Question

Vectors question?

With respect to a fixed origin `O`, the lines `l_1` and `l_2` are given by the equations:
`l_1: bb(r)=(9bb(i)+13bb(j)-3bb(k))+lambda(bb(i)+4bb(j)-2bb(k))`
`l_2: bb(r)=(2bb(i)-bb(j)+bb(k))+mu(2bb(i)+bb(j)+bb(k))`

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 `4bbi+16bbj-3bbk` and that the point `P` lies on `l_1` such that `AP` is perpendicular to `l_1`, find the exact coordinates of `P`.

Answer 1

`a.` `6 bb(i) + 13 bb(j) + 3 bb(k)`

`b.` `69.1^(circ)`

`c.` `P (frac(28)(3), frac(43)(3), - frac(11)(3))`

Explanation:

`a.`

First, let's express `l_(1)` and `l_(2)` in component form:

`l_(1): (9 + lambda) bb(i) + (13 + 4 lambda) bb(j) + (- 3 - 2 lambda) bb(k)`

`l_(2): (2 + 2 mu) bb(i) + (- 1 + mu) bb(j) + (1 + mu) bb(k)`

Then, for intersection, the individual components of `l_(1)` and `l_(2)` must be equal to each other:

`Rightarrow 9 + lambda = 2 + 2 mu " " " " " " " " (i)`

`Rightarrow 13 + 4 lambda = - 1 + mu " " " " (ii)`

`Rightarrow - 3 - 2 lambda = 1 + mu " " " " (iii)`

Let's subtract `(ii)` from `(iii)`:

`Rightarrow (- 3 - 2 lambda) - (13 + 4 lambda) = (1 + mu) - (- 1 + mu)`

`Rightarrow - 16 - 6 lamda = 2`

`Rightarrow 6 lamda = - 18`

`Rightarrow lambda = - 3`

Using `(ii)`:

`Rightarrow 13 + 4 (- 3) = - 1 + mu`

`Rightarrow 13 - 12 = - 1 + mu`

`Rightarrow 1 = - 1 + mu`

`Rightarrow mu = 2`

Now, we need to check for consistency using `(i)`:

`Rightarrow 9 + (- 3) = 2 + 2 (2)`

`Rightarrow 9 - 3 = 2 + 4`

`Rightarrow 6 = 6`

Then, let's use `l_(1)` to find the position vector of the point of intersection:

`Rightarrow "Position vector" = (9 + (-3)) bb(i) + (13 + 4 (- 3)) bb(j) + (- 3 - 2 (- 3)) bb(k)`

`therefore "Position vector" = 6 bb(i) + 13 bb(j) + 3 bb(k)`

`"`

`b.`

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

In this case, the two direction vectors are `bb(i) + 4 bb(j) - 2 bb(k)` and `2 bb(i) + bb(j) + bb(k)`.

The angle can be found using `cos(theta) = frac(A cdot B)(|A| |B|)`:

`Rightarrow cos(theta) = frac((bb(i) + 4 bb(j) - 2 bb(k)) cdot (2 bb(i) + bb(j) + bb(k)))(|bb(i) + 4 bb(j) - 2 bb(k)| |2 bb(i) + bb(j) + bb(k)|)`

`Rightarrow cos(theta) = frac((1)(2) + (4)(1) + (- 2)(1))(sqrt(1^(2) + 4^(2) + (- 2)^(2)) cdot sqrt(2^(2) + 1^(2) + 1^(2)))`

`Rightarrow cos(theta) = frac(4)(sqrt(21) cdot sqrt(6)) = frac(4)(sqrt(126))`

`Rightarrow theta = arccos(frac(4)(sqrt(126)))`

`therefore theta = 69.1^(circ)`

`"`

`c.`

Let's consider the point `P` to have coordinates `(x, y, z)`.

Then, the line `AP` will have direction vector `(x - 4) bb(i) + (y - 16) bb(j) + (z + 3) bb(k)`.

Since `AP` is perpendicular to `l_(1)`, the dot product of this direction vector and the direction vector of `l_(1)` will be equal to zero:

`Rightarrow ((x - 4) bb(i) + (y - 16) bb(j) + (z + 3) bb(k)) cdot (bb(i) + 4 bb(j) - 2 bb(k)) = 0`

`Rightarrow x - 4 + 4 y - 64 - 2 z - 6 = 0`

`Rightarrow 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:

`Rightarrow (9 + lambda) + 4 (13 + 4 lambda) - 2 (- 3 - 2 lambda) = 74`

`Rightarrow 9 + lambda + 52 + 16 lambda + 6 + 4 lambda = 74`

`Rightarrow 21 lambda = 7`

`Rightarrow lambda = frac(1)(3)`

So the coordinates are given by:

`x = 9 + (frac(1)(3)) = frac(28)(3)`

`y = 13 + 4 (frac(1)(3)) = frac(43)(3)`

`z = - 3 - 2 (frac(1)(3)) = - frac(11)(3)`

Therefore, the coordinates of the point `P` are `(frac(28)(3), frac(43)(3), - frac(11)(3))`.