Vector worksheet

1. Magnitude |A|: 4.36

1. Find the length of A: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|A|}$$=\sqrt{3^2+3^2+1^2}$$=\sqrt{19}$$=4.358899$

2. Unit vector Â: (0.707107, 0.707107)

1. Find the length of A: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|A|}$$=\sqrt{2^2+2^2}$$=\sqrt{8}$$=2.828427$

2. Divide each component by the magnitude to get a vector of length 1

   $Â$$=\frac{\text{(2, 2)}}{2.828427}$$=\text{(0.707107, 0.707107)}$

3. Angle between A and B: 68.55 °

1. Find the dot product: multiply matching components and add the results

   $\mathrm{A\; ·\; B}$$=2\times 1+3\times \left(\mathrm{-1}\right)+2\times 3$$=5$

2. Find the length of A: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|A|}$$=\sqrt{2^2+3^2+2^2}$$=\sqrt{17}$$=4.123106$

3. Find the length of B: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|B|}$$=\sqrt{1^2+{\left(\mathrm{-1}\right)}^2+3^2}$$=\sqrt{11}$$=3.316625$

4. Divide the dot product by both lengths. This gives the cosine of the angle θ between the vectors

   $\mathrm{cos\; \theta }$$=\frac{5}{4.123106\times 3.316625}$$=0.36563621$

5. Undo the cosine (take the inverse cosine) to get the angle itself

   $\theta$$=\arccos \left(0.36563621\right)$$=68.553258\text{°}$

4. A × B: (0, 0, -12)

1. Treat the 2D vectors as 3D vectors lying flat in the xy-plane (z = 0)

   $A$$=\text{(2, 4, 0)}$

2. Find the cross product, a vector at right angles to both, by cross-multiplying pairs of components: (a_y b_z − a_z b_y, a_z b_x − a_x b_z, a_x b_y − a_y b_x)

   $\mathrm{A\; \times \; B}$$=\text{(4×0 − 0×(-2), 0×2 − 2×0, 2×(-2) − 4×2)}$$=\text{(0, 0, -12)}$

3. Find its length. This equals the area of the slanted rectangle (parallelogram) that A and B make

   $\mathrm{|A\; \times \; B|}$$=\sqrt{0^2+0^2+{\left(\mathrm{-12}\right)}^2}$$=\sqrt{144}$$=12$

5. Dot product A · B: 2

1. Find the dot product: multiply matching components and add the results

   $\mathrm{A\; ·\; B}$$=2\times 1+3\times \left(\mathrm{-1}\right)+1\times 3$$=2$

6. Magnitude |A|: 3.16

1. Find the length of A: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|A|}$$=\sqrt{1^2+3^2}$$=\sqrt{10}$$=3.162278$

7. Angle between A and B: 78.69 °

1. Find the dot product: multiply matching components and add the results

   $\mathrm{A\; ·\; B}$$=3\times 2+2\times \left(\mathrm{-2}\right)$$=2$

2. Find the length of A: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|A|}$$=\sqrt{3^2+2^2}$$=\sqrt{13}$$=3.605551$

3. Find the length of B: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|B|}$$=\sqrt{2^2+{\left(\mathrm{-2}\right)}^2}$$=\sqrt{8}$$=2.828427$

4. Divide the dot product by both lengths. This gives the cosine of the angle θ between the vectors

   $\mathrm{cos\; \theta }$$=\frac{2}{3.605551\times 2.828427}$$=0.19611614$

5. Undo the cosine (take the inverse cosine) to get the angle itself

   $\theta$$=\arccos \left(0.19611614\right)$$=78.690068\text{°}$

8. Magnitude |A|: 4.47

1. Find the length of A: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|A|}$$=\sqrt{2^2+4^2}$$=\sqrt{20}$$=4.472136$

9. Dot product A · B: 2

1. Find the dot product: multiply matching components and add the results

   $\mathrm{A\; ·\; B}$$=2\times 1+4\times \left(\mathrm{-1}\right)+1\times 4$$=2$

10. Angle between A and B: 71.68 °

1. Find the dot product: multiply matching components and add the results

   $\mathrm{A\; ·\; B}$$=2\times 1+2\times \left(\mathrm{-1}\right)+1\times 4$$=4$

2. Find the length of A: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|A|}$$=\sqrt{2^2+2^2+1^2}$$=\sqrt{9}$$=3$

3. Find the length of B: square each component, add them, then take the square root (the Pythagorean theorem)

   $\mathrm{|B|}$$=\sqrt{1^2+{\left(\mathrm{-1}\right)}^2+4^2}$$=\sqrt{18}$$=4.242641$

4. Divide the dot product by both lengths. This gives the cosine of the angle θ between the vectors

   $\mathrm{cos\; \theta }$$=\frac{4}{3\times 4.242641}$$=0.31426968$

5. Undo the cosine (take the inverse cosine) to get the angle itself

   $\theta$$=\arccos \left(0.31426968\right)$$=71.683269\text{°}$