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1. The lines l and m have vector equations

r = i −2k +s(2i +j +3k) and r = 6i −5j +4k +t(i −2j +k)

respectively.

(i) Show that l and m intersect, and find the position vector of their point of intersection. [5]

(ii) Find the equation of the plane containing l and m, giving your answer in the format ax +by +cz=d.

[6]

2. Two planes have equations x +2y −2z = 2 and 2x − 3y + 6z = 3. The planes intersect in the straight

line l.

(i) Calculate the acute angle between the two planes. [4]

(ii) Find a vector equation for the line l. [6]

r = i −2k +s(2i +j +3k) and r = 6i −5j +4k +t(i −2j +k)

respectively.

(i) Show that l and m intersect, and find the position vector of their point of intersection. [5]

(ii) Find the equation of the plane containing l and m, giving your answer in the format ax +by +cz=d.

[6]

2. Two planes have equations x +2y −2z = 2 and 2x − 3y + 6z = 3. The planes intersect in the straight

line l.

(i) Calculate the acute angle between the two planes. [4]

(ii) Find a vector equation for the line l. [6]

(1i)

i - 2k + s(2i + j + 3k) = 6i - 5j + 4k + t(i - 2j + k)

<=> s(2i + j + 3k) - t(i - 2j + k) = 5i - 5j + 6k

<=> 2s - t = 5 and s + 2t = -5 and 3s - t = 6.

The three equations on the last line are satisfied by s = 1, t = -3. So the two lines intersect. The point of intersection has position vector

i - 2k + (2i + j + 3k) = 3i + j + k.

[Check: 6i - 5j + 4k -3(i - 2j + k) also equals 3i + j + k.]

(1ii)

Normal vector

n

= (2i + j + 3k) x (i - 2j + k)

= 7i + j - 5k.

The plane has equation r.n = d for some number d. The plane goes through i - 2k, so d = (7i + j - 5k).(i - 2k) = 17. So the equation of the plane is

7x + y - 5z = 17.

[Check: (6, -5, 4) and (3, 1, 1) also satisfy the equation.]

(2i)

Calculate the angle between the normal vectors.

cos(theta)

= (1, 2, -2).(2, -3, 6)/[length(1, 2, -2) * length(2, -3, 6)]

= -16/[3*7].

theta = 2.43705.

The acute angle is pi - theta = 0.704547.

(2ii)

We need two points on l.

First point. Put x = 0 and solve

2y - 2z = 2,

-3y + 6z = 3

to get y = 3, z = 2.

Second point. Put y = 0 and solve

x - 2z = 2,

2x + 6z = 3

to get x = 9/5, z = -1/10.

So (0, 3, 2) and (9/5, 0, -1/10) are points on l. A vector equation of the line through the two points is

r = 3j + 2k + s((9/5)i - 3j - (21/10)k).

[Check.]

i - 2k + s(2i + j + 3k) = 6i - 5j + 4k + t(i - 2j + k)

<=> s(2i + j + 3k) - t(i - 2j + k) = 5i - 5j + 6k

<=> 2s - t = 5 and s + 2t = -5 and 3s - t = 6.

The three equations on the last line are satisfied by s = 1, t = -3. So the two lines intersect. The point of intersection has position vector

i - 2k + (2i + j + 3k) = 3i + j + k.

[Check: 6i - 5j + 4k -3(i - 2j + k) also equals 3i + j + k.]

(1ii)

Normal vector

n

= (2i + j + 3k) x (i - 2j + k)

= 7i + j - 5k.

The plane has equation r.n = d for some number d. The plane goes through i - 2k, so d = (7i + j - 5k).(i - 2k) = 17. So the equation of the plane is

7x + y - 5z = 17.

[Check: (6, -5, 4) and (3, 1, 1) also satisfy the equation.]

(2i)

Calculate the angle between the normal vectors.

cos(theta)

= (1, 2, -2).(2, -3, 6)/[length(1, 2, -2) * length(2, -3, 6)]

= -16/[3*7].

theta = 2.43705.

The acute angle is pi - theta = 0.704547.

(2ii)

We need two points on l.

First point. Put x = 0 and solve

2y - 2z = 2,

-3y + 6z = 3

to get y = 3, z = 2.

Second point. Put y = 0 and solve

x - 2z = 2,

2x + 6z = 3

to get x = 9/5, z = -1/10.

So (0, 3, 2) and (9/5, 0, -1/10) are points on l. A vector equation of the line through the two points is

r = 3j + 2k + s((9/5)i - 3j - (21/10)k).

[Check.]

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