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Find the position vector of the point of intersection of the lines l1: oldsymbolr=(6−2 oldsymbollambda) oldsymboli+(oldsymbollambda-5) oldsymbolj and l2: oldsymbolr=oldsymbolmu oldsymboli+3(1-

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2 parts
a

2i−3j2\mathbf{i} - 3\mathbf{j}

b

45∘45^\circ

Step-by-step solution

8 steps

Part a· 3 steps

To find the position vector of the point of intersection, we set the two vector equations equal to each other.

Given lines:
l1:r=(6−2λ)i+(λ−5)jl_1: \mathbf{r} = (6 - 2\lambda)\mathbf{i} + (\lambda - 5)\mathbf{j}
l2:r=μi+3(1−μ)jl_2: \mathbf{r} = \mu\mathbf{i} + 3(1 - \mu)\mathbf{j}

  1. Equate the i\mathbf{i} and j\mathbf{j} components.

    Equating i\mathbf{i} components:

    6−2λ=μ(1)6 - 2\lambda = \mu \quad (1)

    Equating j\mathbf{j} components:

    λ−5=3(1−μ)(2)\lambda - 5 = 3(1 - \mu) \quad (2)

  2. Solve the system of equations for λ\lambda and μ\mu.

    Substitute (1) into (2):

    λ−5=3(1−(6−2λ))\lambda - 5 = 3(1 - (6 - 2\lambda)) λ−5=3(1−6+2λ)\lambda - 5 = 3(1 - 6 + 2\lambda) λ−5=3(−5+2λ)\lambda - 5 = 3(-5 + 2\lambda) λ−5=−15+6λ\lambda - 5 = -15 + 6\lambda 10=5λ10 = 5\lambda λ=2\lambda = 2

    Substitute λ=2\lambda = 2 back into (1):

    μ=6−2(2)\mu = 6 - 2(2) μ=6−4\mu = 6 - 4 μ=2\mu = 2

  3. Substitute the value of λ\lambda (or μ\mu) into one of the line equations to find the position vector.

    Using l1l_1 with λ=2\lambda = 2:

    r=(6−2(2))i+(2−5)j\mathbf{r} = (6 - 2(2))\mathbf{i} + (2 - 5)\mathbf{j} r=(6−4)i+(−3)j\mathbf{r} = (6 - 4)\mathbf{i} + (-3)\mathbf{j} r=2i−3j\mathbf{r} = 2\mathbf{i} - 3\mathbf{j}

    The position vector of the point of intersection is 2i−3j\boxed{2\mathbf{i} - 3\mathbf{j}}.

Part b

45∘45^\circ

To find the angle between l1l_1 and l2l_2, we use their direction vectors.

  1. Identify the direction vectors.

    For l1:r=(6−2λ)i+(λ−5)j=(6i−5j)+λ(−2i+j)l_1: \mathbf{r} = (6 - 2\lambda)\mathbf{i} + (\lambda - 5)\mathbf{j} = (6\mathbf{i} - 5\mathbf{j}) + \lambda(-2\mathbf{i} + \mathbf{j}).
    The direction vector for l1l_1 is d1=−2i+j\mathbf{d}_1 = -2\mathbf{i} + \mathbf{j}.
    For l2:r=μi+3(1−μ)j=3j+μ(i−3j)l_2: \mathbf{r} = \mu\mathbf{i} + 3(1 - \mu)\mathbf{j} = 3\mathbf{j} + \mu(\mathbf{i} - 3\mathbf{j}).
    The direction vector for l2l_2 is d2=i−3j\mathbf{d}_2 = \mathbf{i} - 3\mathbf{j}.

  2. Calculate the dot product of the direction vectors.

    d1⋅d2=(−2)(1)+(1)(−3)=−2−3=−5\mathbf{d}_1 \cdot \mathbf{d}_2 = (-2)(1) + (1)(-3) = -2 - 3 = -5

  3. Calculate the magnitudes of the direction vectors.

    ∣d1∣=(−2)2+(1)2=4+1=5|\mathbf{d}_1| = \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} ∣d2∣=(1)2+(−3)2=1+9=10|\mathbf{d}_2| = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}

  4. Use the dot product formula to find the angle θ\theta.

    cos⁡θ=d1⋅d2∣d1∣∣d2∣\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|} cos⁡θ=−5510\cos\theta = \frac{-5}{\sqrt{5} \sqrt{10}} cos⁡θ=−550\cos\theta = \frac{-5}{\sqrt{50}} cos⁡θ=−552\cos\theta = \frac{-5}{5\sqrt{2}} cos⁡θ=−12\cos\theta = -\frac{1}{\sqrt{2}}

  5. Find θ\theta.

    The angle between two lines is usually taken as the acute angle.

    ∣cos⁡θ∣=∣−12∣=12|\cos\theta| = \left|-\frac{1}{\sqrt{2}}\right| = \frac{1}{\sqrt{2}} θ=arccos⁡(12)\theta = \arccos\left(\frac{1}{\sqrt{2}}\right) θ=45∘\theta = 45^\circ

    The angle between l1l_1 and l2l_2 is 45∘\boxed{45^\circ}.

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