# Vectors question?

## With respect to a fixed origin $O$, the lines ${l}_{1}$ and ${l}_{2}$ are given by the equations: ${l}_{1} : \boldsymbol{r} = \left(9 \boldsymbol{i} + 13 \boldsymbol{j} - 3 \boldsymbol{k}\right) + \lambda \left(\boldsymbol{i} + 4 \boldsymbol{j} - 2 \boldsymbol{k}\right)$ ${l}_{2} : \boldsymbol{r} = \left(2 \boldsymbol{i} - \boldsymbol{j} + \boldsymbol{k}\right) + \mu \left(2 \boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k}\right)$ where $\lambda$ and $\mu$ are scalar parameters. a. Given ${l}_{1}$ and ${l}_{2}$ meet, find the position vector of their point of intersection. b. Find the acute angle between ${l}_{1}$ and ${l}_{2}$, giving your answer in degrees to 1 decimal place. c. Given that the point $A$ has position vector $4 \boldsymbol{i} + 16 \boldsymbol{j} - 3 \boldsymbol{k}$ and that the point $P$ lies on ${l}_{1}$ such that $A P$ is perpendicular to ${l}_{1}$, find the exact coordinates of $P$.

May 23, 2018

$a .$ $6 \boldsymbol{i} + 13 \boldsymbol{j} + 3 \boldsymbol{k}$

$b .$ ${69.1}^{\circ}$

$c .$ $P \left(\frac{28}{3} , \frac{43}{3} , - \frac{11}{3}\right)$

#### Explanation:

$a .$

First, let's express ${l}_{1}$ and ${l}_{2}$ in component form:

${l}_{1} : \left(9 + \lambda\right) \boldsymbol{i} + \left(13 + 4 \lambda\right) \boldsymbol{j} + \left(- 3 - 2 \lambda\right) \boldsymbol{k}$

${l}_{2} : \left(2 + 2 \mu\right) \boldsymbol{i} + \left(- 1 + \mu\right) \boldsymbol{j} + \left(1 + \mu\right) \boldsymbol{k}$

Then, for intersection, the individual components of ${l}_{1}$ and ${l}_{2}$ must be equal to each other:

$R i g h t a r r o w 9 + \lambda = 2 + 2 \mu \text{ " " " " " " } \left(i\right)$

$R i g h t a r r o w 13 + 4 \lambda = - 1 + \mu \text{ " " } \left(i i\right)$

$R i g h t a r r o w - 3 - 2 \lambda = 1 + \mu \text{ " " } \left(i i i\right)$

Let's subtract $\left(i i\right)$ from $\left(i i i\right)$:

$R i g h t a r r o w \left(- 3 - 2 \lambda\right) - \left(13 + 4 \lambda\right) = \left(1 + \mu\right) - \left(- 1 + \mu\right)$

$R i g h t a r r o w - 16 - 6 l a m \mathrm{da} = 2$

$R i g h t a r r o w 6 l a m \mathrm{da} = - 18$

$R i g h t a r r o w \lambda = - 3$

Using $\left(i i\right)$:

$R i g h t a r r o w 13 + 4 \left(- 3\right) = - 1 + \mu$

$R i g h t a r r o w 13 - 12 = - 1 + \mu$

$R i g h t a r r o w 1 = - 1 + \mu$

$R i g h t a r r o w \mu = 2$

Now, we need to check for consistency using $\left(i\right)$:

$R i g h t a r r o w 9 + \left(- 3\right) = 2 + 2 \left(2\right)$

$R i g h t a r r o w 9 - 3 = 2 + 4$

$R i g h t a r r o w 6 = 6$

Then, let's use ${l}_{1}$ to find the position vector of the point of intersection:

$R i g h t a r r o w \text{Position vector} = \left(9 + \left(- 3\right)\right) \boldsymbol{i} + \left(13 + 4 \left(- 3\right)\right) \boldsymbol{j} + \left(- 3 - 2 \left(- 3\right)\right) \boldsymbol{k}$

$\therefore \text{Position vector} = 6 \boldsymbol{i} + 13 \boldsymbol{j} + 3 \boldsymbol{k}$

"

$b .$

The angle between two lines is the angle between their direction vectors.

In this case, the two direction vectors are $\boldsymbol{i} + 4 \boldsymbol{j} - 2 \boldsymbol{k}$ and $2 \boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k}$.

The angle can be found using $\cos \left(\theta\right) = \frac{A \cdot B}{| A | | B |}$:

$R i g h t a r r o w \cos \left(\theta\right) = \frac{\left(\boldsymbol{i} + 4 \boldsymbol{j} - 2 \boldsymbol{k}\right) \cdot \left(2 \boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k}\right)}{| \boldsymbol{i} + 4 \boldsymbol{j} - 2 \boldsymbol{k} | | 2 \boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k} |}$

$R i g h t a r r o w \cos \left(\theta\right) = \frac{\left(1\right) \left(2\right) + \left(4\right) \left(1\right) + \left(- 2\right) \left(1\right)}{\sqrt{{1}^{2} + {4}^{2} + {\left(- 2\right)}^{2}} \cdot \sqrt{{2}^{2} + {1}^{2} + {1}^{2}}}$

$R i g h t a r r o w \cos \left(\theta\right) = \frac{4}{\sqrt{21} \cdot \sqrt{6}} = \frac{4}{\sqrt{126}}$

$R i g h t a r r o w \theta = \arccos \left(\frac{4}{\sqrt{126}}\right)$

$\therefore \theta = {69.1}^{\circ}$

"

$c .$

Let's consider the point $P$ to have coordinates $\left(x , y , z\right)$.

Then, the line $A P$ will have direction vector $\left(x - 4\right) \boldsymbol{i} + \left(y - 16\right) \boldsymbol{j} + \left(z + 3\right) \boldsymbol{k}$.

Since $A P$ is perpendicular to ${l}_{1}$, the dot product of this direction vector and the direction vector of ${l}_{1}$ will be equal to zero:

$R i g h t a r r o w \left(\left(x - 4\right) \boldsymbol{i} + \left(y - 16\right) \boldsymbol{j} + \left(z + 3\right) \boldsymbol{k}\right) \cdot \left(\boldsymbol{i} + 4 \boldsymbol{j} - 2 \boldsymbol{k}\right) = 0$

$R i g h t a r r o w x - 4 + 4 y - 64 - 2 z - 6 = 0$

$R i g h t a r r o w x + 4 y - 2 z = 74$

Also, we have that:

$x = 9 + \lambda$

$y = 13 + 4 \lambda$

$z = - 3 - 2 \lambda$

Substituting these into the above equation we get:

$R i g h t a r r o w \left(9 + \lambda\right) + 4 \left(13 + 4 \lambda\right) - 2 \left(- 3 - 2 \lambda\right) = 74$

$R i g h t a r r o w 9 + \lambda + 52 + 16 \lambda + 6 + 4 \lambda = 74$

$R i g h t a r r o w 21 \lambda = 7$

$R i g h t a r r o w \lambda = \frac{1}{3}$

So the coordinates are given by:

$x = 9 + \left(\frac{1}{3}\right) = \frac{28}{3}$

$y = 13 + 4 \left(\frac{1}{3}\right) = \frac{43}{3}$

$z = - 3 - 2 \left(\frac{1}{3}\right) = - \frac{11}{3}$

Therefore, the coordinates of the point $P$ are $\left(\frac{28}{3} , \frac{43}{3} , - \frac{11}{3}\right)$.