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A2 Vector Questions
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]
Reply 1
(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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