Write the vector equations in component form.

Question

Write the vector equations in component form.

Asked on April 26, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-26T06:56:33.311Z

Here's the solution for question 7: a) Find the value of $a$. Step 1: Write the vector equations in component form. For $L_1$: $\mathbf{r}_1 = (1 - 4\mu)\mathbf{i} + (1 + \mu)\mathbf{j} - 3\mathbf{k}$ For $L_2$: $\mathbf{r}_2 = (9 + 2\lambda)\mathbf{i} + (2 + a\lambda)\mathbf{j} + (1 + 2\lambda)\mathbf{k}$ Step 2: Since the lines intersect, their position vectors at the point of intersection are equal. Equate the corresponding components. $$ 1 - 4\mu = 9 + 2\lambda \quad (1) $$ $$ 1 + \mu = 2 + a\lambda \quad (2) $$ $$ -3 = 1 + 2\lambda \quad (3) $$ Step 3: Solve equation (3) for $\lambda$. $$ -3 = 1 + 2\lambda $$ $$ 2\lambda = -4 $$ $$ \lambda = -2 $$ Step 4: Substitute $\lambda = -2$ into equation (1) to find $\mu$. $$ 1 - 4\mu = 9 + 2(-2) $$ $$ 1 - 4\mu = 9 - 4 $$ $$ 1 - 4\mu = 5 $$ $$ -4\mu = 4 $$ $$ \mu = -1 $$ Step 5: Substitute $\mu = -1$ and $\lambda = -2$ into equation (2) to find $a$. $$ 1 + (-1) = 2 + a(-2) $$ $$ 0 = 2 - 2a $$ $$ 2a = 2 $$ $$ a = 1 $$ The value of $a$ is: $$\boxed{\text{1}}$$ b) Find the position vector of the point of intersection of $L_1$ and $L_2$. Step 1: Substitute $\mu = -1$ into the equation for $L_1$. $$ \mathbf{r} = \mathbf{i} + \mathbf{j} - 3\mathbf{k} + (-1)(-4\mathbf{i} + \mathbf{j}) $$ $$ \mathbf{r} = \mathbf{i} + \mathbf{j} - 3\mathbf{k} + 4\mathbf{i} - \mathbf{j} $$ $$ \mathbf{r} = (1+4)\mathbf{i} + (1-1)\mathbf{j} - 3\mathbf{k} $$ $$ \mathbf{r} = 5\mathbf{i} + 0\mathbf{j} - 3\mathbf{k} $$ The position vector of the point of intersection is: $$\boxed{\text{5\mathbf{i} - 3\mathbf{k}}}$$ c) Find the cosine of the angle between $L_1$ and $L_2$. Step 1: Identify the direction vectors of the lines. For $L_1$, the direction vector is $\mathbf{d}_1 = -4\mathbf{i} + \mathbf{j}$. For $L_2$, the direction vector is $\mathbf{d}_2 = 2\mathbf{i} + a\mathbf{j} + 2\mathbf{k}$. Since $a=1$, $\mathbf{d}_2 = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$. Step 2: Calculate the dot product of the direction vectors. $$ \mathbf{d}_1 \cdot \mathbf{d}_2 = (-4)(2) + (1)(1) + (0)(2) $$ $$ \mathbf{d}_1 \cdot \mathbf{d}_2 = -8 + 1 + 0 = -7 $$ Step 3: Calculate the magnitudes of the direction vectors. $$ |\mathbf{d}_1| = \sqrt{(-4)^2 + (1)^2 + (0)^2} = \sqrt{16 + 1 + 0} = \sqrt{17} $$ $$ |\mathbf{d}_2| = \sqrt{(2)^2 + (1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 $$ Step 4: Use the formula for the cosine of the angle between two vectors, $\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|}$. $$ \cos\theta = \frac{-7}{\sqrt{17} \cdot 3} $$ $$ \cos\theta = \frac{-7}{3\sqrt{17}} $$ The cosine of the angle between $L_1$ and $L_2$ is: $$\boxed{\text{}\frac{-7}{3\sqrt{17}}\text{}}$$ 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-04-26T06:56:33.311Z

