[naturesenshi]

Can some please tell me how I can solve this and explain why?. I've looked at the marks scheme but that also didn't help much

8) The line l1 passes through points A and B with position vectors (-3i + 3j + 2k) and (7i -j +12k) respectively, relative to a fixed origin.

i) Find a vector equation for l1. [2]

The line l2 has the equation
r=(5j -7k) + t(i -2j + 7k)

The point C lies on l2 and is such that AC is perpendicular to BC.

ii) Show that one possible vector for C is (i + 3j) and find the other. [8]

Assuming that C has the position vector (i + 3j),

ii) find the area of the triangle ABC, giving your answer in the form k(5)^1/2 [3]

Thank you!!

[aznkid66]

Do you have a problem with all three parts?

i)
The vector equation of a line is r=[point]+(parameter)[direction].
So when you're given two points (for example A and B), you have a point on the line (for example A) and you can find the direction by finding the vector that connects the two points (for example vector AB).

So l1: r = (-3i + 3j + 2k) + s (10i - 4j + 10k)

ii) Perpendicular means the angle between the two vectors equals 90. You should always think perpendicular = dot product because the dot product is one of the simplest formulae that includes the angle between two vectors.

AC is perpendicular to BC if:
AC•BC=|AC||BC|cos90=0
(A-C)•(B-C)=0
(-3i+3j+2k-i-3j)•(7i-j+12k-i-3j)=0
(-4i+2k)•(6i-4j+12k)=0
-24+0+24=0 QED

Spoiler:

Show

Oh, I forgot that you have to show that C is on l2.

x=0+t=1
y=5-2t=3
z=-7+7t=0

All true when t=1, so there exists some t (t=1) such that r=C.

Let's go back a few steps, replacing the components of C with xi+yj+zk

(-3i+3j+2k-xi-yj-zk)•(7i-j+12k-xi-yj-zk)=0
((-3-x)i+(3-y)j+(2-z)k)•((7-x)i+(-1-y)j+(12-z)k)=0
(-3-x)(7-x)+(3-y)(-1-y)+(2-z)(12-z)=0

A quadratic with 3 variables...seems impossible to solve, right?
Well, have you started to wonder why they gave you l2?

l2: r = (5j-7k) + t (i-2j+7k)

Put it into parametric, and you get values for all x, y, and z on l2.

x=0+t
y=5-2t
z=-7+7t

You know that C is on l2, so its components must also follow these equations, so substitute these values in for x, y, and z. You end up with a quadratic with one variable: the parameter t.

(-3-x)(7-x)+(3-y)(-1-y)+(2-z)(12-z)=0
(-3-(0+t))(7-(0+t)) + ((3-(5-2t)(-1-(5-2t)) + (2-(-7+7t))(12-(-7+7t)) = 0
(-3-t)(7-t) + (-2+2t)(-6+2t) + (9-7t)(19-7t) = 0
-21 + -4t + t^2 + 12 + -16t + 4t^2 + 171 + -196t + 49t^2 = 0
54t^2 - 216t + 162 = 0
(54t + 162)(t-1) = 0
54(t + 3)(t-1) = 0
t = -3, 1
(Note: If you couldn't find the common factor of 54, you could use that you already know one root of t; t=1 is one of the zeros because t=1 for the given C.)

So the position vectors for C are (5j-7k) + t(i-2j+7k), t=1 and t=-3
so the position vector for the C not given is (5j-7k) + (-3)(i-2j+7k) = (5j-7k)+(-3i+6j+21k) = -3i + 11j + 14k.

iii) You should know either that the modulus of the cross product is the area of parallelogram created by two vectors as its diagonals, or know that |u cross v|=|u||v|sinC, where C is the angle between u and v.

Thus, where u and v are the sides of the triangle/parallelogram, half the modulus of (u cross v) is the area of the triangle.

Set u and v as two vectors whose tails start at one of the triangle's vertex (e.g. u = AB, v = AC) and find the cross product using the "determinant-of-a-3-by-3" method.

[aznkid66]

Cross product:

Spoiler:

Show

Oh yeah, since we know ACB is a right-angle, we can just use 1/2bh where b and h are |AC| and |BC|.
|AC|=sqrt(20)=2sqrt(5), |BC|=sqrt(196)=14

|AC||BC|=28(5^.5) => 1/2|AC||BC|=14(5^.5) => k=14

[otrivine]

(Original post by aznkid66)
Cross product:



Oh yeah, since we know ACB is a right-angle, we can just use 1/2bh where b and h are |AC| and |BC|.

for integrating like sin3xcos3x or (secx+tanx)^2*




can we use integration by parts?