Find the direction angles of the vector given below.
Then write each vector in the form
v = ||v||[(cos A) i + (cos B) i + (cos Y) k ].
v = -6i + 12j + 4k
the maginitued of vector v = √((-6)^2+12^2+4^2)
= 14
so the directin cosines are (-6/14,12/14,4/14)
and the angles are 115.4º, 31º, and 73.4º
To find the direction angles of the vector v = -6i + 12j + 4k, we can use the formulas:
cos A = (v_x) / ||v||
cos B = (v_y) / ||v||
cos Y = (v_z) / ||v||
where v_x, v_y, and v_z are the components of the vector v, and ||v|| is the magnitude of the vector v.
First, let's calculate the magnitude of the vector v:
||v|| = sqrt((-6)^2 + 12^2 + 4^2)
= sqrt(36 + 144 + 16)
= sqrt(196)
= 14
Now, let's calculate the direction angles:
cos A = (-6) / 14
= -3/7
cos B = 12 / 14
= 6/7
cos Y = 4 / 14
= 2/7
Therefore, the direction angles of the vector v are A = cos^(-1)(-3/7), B = cos^(-1)(6/7), and Y = cos^(-1)(2/7).
Now, let's write the vector v in the form v = ||v||[(cos A) i + (cos B) i + (cos Y) k]:
v = 14[(-3/7)i + (6/7)j + (2/7)k]
To find the direction angles of a vector, we can use the following formulas:
θ = arccos(v_z / ||v||),
φ = arccos(v_xy / ||v||),
where v_z is the z-component of the vector, v_xy is the magnitude of the vector projected onto the xy-plane, and ||v|| is the magnitude of the vector.
Now, let's apply these formulas to find the direction angles of the vector v = -6i + 12j + 4k:
1. Find the magnitude of the vector:
||v|| = sqrt((-6)^2 + 12^2 + 4^2) = sqrt(36 + 144 + 16) = sqrt(196) = 14.
2. Find the z-component of the vector:
v_z = 4.
3. Find the magnitude of the vector projected onto the xy-plane:
v_xy = sqrt((-6)^2 + 12^2) = sqrt(36 + 144) = sqrt(180) = 6*sqrt(5).
4. Calculate the direction angles:
θ = arccos(4 / 14) = arccos(2/7),
φ = arccos(6*sqrt(5) / 14) = arccos(3*sqrt(5)/7).
Therefore, the direction angles of the vector v = -6i + 12j + 4k are θ = arccos(2/7) and φ = arccos(3*sqrt(5)/7).
Finally, to write the vector v in the form v = ||v||[(cos A) i + (cos B) j + (cos Y) k], we substitute the calculated values:
v = 14[(cos(arccos(2/7)))i + (cos(arccos(3*sqrt(5)/7)))j + (cos Y)k].
Since the direction angles, A and B, are already known, the final form of the vector v is:
v = 14[(2/7)i + (3*sqrt(5)/7)j + (cos Y)k].