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#.

1 Answer
May 23, 2018

#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))#.