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3D Vectors & Dot Product - IB Mathematics Applications and Interpretation

  1. Find the angle between the space diagonal OP⃗\vec{OP} and the face diagonal AB⃗\vec{AB}.
    Answer: 90.0° First, define the vectors from the given coordinates. OP⃗=p−o=(543)\vec{OP} = \mathbf{p} - \mathbf{o} = \begin{pmatrix} 5 \\ 4 \\ 3 \end{pmatrix}. AB⃗=b−a=(0−54−03−0)=(−543)\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 0-5 \\ 4-0 \\ 3-0 \end{pmatrix} = \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix}. Next, calculate the dot product of these two vectors: OP⃗⋅AB⃗=(5)(−5)+(4)(4)+(3)(3)=−25+16+9=0\vec{OP} \cdot \vec{AB} = (5)(-5) + (4)(4) + (3)(3) = -25 + 16 + 9 = 0. Since the dot product of the two vectors is 0, they are perpendicular. Therefore, the angle between them is 90.0∘90.0^{\circ}.
  2. A rocket's trajectory after launch is modelled by the vector r=4i+7j+2k\mathbf{r} = 4\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}. Find the angle, to the nearest degree, that the rocket's path makes with the positive z-axis.
    Answer: 76° The positive z-axis can be represented by the unit vector k=(001)\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}. We need to find the angle θ\theta between r\mathbf{r} and k\mathbf{k}. First, calculate the dot product: r⋅k=(4)(0)+(7)(0)+(2)(1)=2\mathbf{r} \cdot \mathbf{k} = (4)(0) + (7)(0) + (2)(1) = 2. Next, find the magnitudes: ∣r∣=42+72+22=16+49+4=69|\mathbf{r}| = \sqrt{4^2 + 7^2 + 2^2} = \sqrt{16+49+4} = \sqrt{69}. ∣k∣=1|\mathbf{k}| = 1. Use the formula cos⁡θ=r⋅k∣r∣∣k∣\cos\theta = \frac{\mathbf{r} \cdot \mathbf{k}}{|\mathbf{r}||\mathbf{k}|}: cos⁡θ=269×1=269\cos\theta = \frac{2}{\sqrt{69} \times 1} = \frac{2}{\sqrt{69}}. θ=arccos⁡(269)≈76.08∘\theta = \arccos\left(\frac{2}{\sqrt{69}}\right) \approx 76.08^{\circ}. To the nearest degree, this is 76°.

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