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With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = (-9i + 10k) + λ(2i + j - k) l₂: r = (3i + j + 17k) + μ(3i - j + 5k) where λ and μ are scalar parameters - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 7
Question 7
With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = (-9i + 10k) + λ(2i + j - k) l₂: r = (3i + j + 17k) + μ(3i - j + 5k) where... show full transcript
Worked Solution & Example Answer:With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = (-9i + 10k) + λ(2i + j - k) l₂: r = (3i + j + 17k) + μ(3i - j + 5k) where λ and μ are scalar parameters - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 7
Step 1
Show that l₁ and l₂ meet and find the position vector of their point of intersection.
Answer
To find if lines l₁ and l₂ intersect, we set their parametric equations equal:
For l₁: r = (-9, 0, 10) + λ(2, 1, -1) → (-9 + 2λ, 0 + λ, 10 - λ)
For l₂: r = (3, 1, 17) + μ(3, -1, 5) → (3 + 3μ, 1 - μ, 17 + 5μ)
Now, equate the components:
- -9 + 2λ = 3 + 3μ
- λ = 1 - μ
- 10 - λ = 17 + 5μ
From equation (2), we can express μ: λ = 1 - μ → μ = 1 - λ.
Substituting this into equations (1) and (3):
- -9 + 2λ = 3 + 3(1 - λ) → -9 + 2λ = 3 + 3 - 3λ → 5λ = 15 → λ = 3.
Then substitute λ = 3 back into μ: μ = 1 - 3 = -2.
Now substituting λ = 3 into l₁ gives: r = (-9i + 10k) + 3(2i + j - k) → r = (-9 + 6)i + 3j + (10 - 3)k → r = (-3i + 3j + 7k).
Hence, the point of intersection is at (-3, 3, 7).
Step 2
Show that l₁ and l₂ are perpendicular to each other.
Answer
To show that lines l₁ and l₂ are perpendicular, we examine their direction vectors:
For l₁, the direction vector is (2, 1, -1). For l₂, the direction vector is (3, -1, 5).
We compute the dot product:
d = (2, 1, -1) • (3, -1, 5) = 2 * 3 + 1 * (-1) + (-1) * 5 = 6 - 1 - 5 = 0.
Since the dot product equals 0, the lines l₁ and l₂ are indeed perpendicular.
Step 3
Show that A lies on l₁.
Answer
To show that the point A (5i + 7j + 3k) lies on l₁, we substitute A's coordinates into the equation of l₁:
Set (-9 + 2λ, 0 + λ, 10 - λ) = (5, 7, 3).
Solving for λ:
- -9 + 2λ = 5 → 2λ = 14 → λ = 7.
- λ = 7 → 0 + λ = 7, which is consistent.
- 10 - λ = 3 → λ = 7, consistent as well.
Since λ consistently gives the same result, point A lies on line l₁.
Step 4
Find the position vector of B.
Answer
To find the image of point A in line l₂ after reflection:
Let the point of intersection for line l₂ be P. We have already found P as (-3, 3, 7).
Using the formula for reflection:
the image point B is given by:
OB = OA + 2OP → OB = A + 2(P - A).
Calculating the coordinates: A = (5, 7, 3), P = (-3, 3, 7),
OB = (5, 7, 3) + 2((-3, 3, 7) - (5, 7, 3)) = (5, 7, 3) + 2((-8, -4, 4)) = (5, 7, 3) + (-16, -8, 8) = (-11, -1, 11).
Thus, the position vector of B is -11i - j + 11k.