1. **Problem statement:** We have two lines in 3D space: - Line $l_1$ passes through points $(0,1,5)$ and $(2,-2,1)$. - Line $l_2$ is given by the vector equation $$\mathbf{r} = 7\mathbf{i} + \mathbf{j} + \mathbf{k} + \mu(\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}).$$ We need to: (i) Show that $l_1$ and $l_2$ are skew lines. (ii) Find the acute angle between the direction of $l_2$ and the x-axis. --- 2. **Step (i): Show lines are skew** - Two lines are skew if they are not parallel and do not intersect. - Find direction vector of $l_1$: $$\vec{d_1} = (2-0, -2-1, 1-5) = (2, -3, -4).$$ - Direction vector of $l_2$ is the coefficient of $\mu$: $$\vec{d_2} = (1, 2, 5).$$ - Check if $\vec{d_1}$ and $\vec{d_2}$ are parallel: They are parallel if $\vec{d_1} = k \vec{d_2}$ for some scalar $k$. Check ratios: $$\frac{2}{1} = 2, \quad \frac{-3}{2} = -1.5, \quad \frac{-4}{5} = -0.8,$$ which are not equal, so lines are not parallel. - Check if lines intersect: Parametrize $l_1$: $$\mathbf{r_1} = (0,1,5) + \lambda(2,-3,-4) = (2\lambda, 1-3\lambda, 5-4\lambda).$$ Parametrize $l_2$: $$\mathbf{r_2} = (7,1,1) + \mu(1,2,5) = (7+\mu, 1+2\mu, 1+5\mu).$$ Set $\mathbf{r_1} = \mathbf{r_2}$: $$2\lambda = 7 + \mu,$$ $$1 - 3\lambda = 1 + 2\mu,$$ $$5 - 4\lambda = 1 + 5\mu.$$ From first equation: $$\mu = 2\lambda - 7.$$ Substitute into second: $$1 - 3\lambda = 1 + 2(2\lambda - 7) \Rightarrow 1 - 3\lambda = 1 + 4\lambda - 14,$$ $$-3\lambda = 4\lambda - 14,$$ $$-3\lambda - 4\lambda = -14,$$ $$-7\lambda = -14,$$ $$\lambda = 2.$$ Then $\mu = 2(2) - 7 = 4 - 7 = -3.$ Check third equation: $$5 - 4(2) = 1 + 5(-3) \Rightarrow 5 - 8 = 1 - 15,$$ $$-3 = -14,$$ which is false. Since the third equation is not satisfied, lines do not intersect. Therefore, lines are not parallel and do not intersect, so they are skew. --- 3. **Step (ii): Find acute angle between $l_2$ direction and x-axis** - Direction vector of $l_2$ is $\vec{d_2} = (1,2,5)$. - Direction vector of x-axis is $\vec{i} = (1,0,0)$. - The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by: $$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}.$$ Calculate dot product: $$\vec{d_2} \cdot \vec{i} = 1 \times 1 + 2 \times 0 + 5 \times 0 = 1.$$ Calculate magnitudes: $$|\vec{d_2}| = \sqrt{1^2 + 2^2 + 5^2} = \sqrt{1 + 4 + 25} = \sqrt{30},$$ $$|\vec{i}| = 1.$$ Calculate cosine: $$\cos \theta = \frac{1}{\sqrt{30} \times 1} = \frac{1}{\sqrt{30}}.$$ Calculate angle: $$\theta = \cos^{-1} \left( \frac{1}{\sqrt{30}} \right).$$ This is the acute angle between $l_2$ direction and x-axis. --- **Final answers:** - (i) Lines $l_1$ and $l_2$ are skew because they are not parallel and do not intersect. - (ii) The acute angle between $l_2$ direction and x-axis is $$\theta = \cos^{-1} \left( \frac{1}{\sqrt{30}} \right).$$