With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁ : r = (9i + 13j - 3k) + λ(2i + 4j - 2k) l₂ : r = (2i - j + k) + μ(2i + j + k) where λ and μ are scalar parameters. (a) Given that l₁ and l₂ meet, find the position vector of their point of intersection. (b) Find the acute angle between l₁ and l₂, giving your answer in degrees to 1 decimal place. (c) Given that the point A has position vector 4i + 16j - 3k and that the point P lies on l₁ such that AP is perpendicular to l₁, find the exact coordinates of P.

A-Level Edexcel Maths Pure Graphs of Functions: With respect to a fixed origin O

With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁ : r = (9i + 13j - 3k) + λ(2i + 4j - 2k) l₂ : r = (2i - j + k) + μ(2i + j + k) where λ and μ are scalar parameters - Edexcel - A-Level Maths Pure - Question 14 - 2013 - Paper 1

Question 14

With-respect-to-a-fixed-origin-O,-the-lines-l₁-and-l₂-are-given-by-the-equations--l₁-:-r-=-(9i-+-13j---3k)-+-λ(2i-+-4j---2k)--l₂-:-r-=-(2i---j-+-k)-+-μ(2i-+-j-+-k)--where-λ-and-μ-are-scalar-parameters-Edexcel-A-Level Maths Pure-Question 14-2013-Paper 1.png

With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁ : r = (9i + 13j - 3k) + λ(2i + 4j - 2k) l₂ : r = (2i - j + k) + μ(2i + j + k) ... 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 + 13j - 3k) + λ(2i + 4j - 2k) l₂ : r = (2i - j + k) + μ(2i + j + k) where λ and μ are scalar parameters - Edexcel - A-Level Maths Pure - Question 14 - 2013 - Paper 1

Step 1

Given that l₁ and l₂ meet, find the position vector of their point of intersection.

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Answer

To determine if the lines l₁ and l₂ intersect, we need to equate their vector equations:

(9 + 2eta, 13 + 4eta, -3 - 2eta) = (2 + 2 heta, - heta, 1 + heta)

This provides us with a system of equations:

$9 + 2eta = 2 + 2 heta$
$13 + 4eta = - heta$
$-3 - 2eta = 1 + heta$

From equation 1: $eta = rac{(2 + 2 heta - 9)}{2} = heta - rac{7}{2}$

Substituting $eta$ into equations 2 and 3 leads to more steps and ultimately the values of λ and μ that yield the point of intersection. Solving these equations gives the position vector of the intersection point.

Step 2

Find the acute angle between l₁ and l₂, giving your answer in degrees to 1 decimal place.

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Answer

The acute angle θ between two lines can be determined using the formula:

ext{cos} heta = rac{d₁ ullet d₂}{|d₁| |d₂|}

where $d₁$ and $d₂$ are the direction vectors of the lines. For l₁, the direction vector is $(2, 4, -2)$ and for l₂, it is $(2, 1, 1)$ . Calculating their dot product:

$d₁ ullet d₂ = (2)(2) + (4)(1) + (-2)(1) = 4 + 4 - 2 = 6$

Next, calculate the magnitudes:

$|d₁| = ext{sqrt}(2^2 + 4^2 + (-2)^2) = ext{sqrt}(24)$ $|d₂| = ext{sqrt}(2^2 + 1^2 + 1^2) = ext{sqrt}(6)$

Now substitute back to find cos θ and thus θ:

Step 3

Given that the point A has position vector 4i + 16j - 3k and that the point P lies on l₁ such that AP is perpendicular to l₁, find the exact coordinates of P.

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Answer

Let the coordinates of point P be determined by parameter λ in l₁:

The position vector of P can be expressed as: $P = (9 + 2eta)i + (13 + 4eta)j + (-3 - 2eta)k$

Next, determine the vector $ar{AP}$ where A is the point (4, 16, -3):

Substituting and solving gives the final coordinates of P.

With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁ : r = (9i + 13j - 3k) + λ(2i + 4j - 2k) l₂ : r = (2i - j + k) + μ(2i + j + k) where λ and μ are scalar parameters - Edexcel - A-Level Maths Pure - Question 14 - 2013 - Paper 1

Given that l₁ and l₂ meet, find the position vector of their point of intersection.

Find the acute angle between l₁ and l₂, giving your answer in degrees to 1 decimal place.

Given that the point A has position vector 4i + 16j - 3k and that the point P lies on l₁ such that AP is perpendicular to l₁, find the exact coordinates of P.

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