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1) State a vector equation of the line passing through the points A and B whose position vectors are i-j+3k and i+2j+2k respectively. Determine the position vector of the point C which divides the line segment AB internally such that AC=2CB.

2) The point P lies on the line which is parallel to the vector 2i+j-k and which passes through the point with position vector i+j+2k. The point Q lies on another line which is parallel to the vector i+j-2k and which passes through the point with position vector i+j+4k. The line PQ is perpendicular to both these lines. Find a vector equation of the line PQ and the coordinates of the mid-point of PQ.

3) The position vectors of three points A, B and C with respect to a fixed origin O are

(2i-2j+k), (4i+2j+k) and (i+j+3k)

respectively.

a) Write down unit vectors in the direction of the lines CA and CB and calculate the size of ACB in degrees, to 1 decimal place.

The mid-point of AB is M.

b) Find a vector equation of the straight line passing through C and M.

c) Show that AB and CM are perpendicular.

d) Find the position vector of the point N on the line CM such that ONC=90 degrees.

3) The lines L1 and L2 intersect at the point B. Given that the equations of the lines are

L1: r = (2+3t)i + (3+4t)j + (4+2t)k

L2: r = (1+2s)i + (1+3s)j + (4s-2)k

where t and s are scalar parameters, find

a) the coordinates of B.

b) the acute angle, in degrees to 1 decimal place, between the lines.

4) Referred to an origin O, the points A, B, C and D are (3,1,-1), (6,7,8), (2,5,0) and (0,7,-2) respectively.

a) Find the vector equations for AB and CD.

b) Show that AB and CD intersect at the point (4,3,2).

c) Calculate the size of ACD to the nearest degree.

Answers:

1) r = 25i/11 + 25j/11 + 16k/11 + @(-i+3j+k).

26i/11 + 22j/11 + 15k/11.

2a) (1/rt14)(i-3j-2k), (1/rt14)(3i+j-2k), 73.4 degrees.

b) r = i+j+3k + @(2i-j-2k).

d) (19i/9 + 4j/9 + 17k/9).

3a) (5,7,6).

b) 26.3 degrees.

4a) Equation of AB is r=3i+j-k + @(i+2j+3k)

Equation of CD is r=2i+5j+ lambda(-i+j-k).

c) 123 degrees.

Cheers

2) The point P lies on the line which is parallel to the vector 2i+j-k and which passes through the point with position vector i+j+2k. The point Q lies on another line which is parallel to the vector i+j-2k and which passes through the point with position vector i+j+4k. The line PQ is perpendicular to both these lines. Find a vector equation of the line PQ and the coordinates of the mid-point of PQ.

3) The position vectors of three points A, B and C with respect to a fixed origin O are

(2i-2j+k), (4i+2j+k) and (i+j+3k)

respectively.

a) Write down unit vectors in the direction of the lines CA and CB and calculate the size of ACB in degrees, to 1 decimal place.

The mid-point of AB is M.

b) Find a vector equation of the straight line passing through C and M.

c) Show that AB and CM are perpendicular.

d) Find the position vector of the point N on the line CM such that ONC=90 degrees.

3) The lines L1 and L2 intersect at the point B. Given that the equations of the lines are

L1: r = (2+3t)i + (3+4t)j + (4+2t)k

L2: r = (1+2s)i + (1+3s)j + (4s-2)k

where t and s are scalar parameters, find

a) the coordinates of B.

b) the acute angle, in degrees to 1 decimal place, between the lines.

4) Referred to an origin O, the points A, B, C and D are (3,1,-1), (6,7,8), (2,5,0) and (0,7,-2) respectively.

a) Find the vector equations for AB and CD.

b) Show that AB and CD intersect at the point (4,3,2).

c) Calculate the size of ACD to the nearest degree.

Answers:

1) r = 25i/11 + 25j/11 + 16k/11 + @(-i+3j+k).

26i/11 + 22j/11 + 15k/11.

2a) (1/rt14)(i-3j-2k), (1/rt14)(3i+j-2k), 73.4 degrees.

b) r = i+j+3k + @(2i-j-2k).

d) (19i/9 + 4j/9 + 17k/9).

3a) (5,7,6).

b) 26.3 degrees.

4a) Equation of AB is r=3i+j-k + @(i+2j+3k)

Equation of CD is r=2i+5j+ lambda(-i+j-k).

c) 123 degrees.

Cheers

Are you sure about those answers ??