Here's the solution for question 7: a) Find the value of $a$. Step 1: Write the vector equations in component form. For $L_1$: $\mathbf{r}_1 = (1 - 4\mu)\mathbf{i} + (1 + \mu)\mathbf{j} - 3\mathbf{k}$ For $L_2$: $\mathbf{r}_2 = (9 + 2\lambda)\mathbf{i} + (2 + a\lambda)\mathbf{j} + (1 + 2\lambda)\mathbf{k}$ Step 2: Since the lines intersect, their position vectors at the point of intersection are equal. Equate the corresponding components. $$ 1 - 4\mu = 9 + 2\lambda \quad (1) $$ $$ 1 + \mu = 2 + a\lambda \quad (2) $$ $$ -3 = 1 + 2\lambda \quad (3) $$ Step 3: Solve equation (3) for $\lambda$. $$ -3 = 1 + 2\lambda $$ $$ 2\lambda = -4 $$ $$ \lambda = -2 $$ Step 4: Substitute $\lambda = -2$ into equation (1) to find $\mu$. $$ 1 - 4\mu = 9 + 2(-2) $$ $$ 1 - 4\mu = 9 - 4 $$ $$ 1 - 4\mu = 5 $$ $$ -4\mu = 4 $$ $$ \mu = -1 $$ Step 5: Substitute $\mu = -1$ and $\lambda = -2$ into equation (2) to find $a$. $$ 1 + (-1) = 2 + a(-2) $$ $$ 0 = 2 - 2a $$ $$ 2a = 2 $$ $$ a = 1 $$ The value of $a$ is: $$\boxed{\text{1}}$$ b) Find the position vector of the point of intersection of $L_1$ and $L_2$. Step 1: Substitute $\mu = -1$ into the equation for $L_1$. $$ \mathbf{r} = \mathbf{i} + \mathbf{j} - 3\mathbf{k} + (-1)(-4\mathbf{i} + \mathbf{j}) $$ $$ \mathbf{r} = \mathbf{i} + \mathbf{j} - 3\mathbf{k} + 4\mathbf{i} - \mathbf{j} $$ $$ \mathbf{r} = (1+4)\mathbf{i} + (1-1)\mathbf{j} - 3\mathbf{k} $$ $$ \mathbf{r} = 5\mathbf{i} + 0\mathbf{j} - 3\mathbf{k} $$ The position vector of the point of intersection is: $$\boxed{\text{5\mathbf{i} - 3\mathbf{k}}}$$ c) Find the cosine of the angle between $L_1$ and $L_2$. Step 1: Identify the direction vectors of the lines. For $L_1$, the direction vector is $\mathbf{d}_1 = -4\mathbf{i} + \mathbf{j}$. For $L_2$, the direction vector is $\mathbf{d}_2 = 2\mathbf{i} + a\mathbf{j} + 2\mathbf{k}$. Since $a=1$, $\mathbf{d}_2 = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$. Step 2: Calculate the dot product of the direction vectors. $$ \mathbf{d}_1 \cdot \mathbf{d}_2 = (-4)(2) + (1)(1) + (0)(2) $$ $$ \mathbf{d}_1 \cdot \mathbf{d}_2 = -8 + 1 + 0 = -7 $$ Step 3: Calculate the magnitudes of the direction vectors. $$ |\mathbf{d}_1| = \sqrt{(-4)^2 + (1)^2 + (0)^2} = \sqrt{16 + 1 + 0} = \sqrt{17} $$ $$ |\mathbf{d}_2| = \sqrt{(2)^2 + (1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 $$ Step 4: Use the formula for the cosine of the angle between two vectors, $\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|}$. $$ \cos\theta = \frac{-7}{\sqrt{17} \cdot 3} $$ $$ \cos\theta = \frac{-7}{3\sqrt{17}} $$ The cosine of the angle between $L_1$ and $L_2$ is: $$\boxed{\text{}\frac{-7}{3\sqrt{17}}\text{}}$$ 3 done, 2 left today. You're making progress.

Write the vector equations in component form.