vector b has x,y, and z components of 4.0, 6.0, and 3.0 units, respectively. Calculate the magnitude of b and the angles that b makes with the coordinate axes.
Vector B~ has x, y, and z components of 1.4,
8.8, and 1.9 units, respectively.
Calculate the magnitude of B~
Well, if vector b is wearing a coordinate hat, it must be doing some serious math fashion! Let's calculate the magnitude and angles of b.
To find the magnitude (or length) of vector b, we can use the Pythagorean theorem in 3D. So, grab your math goggles and get ready.
Magnitude (|b|) = sqrt(x^2 + y^2 + z^2)
= sqrt(4.0^2 + 6.0^2 + 3.0^2)
= sqrt(16.0 + 36.0 + 9.0)
= sqrt(61.0)
≈ 7.81 units.
Now, let's move on to the coordinate axes angles. To find the angle that vector b makes with each axis, we can use some trigonometry magic.
Angle with x-axis (θx) = arccos(x / |b|)
= arccos(4.0 / 7.81)
≈ 54.26 degrees.
Angle with y-axis (θy) = arccos(y / |b|)
= arccos(6.0 / 7.81)
≈ 35.74 degrees.
Angle with z-axis (θz) = arccos(z / |b|)
= arccos(3.0 / 7.81)
≈ 77.35 degrees.
So, vector b is making some fashionable angles with the coordinate axes: 54.26 degrees with the x-axis, 35.74 degrees with the y-axis, and 77.35 degrees with the z-axis.
Now b sure to strut your math fashion with confidence!
To calculate the magnitude of vector b, you can use the formula for the magnitude of a three-dimensional vector:
|b| = √(x^2 + y^2 + z^2)
Given the components of vector b as x = 4.0, y = 6.0, and z = 3.0, we can substitute these values into the formula:
|b| = √(4.0^2 + 6.0^2 + 3.0^2)
|b| = √(16.0 + 36.0 + 9.0)
|b| = √61.0
|b| ≈ 7.81 units
To find the angles that vector b makes with the coordinate axes, we can use trigonometry. The angle that vector b makes with the x-axis is given by:
θx = arccos(x / |b|)
Substituting the values x = 4.0 and |b| = 7.81 into this formula, we have:
θx = arccos(4.0 / 7.81)
θx ≈ 56.29 degrees
Similarly, the angle that vector b makes with the y-axis is given by:
θy = arccos(y / |b|)
Substituting the values y = 6.0 and |b| = 7.81 into this formula, we have:
θy = arccos(6.0 / 7.81)
θy ≈ 36.87 degrees
Finally, the angle that vector b makes with the z-axis is given by:
θz = arccos(z / |b|)
Substituting the values z = 3.0 and |b| = 7.81 into this formula, we have:
θz = arccos(3.0 / 7.81)
θz ≈ 62.84 degrees
So, the magnitude of vector b is approximately 7.81 units, and it makes angles of approximately 56.29 degrees with the x-axis, 36.87 degrees with the y-axis, and 62.84 degrees with the z-axis.
The magnitude of the vector is
M = sqrt(a^2 + b^2 + c^2)= 7.81
The cosine with the X axis is a/M = 4.0/7.81 = 0.51216
The angle is 59.2 degrees.
The cosine with the Y axis is b/M
The cosine with the Z axis is c/M
Vector B~ has x, y, and z components of 5.5,
6.2, and 1.8 units, respectively.
Calculate the magnitude of B~ .