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Write the vector equations in component form.

Mathematics

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Here's the solution for question 7:

Part a· 5 steps

Find the value of aa.

  1. Write the vector equations in component form.

    For L1L_1: r1=(1−4μ)i+(1+μ)j−3k\mathbf{r}_1 = (1 - 4\mu)\mathbf{i} + (1 + \mu)\mathbf{j} - 3\mathbf{k}
    For L2L_2: r2=(9+2λ)i+(2+aλ)j+(1+2λ)k\mathbf{r}_2 = (9 + 2\lambda)\mathbf{i} + (2 + a\lambda)\mathbf{j} + (1 + 2\lambda)\mathbf{k}

  2. Since the lines intersect, their position vectors at the point of intersection are equal.

    Equate the corresponding components.

    1−4μ=9+2λ(1)1 - 4\mu = 9 + 2\lambda \quad (1) 1+μ=2+aλ(2)1 + \mu = 2 + a\lambda \quad (2) −3=1+2λ(3)-3 = 1 + 2\lambda \quad (3)

  3. Solve equation (3) for λ\lambda.

    −3=1+2λ-3 = 1 + 2\lambda 2λ=−42\lambda = -4 λ=−2\lambda = -2

  4. Substitute λ=−2\lambda = -2 into equation (1) to find μ\mu.

    1−4μ=9+2(−2)1 - 4\mu = 9 + 2(-2) 1−4μ=9−41 - 4\mu = 9 - 4 1−4μ=51 - 4\mu = 5 −4μ=4-4\mu = 4 μ=−1\mu = -1

  5. Substitute μ=−1\mu = -1 and λ=−2\lambda = -2 into equation (2) to find aa.

    1+(−1)=2+a(−2)1 + (-1) = 2 + a(-2) 0=2−2a0 = 2 - 2a 2a=22a = 2 a=1a = 1

    The value of aa is:

    1\boxed{\text{1}}

Part b· 1 step

Find the position vector of the point of intersection of L1L_1 and L2L_2.

  1. Substitute μ=−1\mu = -1 into the equation for L1L_1.

    r=i+j−3k+(−1)(−4i+j)\mathbf{r} = \mathbf{i} + \mathbf{j} - 3\mathbf{k} + (-1)(-4\mathbf{i} + \mathbf{j}) r=i+j−3k+4i−j\mathbf{r} = \mathbf{i} + \mathbf{j} - 3\mathbf{k} + 4\mathbf{i} - \mathbf{j} r=(1+4)i+(1−1)j−3k\mathbf{r} = (1+4)\mathbf{i} + (1-1)\mathbf{j} - 3\mathbf{k} r=5i+0j−3k\mathbf{r} = 5\mathbf{i} + 0\mathbf{j} - 3\mathbf{k}

    The position vector of the point of intersection is:

    \boxed{\text{5\mathbf{i} - 3\mathbf{k}}}

Part c· 4 steps

Find the cosine of the angle between L1L_1 and L2L_2.

  1. Identify the direction vectors of the lines.

    For L1L_1, the direction vector is d1=−4i+j\mathbf{d}_1 = -4\mathbf{i} + \mathbf{j}.
    For L2L_2, the direction vector is d2=2i+aj+2k\mathbf{d}_2 = 2\mathbf{i} + a\mathbf{j} + 2\mathbf{k}. Since a=1a=1, d2=2i+j+2k\mathbf{d}_2 = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}.

  2. Calculate the dot product of the direction vectors.

    d1⋅d2=(−4)(2)+(1)(1)+(0)(2)\mathbf{d}_1 \cdot \mathbf{d}_2 = (-4)(2) + (1)(1) + (0)(2) d1⋅d2=−8+1+0=−7\mathbf{d}_1 \cdot \mathbf{d}_2 = -8 + 1 + 0 = -7

  3. Calculate the magnitudes of the direction vectors.

    ∣d1∣=(−4)2+(1)2+(0)2=16+1+0=17|\mathbf{d}_1| = \sqrt{(-4)^2 + (1)^2 + (0)^2} = \sqrt{16 + 1 + 0} = \sqrt{17} ∣d2∣=(2)2+(1)2+(2)2=4+1+4=9=3|\mathbf{d}_2| = \sqrt{(2)^2 + (1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

  4. Use the formula for the cosine of the angle between two vectors, cos⁡θ=d1⋅d2∣d1∣∣d2∣\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|}.

    cos⁡θ=−717⋅3\cos\theta = \frac{-7}{\sqrt{17} \cdot 3} cos⁡θ=−7317\cos\theta = \frac{-7}{3\sqrt{17}}

    The cosine of the angle between L1L_1 and L2L_2 is:

    −7317\boxed{\text{}\frac{-7}{3\sqrt{17}}\text{}}
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