given vectors ,
Vector a= 2i + 3j, then the angle between vector a and y axis is?
Plz solve
cosine theta = (a*i)/|a|
|a| is the magnitude of the a vector,
which is sqrt13 = 3.6056
a*1 is the dot product of a and the i vector, which is 2.
Therefore cos theta = 2/3.6056 = 0.5547
theta = 56.3 degrees
To find the angle between vector a and the y-axis, you can use the dot product between the two vectors. The dot product of two vectors can be calculated using the formula:
a · b = |a| |b| cos(theta)
Where a · b is the dot product of vectors a and b, |a| and |b| are the magnitudes of vectors a and b, and theta is the angle between the two vectors.
In this case, vector a is given as 2i + 3j. The y-axis can be represented by the vector 0i + 1j since it has a unit magnitude in the positive y-direction.
Now, let's calculate the dot product:
a · y-axis = (2i + 3j) · (0i + 1j)
= (2 * 0) + (3 * 1)
= 3
To find the magnitudes of each vector:
|a| = sqrt((2^2) + (3^2)) = sqrt(4 + 9) = sqrt(13)
|y-axis| = sqrt((0^2) + (1^2)) = sqrt(1) = 1
Now, substituting these values into the dot product formula:
3 = sqrt(13) * 1 * cos(theta)
cos(theta) = 3 / sqrt(13)
To find the angle, you can take the inverse cosine (arccos) of this value:
theta = arccos(3 / sqrt(13))
Using a calculator, the value of theta can be found, which is approximately 49.04 degrees.
Therefore, the angle between vector a and the y-axis is approximately 49.04 degrees.