More practice questions

4. Q4. Find the component of A along B, where

   $$\mathbf{A} = 2\hat i - \hat j + 2\hat k$$ $$\mathbf{B} = 3\hat i + 4\hat k$$
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   The dot product is 6 + 0 + 8 = 14 and $B = \sqrt{9 + 16} = 5$, so the component of A along B is 14/5, that is 2.8.

5. Q5. The magnitude of A + B equals the magnitude of A − B. What is the angle between A and B?

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   Square both sides. The square of a magnitude is the dot product of the vector with itself (see the identity in the properties section), so

   $$\begin{aligned}&A^{2} + 2\,\mathbf{A} \cdot \mathbf{B} + B^{2} \\ &= A^{2} - 2\,\mathbf{A} \cdot \mathbf{B} + B^{2}\end{aligned}$$

   So four times the dot product is zero. If neither vector is zero, they are perpendicular, and the angle is 90°.

6. Q6. A force F of $(2\hat i + 3\hat j)$ N acts on a particle that moves from the point (1, 2) m to the point (4, 6) m. Find the work done by the force.

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   The displacement, in metres, is

   $$\begin{aligned} \mathbf{d} &= (4 - 1)\hat i + (6 - 2)\hat j \\ &= 3\hat i + 4\hat j \end{aligned}$$

   So the work is

   $$\begin{aligned} W &= \mathbf{F} \cdot \mathbf{d} = 2(3) + 3(4) \\ &= 18\ \text{J} \end{aligned}$$

7. Q7. The angle between two vectors of equal magnitude is 120°. Show that their sum has the same magnitude as each of them.

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   For vectors of magnitude A with θ = 120°, |A + B|^² = A^² + B^² + 2AB cos θ, and with B = A:

   $$\begin{aligned} |\mathbf{A} + \mathbf{B}|^{2} &= 2A^{2}(1+\cos 120^\circ) \\ &= 2A^{2} - A^{2} = A^{2} \end{aligned}$$

   so $|\mathbf{A} + \mathbf{B}| = A$⁠.