#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))#.