find the magnitude and direction angle of the vector.
3i-4j and -3i-5j
Givem u,v I can find
u+v, u-v, u•v, u×v
I don't know what u and v is.
If all you want is each vector, then
|3-4j| = √(3^2+4^2) = 5
The direction θ is such that
tanθ = y/x = -4/3 so it is in QIV
Do -3-5j the same way.
To find the magnitude and direction angle of a vector, we can use the following formulas:
Magnitude (or length) of a vector:
|v| = √(v₁² + v₂²)
where v₁ and v₂ represent the components of the vector.
Direction angle (θ) of a vector:
θ = arctan(v₂/v₁)
where v₁ and v₂ represent the components of the vector.
Let's calculate the magnitude and direction angle for the given vectors.
1. Vector v₁ = 3i - 4j:
Magnitude |v₁| = √(3² + (-4)²)
= √(9 + 16)
= √25
= 5
Direction angle θ₁ = arctan((-4)/3)
≈ -53.13° (rounded to two decimal places)
2. Vector v₂ = -3i - 5j:
Magnitude |v₂| = √((-3)² + (-5)²)
= √(9 + 25)
= √34
Direction angle θ₂ = arctan((-5)/(-3))
≈ 59.04° (rounded to two decimal places)
Therefore, the magnitude and direction angles for the given vectors are:
1. Magnitude |v₁| = 5, direction angle θ₁ ≈ -53.13°
2. Magnitude |v₂| = √34, direction angle θ₂ ≈ 59.04°
To find the magnitude and direction angle of a vector, we can use the following formulas:
Magnitude of a vector = √(x^2 + y^2)
Direction angle (θ) = arctan(y / x)
Let's calculate the magnitude and direction angle for the given vectors:
For the vector 3i - 4j:
x = 3 and y = -4
Magnitude = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5
Direction angle = arctan((-4) / 3) = -53.13° (rounded to two decimal places)
For the vector -3i - 5j:
x = -3 and y = -5
Magnitude = √((-3)^2 + (-5)^2) = √(9 + 25) = √34
Direction angle = arctan((-5) / (-3)) = 59.04° (rounded to two decimal places)
Therefore, the magnitude and direction angle of the vector 3i - 4j are 5 and -53.13°, respectively.
And the magnitude and direction angle of the vector -3i - 5j are √34 and 59.04°.