122k views
0 votes
9 a The vectors p and q are such that p - 11i-24j and q = 2i + aj. 1 i Find the value of each of the constants a and such that p + 2q = (a + B)i-20j. ii Using the values of a and ß found in part i, find the unit vector in the direction P + 2q. B The points A and B have position vectors a and b with respect to an origin O. The point C lies on AB and is such that AB: AC is 1:A. Find an expression for OC in terms of a, b and 2. [3] [2] [3] c The points S and Thave position vectors s and t with respect to an origin O. The points O, S and 7 do not lie in a straight line. Given that the vector 2s + ut is parallel to the vector (+3)s +9t where is a positive constant, find the value of μ. Cambridge IGCSE Additional Mathematics 0606 Paper 22 Q10 Mar 2016 [3] 335​

9 a The vectors p and q are such that p - 11i-24j and q = 2i + aj. 1 i Find the value-example-1
User Slipset
by
8.3k points

1 Answer

4 votes

Answer:

Therefore, the value of μ is 6.

Explanation:

a)

i) To find the value of constants a and b, we'll equate the components of p + 2q to the given expression (a + b)i - 20j.

From p - 11i - 24j, we can determine that the i-component of p is -11 and the j-component is -24.

From q = 2i + aj, we can determine that the i-component of q is 2 and the j-component is a.

Now, let's calculate the components of p + 2q:

(i-component) = (-11) + 2(2) = -11 + 4 = -7

(j-component) = (-24) + 2(a) = -24 + 2a

We can set these components equal to the given expression:

-7 = a + b

-24 + 2a = -20

Solving the second equation, we have:

2a = -20 + 24

2a = 4

a = 2

Substituting the value of a in the first equation, we get:

-7 = 2 + b

b = -9

Therefore, the values of a and b are a = 2 and b = -9.

ii) Using the values of a = 2 and b = -9, we can find the unit vector in the direction of p + 2q.

The magnitude of p + 2q is:

√[(-7)^2 + (-24 + 2(2))^2] = √[49 + (-24 + 4)^2] = √[49 + (-20)^2] = √[49 + 400] = √449

The unit vector in the direction of p + 2q is:

[(a + b)i - 20j] / √449

= [(2 - 9)i - 20j] / √449

= (-7i - 20j) / √449

b)

The position vector of C is given by:

OC = OA + AC

Since AB:AC is 1:A, we can express AC as (1/A)AB.

Therefore, OC = OA + (1/A)AB.

c)

Given that the vector 2s + ut is parallel to the vector (+3)s + 9t, we can set up an equation of proportionality:

2s + ut = k((+3)s + 9t)

where k is the positive constant.

Expanding the equation, we get:

2s + ut = 3ks + 9kt

Comparing the coefficients of s and t, we have:

2 = 3k

u = 9k

From the first equation, we can solve for k:

3k = 2

k = 2/3

Substituting the value of k in the second equation, we get:

u = 9(2/3)

u = 6

Therefore, the value of μ is 6.

User Mhopeng
by
8.4k points