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The cosine of the acute angle between r1 and r2. The direction vector for r1 is d1 = i + 2j + 2k. The direction vector for r2 is d2 = -j + 3k. Find the cosine of the acute angle between the two lines.

Mathematics

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5. (b) The position vector of the point of intersection
✅ Correct.
Your steps and final answer are accurate.

5. (c) The cosine of the acute angle between r1\mathbf{r_1} and r2\mathbf{r_2}
❌ Incorrect.
Your calculation for the dot product d1⋅d2\mathbf{d_1} \cdot \mathbf{d_2} is incorrect. The components were multiplied incorrectly.

Here is the corrected solution:

  1. Identify the direction vectors of the lines.

    The direction vector for r1\mathbf{r_1} is d1=i+2j+2k\mathbf{d_1} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}.
    The direction vector for r2\mathbf{r_2} is d2=−j+3k\mathbf{d_2} = -\mathbf{j} + 3\mathbf{k}.

  2. Calculate the dot product of the direction vectors.

    d1⋅d2=(1)(0)+(2)(−1)+(2)(3)\mathbf{d_1} \cdot \mathbf{d_2} = (1)(0) + (2)(-1) + (2)(3) d1⋅d2=0−2+6\mathbf{d_1} \cdot \mathbf{d_2} = 0 - 2 + 6 d1⋅d2=4\mathbf{d_1} \cdot \mathbf{d_2} = 4

  3. Calculate the magnitudes of the direction vectors.

    ∣d1∣=12+22+22=1+4+4=9=3|\mathbf{d_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3 ∣d2∣=02+(−1)2+32=0+1+9=10|\mathbf{d_2}| = \sqrt{0^2 + (-1)^2 + 3^2} = \sqrt{0+1+9} = \sqrt{10}

  4. Use the dot product formula to find the cosine of the angle.

    cos⁡θ=∣d1⋅d2∣∣d1∣∣d2∣\cos \theta = \frac{|\mathbf{d_1} \cdot \mathbf{d_2}|}{|\mathbf{d_1}||\mathbf{d_2}|} cos⁡θ=∣4∣(3)(10)\cos \theta = \frac{|4|}{(3)(\sqrt{10})} cos⁡θ=4310\cos \theta = \frac{4}{3\sqrt{10}}

    Rationalize the denominator:

    cos⁡θ=41031010=41030=21015\cos \theta = \frac{4\sqrt{10}}{3\sqrt{10}\sqrt{10}} = \frac{4\sqrt{10}}{30} = \frac{2\sqrt{10}}{15}

    The cosine of the acute angle between r1\mathbf{r_1} and r2\mathbf{r_2} is 21015\boxed{\frac{2\sqrt{10}}{15}}.

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