Thread: Need help (vectors and circle geometry)!

Vectors- i and j are unit vectors- i=1 unit along the x axis j=1 unit along the y axis.

1. Given that a= 2i+3j
Find the angle it makes with the positive x axis.

2. Given that a=i + j and b= 2i - j
Find the angle each makes with the positive x axis a) 2a+b and b) 3b-a

3. A chord of a circle (x+2)^2+(y-9)^2=26 has the equation y=x+5.
Find the coordinates of the end points of the chord and hence find the equation of the perpendicular bisector of the chord. Verify that the perpendicular bisector passes through the centre of the circle.

4. The points A(1,5) and B (7,9) are the ends of a diameter of circle. Show that P (2,4) lies on the circle by showing that AP and BP are perpendicular and using the property that the angle in a semicircle is a right angle.

for question 4 I found that AP and BP are perpendicular after substituting the values of A, P and B (gradients are 1 and -1) however what else do I need to prove?

Thanks so much in advance if you can help

Originally Posted by Confuzzled?

1. Given that a= 2i+3j
Find the angle it makes with the positive x axis.

2. Given that a=i + j and b= 2i - j
Find the angle each makes with the positive x axis a) 2a+b and b) 3b-a

Consider a right triangle, you can then use the fact that the tangent of the angle is equal to the opposite side over the adjacent side. So for a general point (a,b), we have that tan(x) = b/a so x = arctan(b/a) where x is the angle you're looking for and "arctan" is the inverse tangent.

Originally Posted by Confuzzled?

3. A chord of a circle (x+2)^2+(y-9)^2=26 has the equation y=x+5.
Find the coordinates of the end points of the chord and hence find the equation of the perpendicular bisector of the chord. Verify that the perpendicular bisector passes through the centre of the circle.

Well, find the intersection points! It's a system of two equations: solve the lineair one for either y or x (it's already solved for y) and substitute that expression in the equations of the circle. You then have a quadratic equation in 1 unknown which you can solve.

To find the perp bisector, you can set up the equation of a line through a point (the point P which is (A+B)/2, the middle of A and B where A and B are the two intersection points you just found) and with a given slope. You can find the slope because for two slopes c and d, we have that cd = -1 if they're perpendicular, and you're give the slope of the line of the chord.

Originally Posted by Confuzzled?

4. The points A(1,5) and B (7,9) are the ends of a diameter of circle. Show that P (2,4) lies on the circle by showing that AP and BP are perpendicular and using the property that the angle in a semicircle is a right angle.

for question 4 I found that AP and BP are perpendicular after substituting the values of A, P and B (gradients are 1 and -1) however what else do I need to prove?

It's a bit unclear to me, are they asking to show that the angle between AP and BP is 90°?

Originally Posted by Confuzzled?

4. The points A(1,5) and B (7,9) are the ends of a diameter of circle. Show that P (2,4) lies on the circle by showing that AP and BP are perpendicular and using the property that the angle in a semicircle is a right angle.

for question 4 I found that AP and BP are perpendicular after substituting the values of A, P and B (gradients are 1 and -1) however what else do I need to prove?

All you need is the converse of:"the angle in a semicircle is a right angle",
which in this case is something like "Given a circle on diameter AB, and a point
P, then if angle APB is a right angle then P is on the circle".

Then what you have shown with this property shows that P is on the
circle.

RonL

hi thanks guys for you help
however for question 1 and 2, I have done that method yet I still can't the right answer (I know what the answers are but I don't know the exact method).

Filling in the correct numbers don't give you the answers you expect?
What are the answers then?

Vectors- i and j are unit vectors- i=1 unit along the x axis j=1 unit along the y axis.

1. Given that a= 2i+3j
Find the angle it makes with the positive x axis

.

This is quite easy... the vector i is pointing x axis direction. So all we need to to is to find dot product of a and i and divide it by |a| (the vector a length) and |i| (it is 1 )

it will be: a \dot i = (2i+3j) \dot i = 2
|a| = sqrt(2^2 + 3^2) = sqrt(13)

This way we have found cosine of the angle:

cos(alpha) = 2/sqrt(13) -----> alpha = arccos(2/sqrt(13))

2. Given that a=i + j and b= 2i - j
Find the angle each makes with the positive x axis a) 2a+b and b) 3b-a

The way will be similar (first dot product, then vectors' lengths, and in the end arccos)

a)

a \dot i = 1 |a| = sqrt(2)

so the angle with x axis: alpha = arccos(1/sqrt(2)) = pi/4 [rad] (oh, what a surprise!)

b \dot i = 2 |b| = sqrt(5)

so the angle with x axis: beta = arccos(2/sqrt(5))


b)
a \dot (2a+b) = 2|a|^2 + a \dot b = 4 + 1 = 5 |2a+b| = |4i+1j| = sqrt(17), |a| = sqrt(2)

thus the angle is: gamma = arccos(5/sqrt(34))

b \dot (2a+b) = 2(a \dot b) + |b|^2 = 2+5 = 7

the angle is: delta = arccos(7/sqrt(85))

c)

a \dot (3b-a) = 3 (a \dot b) - |a|^2 = 3 - 2 = 1 |3b-a| = |5i -4j| = sqrt(41)

the angle is epsilon = arccos(1/sqrt(82))

b \dot (3b-a) = 3 |b|^2 - a \dot b = 15 - 1 = 14

the angle is zeta = arccos(14/sqrt(205))