = 5, where the i-component of v is 3i.