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With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = (6 \- λ, -3 + 2λ, -2)ᵀ l₂: r = (–5 \- 15μ, 3 + 2μ, 1)ᵀ; where λ and μ are scalar parameters - Edexcel - A-Level Maths: Pure - Question 8 - 2011 - Paper 5
Question 8
With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = (6 \- λ, -3 + 2λ, -2)ᵀ l₂: r = (–5 \- 15μ, 3 + 2μ, 1)ᵀ; where λ and μ ar... 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 = (6 \- λ, -3 + 2λ, -2)ᵀ l₂: r = (–5 \- 15μ, 3 + 2μ, 1)ᵀ; where λ and μ are scalar parameters - Edexcel - A-Level Maths: Pure - Question 8 - 2011 - Paper 5
Step 1
Show that l₁ and l₂ meet and find the position vector of their point of intersection A.
Answer
To determine if lines l₁ and l₂ intersect, we equate their parametric equations:
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For the x-components: [ 6 - \lambda = -5 - 15\mu ] This rewrites to: [ \lambda + 15\mu = 11 \quad (1) ]
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For the y-components: [ -3 + 2\lambda = 3 + 2\mu ] This rewrites to: [ 2\lambda - 2\mu = 6 \quad (2) ]
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For the z-components: [ -2 = 1 ] Since this gives no valid equation, we can conclude that the y-component leads to two solutions for (\lambda) and (\mu): Using equations (1) and (2) we can solve simultaneously: From (2): [ \lambda - \mu = 3 \quad (3) ] Solving (1) and (3):
- Using (1): [ \lambda + 15(\lambda - 3) = 11 ] [ \lambda + 15\lambda - 45 = 11 ] [ 16\lambda = 11 + 45 \rightarrow \lambda = 3 ] Substituting (\lambda = 3) back into (3): [ 3 - \mu = 3 \rightarrow \mu = 0 ]
Thus, substituting these values into either line equation gives the intersection point A:
[ \text{Using l₁: } \ r = \left(6 - 3, -3 + 2(3), -2\right) = \left(3, 3, -2\right)]
Therefore the coordinates of point A are (3, 3, -2).
Step 2
Find, to the nearest 0.1°, the acute angle between l₁ and l₂.
Answer
To find the acute angle (\theta) between the two lines, we use the direction vectors of the lines:
- Direction vector of l₁: ( (-1, 2, 0) ))
- Direction vector of l₂: ( (-15, 2, 0) ))
The cosine of the angle can be found using the dot product formula:
[ \cos \theta = \frac{u \cdot v}{||u|| ||v||} ]
Where:
- (u = (-1, 2, 0))
- (v = (-15, 2, 0))
Calculating the dot product: [ u \cdot v = -1(-15) + 2(2) + 0(0) = 15 + 4 = 19 ]
Calculating the magnitudes: [ ||u|| = \sqrt{(-1)² + (2)² + (0)²} = \sqrt{5} ] [ ||v|| = \sqrt{(-15)² + (2)² + (0)²} = \sqrt{229} ]
Now substituting back into the cosine formula:
[ \cos \theta = \frac{19}{\sqrt{5}\cdot \sqrt{229}} = 0.872]
Thus,
[ \theta = \cos^{-1}(0.872) = 29.0^{\circ} ]
Rounding to the nearest 0.1° gives an acute angle of 69.1°.
Step 3
Show that B lies on l₁.
Answer
To show that point B with position vector ( (5, -1) ) lies on line l₁, we substitute x and y values into the parametric equations of l₁:
The equation of l₁ is given as: [ r = (6 - \lambda, -3 + 2\lambda, -2)^(t) ]
Using the components to find (\lambda):
- From the x-component: [ 6 - \lambda = 5 \implies \lambda = 1 ]
- Substitute (\lambda = 1) in the y-component: [ -3 + 2(1) = -1 \implies \text{{This is valid.}} ]
Since both coordinates of B satisfy the equations of l₁, we conclude that B lies on l₁.
Step 4
Find the shortest distance from B to the line l₂, giving your answer to 3 significant figures.
Answer
To calculate the shortest distance from point B to line l₂, we first need the formula for the distance (d): [ d = |\overrightarrow{AB} \cdot \hat{n}| ] Where (\hat{n}) is the unit normal vector of line l₂.
- Determine point A as the intersection of l₁ and l₂ which we found earlier as (A(3, 3, -2)).
- The vector (\overrightarrow{AB}) from point A to B is: [ \overrightarrow{AB} = B - A = (5 - 3, -1 - 3, 0 + 2) = (2, -4, 2)]
- The direction vector for line l₂ is ( (-15, 2, 0) ). Normalize this vector: [ ||(-15, 2, 0)|| = \sqrt{(-15)² + (2)²} = \sqrt{229} \rightarrow \hat{n} = \left(-\frac{15}{\sqrt{229}}, \frac{2}{\sqrt{229}}, 0 \right) ]
- Now applying the formula: [ d = |\overrightarrow{AB} \cdot \hat{n}| = |(2, -4, 2) \cdot \left(-\frac{15}{\sqrt{229}}, \frac{2}{\sqrt{229}}, 0 \right)| = \frac{|(-30) + (-8)|}{\sqrt{229}} = \frac{38}{\sqrt{229}}]
- Calculating gives: [ d \approx 2.5 ] to 3 significant figures.