I am having some problems, especially getting started. Please let me know if the

Question

I am having some problems, especially getting started. Please let me know if the problem is not readable:

Displacement d_1^vector is in the yz plane 59.0 degree from the positive direction of the y axis, has a positive z component, and has a magnitude of 3.85 m. Displacement d_2^vector is in the xz plane 29.4 degree from the positive direction of the x axis, has a positive z component, and has magnitude 1.05 m. What are (a) d_1^vector middot d_2^vector, (b) the x component or d_1^vector times d_2^vector, (c) the y component of d_1^vector times d_2^vector, (d) the z component of d_1^vector times d_2^vector, and (e) the angle between d_1^vector and d_2^vector?

Explanation / Answer

First you need to calculate the x, y and z components for each vector using trigonometry.
For instance, d1 projects on to the:

x-axis at 0 (the d1 vector is in the yz plane, so no x component)

y-axis at 3.85 * cos(59) = 1.98

z-axis at 3.85 x sin(59) = 3.3
The d1 vector can now be written as d1 = [0 1.98 3.3] i.e [x y z].
Calculating the d2 vector x,y,z components gives d2 = [0.515 0 0.914]

x-axis at 1.05 x sin(29.4) = 0.515

y-axis at 0 (the d2 vector is in the xz plane, so no y component)

z-axis at 1.05 x cos(29.4) = 0.914

Now, d1.d2 is the scalar product. The answer is just a number, not a vector.
d1.d2 = (0 x 515) + (1.98 x 0) + (3.3 x 0.914) = 3.0162 m (Answer).

d1 x d2

| i j k |
| 0 1.98 3.3 |
| 0.515 0 0.914 |

Solving the determinant:

i(1.98*0.914) - j(-3.3*0.515) + k(-1.98*0.515) = 1.81 i + 1.7j - 1.02k

x component = 1.8

y component = 1.7

z component = -1.02

angle between d1 and d2

|d1| = 3.848

|d2| = 1.05

d1 . d2 = |d1| * |d2| * cos(theta) = 3.0162

theta = arccos(0.7465) = 41.7 degrees

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.