1)

A(1,-1,3)

B(1,2,2)

AB = (0,3,-1)

l: r = OA + t.AB

r = (1, -1, 3) + t(0,3,-1)

===================

c(x,y,z)

C is 2/3 of theway along AB

C - A = (2/3)(B - A) = (2/3)BA

x - 1 = (2/3)(0) --> x = 0

y + 1 = (2/3)(3) --> y = 1

z + 3 = (2/3)(-1) --> z = -7/3

C = (1, 1, -7/3)

============

1)

A(1,-1,3)

B(1,2,2)

AB = (0,3,-1)

l: r = OA + t.AB

r = (1, -1, 3) + t(0,3,-1)

===================

c(x,y,z)

C is 2/3 of theway along AB

C - A = (2/3)(B - A) = (2/3)BA

x - 1 = (2/3)(0) --> x = 0

y + 1 = (2/3)(3) --> y = 1

z + 3 = (2/3)(-1) --> z = -7/3

C = (1, 1, -7/3)

============

3.

a)

Solve the simultaneous equations:

2+3t = 1+2s

3+4t = 1+3s

To verify that these lines intersect, plug in the values you got for t and s in:

4+2t = 4s-2

Now substitute these values in either L1 or L2.

b)

Find the direction vectors of both lines:

L1: r = (2+3t)i + (3+4t)j + (4+2t)k = (2, 3, 4) + t(3, 4, 2) -- direction vector = a = (3, 4, 2).

L2: r = (1+2s)i + (1+3s)j + (4s-2)k = (1, 1, -2) + s(2, 3, 4) -- direction vector = b = (2, 3, 4).

Now:

a . b = |a| |b| cosθ

(3, 4, 2) . (2, 3, 4) = (3² + 4² + 2²)^{0.5} (2² + 3² + 4²)^{0.5} cosθ

6 + 12 + 8 = 29 cosθ

cosθ = 26/29

θ = 26.3

4.

a)

AB: r = A + t(B-A)

CD: r = C + s(D-C)

b) & c) are the same as question 3 above.

a)

Solve the simultaneous equations:

2+3t = 1+2s

3+4t = 1+3s

To verify that these lines intersect, plug in the values you got for t and s in:

4+2t = 4s-2

Now substitute these values in either L1 or L2.

b)

Find the direction vectors of both lines:

L1: r = (2+3t)i + (3+4t)j + (4+2t)k = (2, 3, 4) + t(3, 4, 2) -- direction vector = a = (3, 4, 2).

L2: r = (1+2s)i + (1+3s)j + (4s-2)k = (1, 1, -2) + s(2, 3, 4) -- direction vector = b = (2, 3, 4).

Now:

a . b = |a| |b| cosθ

(3, 4, 2) . (2, 3, 4) = (3² + 4² + 2²)

6 + 12 + 8 = 29 cosθ

cosθ = 26/29

θ = 26.3

4.

a)

AB: r = A + t(B-A)

CD: r = C + s(D-C)

b) & c) are the same as question 3 above.

3/ ( the other one)

A(2,-2,1), B(4,2,1), C(1,1,3)

a)

CA = (1,-3,-2), |CA| = √(1 + 9 + 4) = √14

CB = (3,1,-2), |CB| = √(9 + 1 + 4) = √14

unit vectors are:

(1/√14)(i-3j-2k),

(1/√14)(3i+j-2k),

=============

CA.CB = (1,-3,-2).(3,1,-2) = 3 - 3 + 4 = 4

cosθ = CA.CB / |CA||CB| = 4 / 14 = 2/7

θ = 73.4 deg

======

b)

M = ½(B + A)

M = (3,0,1)

C = (1,1,3)

CM = M - C = (2,-1,-2)

r = C + t.CM

r = (1,1,3) + t.(2,-1,-2)

==================

c)

AB = B - A = (4-2,2+2,1-1) = (2,4,-2)

AB.CM = (2,4,0).(2,-1,-2) = 4 - 4 + 0 = 0

:. AB and CM are perpindicular

=======================

d)

N lies on CM, so

N = (1+2t, 1-t, 3-2t) for some value of t.

ON.r = 0

(1+2t, 1-t, 3-2t).(2,-1,-2)

= 2 + 4t - 1 + t - 6 + 4t

= -5 + 9t = 0

t = 5/9

=====

N = (1/9)(19, 4, 17)

===============

A(2,-2,1), B(4,2,1), C(1,1,3)

a)

CA = (1,-3,-2), |CA| = √(1 + 9 + 4) = √14

CB = (3,1,-2), |CB| = √(9 + 1 + 4) = √14

unit vectors are:

(1/√14)(i-3j-2k),

(1/√14)(3i+j-2k),

=============

CA.CB = (1,-3,-2).(3,1,-2) = 3 - 3 + 4 = 4

cosθ = CA.CB / |CA||CB| = 4 / 14 = 2/7

θ = 73.4 deg

======

b)

M = ½(B + A)

M = (3,0,1)

C = (1,1,3)

CM = M - C = (2,-1,-2)

r = C + t.CM

r = (1,1,3) + t.(2,-1,-2)

==================

c)

AB = B - A = (4-2,2+2,1-1) = (2,4,-2)

AB.CM = (2,4,0).(2,-1,-2) = 4 - 4 + 0 = 0

:. AB and CM are perpindicular

=======================

d)

N lies on CM, so

N = (1+2t, 1-t, 3-2t) for some value of t.

ON.r = 0

(1+2t, 1-t, 3-2t).(2,-1,-2)

= 2 + 4t - 1 + t - 6 + 4t

= -5 + 9t = 0

t = 5/9

=====

N = (1/9)(19, 4, 17)

===============

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