With respect to a fixed origin O, the lines $ l_1 $ and $ l_2 $ are given by the equations $ l_1 : r = (9i + 13j - 3k) + \lambda (2i + 4j - 2k) $ $ l_2 : r = (2i - j + k) + \mu(2i + j + k) $ where $ \lambda $ and $ \mu $ are scalar parameters. (a) Given that $ l_1 $ and $ l_2 $ meet, find the position vector of their point of intersection. (b) Find the acute angle between $ l_1 $ and $ l_2 $ 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 $ l_1 $ such that AP is perpendicular to $ l_1 $, find the exact coordinates of P.

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With respect to a fixed origin O, the lines $ l_1 $ and $ l_2 $ are given by the equations $ l_1 : r = (9i + 13j - 3k) + \lambda (2i + 4j - 2k) $ $ l_2 : r = (2i - j + k) + \mu(2i + j + k) $ where $ \lambda $ and $ \mu $ are scalar parameters - Edexcel - A-Level Maths Pure - Question 18 - 2013 - Paper 1

Question 18

$With-respect-to-a-fixed-origin-O,-the-lines-$-l_1-$-and-$-l_2-$-are-given-by-the-equations--$-l_1-:-r-=-(9i-+-13j---3k)-+-\lambda-(2i-+-4j---2k)-$---$-l_2-:-r-=-(2i---j-+-k)-+-\mu(2i-+-j-+-k)-$--where-$-\lambda-$-and-$-\mu-$-are-scalar-parameters-Edexcel-A-Level Maths Pure-Question 18-2013-Paper 1.png$

With respect to a fixed origin O, the lines $ l_1 $ and $ l_2 $ are given by the equations $ l_1 : r = (9i + 13j - 3k) + \lambda (2i + 4j - 2k) $ \( l_2 : r... show full transcript

Worked Solution & Example Answer:With respect to a fixed origin O, the lines $ l_1 $ and $ l_2 $ are given by the equations $ l_1 : r = (9i + 13j - 3k) + \lambda (2i + 4j - 2k) $ $ l_2 : r = (2i - j + k) + \mu(2i + j + k) $ where $ \lambda $ and $ \mu $ are scalar parameters - Edexcel - A-Level Maths Pure - Question 18 - 2013 - Paper 1

Step 1

Given that $ l_1 $ and $ l_2 $ meet, find the position vector of their point of intersection.

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Answer

To find the point of intersection, we set the vector equations of ( l_1 ) and ( l_2 ) equal to each other:

[ (9i + 13j - 3k) + \lambda (2i + 4j - 2k) = (2i - j + k) + \mu(2i + j + k) ]

Expanding both sides:

[ (9 + 2\lambda)i + (13 + 4\lambda)j + (-3 - 2\lambda)k = (2 + 2\mu)i + (-1 + \mu)j + (1 + \mu)k ]

Setting the coefficients equal gives us three equations:

( 9 + 2\lambda = 2 + 2\mu )
( 13 + 4\lambda = -1 + \mu )
( -3 - 2\lambda = 1 + \mu )

From these equations, we can solve for ( \lambda ) and ( \mu ).
Solving equations (1) and (2):
From (1): [ 2\lambda - 2\mu = -7 \quad (1) ]
From (2): [ 4\lambda - \mu = -14 \quad (2) ]

Multiplying (1) by 2: [ 4\lambda - 4\mu = -14 \quad (3) ]

Subtracting (2) from (3): [ 4\mu = -7 \implies \mu = \frac{7}{4} ]

Substituting ( \mu ) back into (1) gives: [ 2\lambda - 2\left(\frac{7}{4}\right) = -7 ]
[ 2\lambda - \frac{14}{4} = -7 \implies 2\lambda = -\frac{14}{4} + 7 = \frac{14 - 14}{4} = 0 \implies \lambda = 0 ]

Substituting ( \lambda ) and ( \mu ) into either equation to find the intersection:
[ r = (9i + 13j - 3k) + 0(2i + 4j - 2k) = (9, 13, -3) ]

Thus, the position vector at the point of intersection is ( \mathbf{r} = 9i + 13j - 3k ).

Step 2

Find the acute angle between $ l_1 $ and $ l_2 $ giving your answer in degrees to 1 decimal place.

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Answer

To find the angle ( \theta ) between the two lines, we can use the direction vectors of the lines. The direction vector of ( l_1 ) is ( (2i + 4j - 2k) ) and of ( l_2 ) is ( (2i + j + k) ).

We calculate the cosine of the angle using the dot product formula:

[ \cos \theta = \frac{a \cdot b}{|a||b|} ]

where:

( a = (2, 4, -2) )
( b = (2, 1, 1) )

Calculating the dot product: [ a \cdot b = 2 \cdot 2 + 4 \cdot 1 + (-2) \cdot 1 = 4 + 4 - 2 = 6 ]

Now we find the magnitudes: [ |a| = \sqrt{2^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6} ] [ |b| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} ]

Now substituting: [ \cos \theta = \frac{6}{(2\sqrt{6})(\sqrt{6})} = \frac{6}{12} = \frac{1}{2} ]

So, ( \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60 \text{ degrees} ).

Step 3

Given that the point A has position vector $ 4i + 16j - 3k $ and that the point P lies on $ l_1 $ such that AP is perpendicular to $ l_1 $, find the exact coordinates of P.

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Answer

Let ( P = (9 + 2\lambda)i + (13 + 4\lambda)j + (-3 - 2\lambda)k ) for some scalar ( \lambda ). The vector from A to P is: [ \overrightarrow{AP} = P - A = ((9 + 2\lambda) - 4)i + ((13 + 4\lambda) - 16)j + ((-3 - 2\lambda) + 3)k ] [ = (5 + 2\lambda)i + (-3 + 4\lambda)j + (-2\lambda)k ]

Since ( AP ) is perpendicular to ( l_1 ), their dot product must be zero: [ (5 + 2\lambda)(2) + (-3 + 4\lambda)(4) + (-2\lambda)(-2) = 0 ]

Expanding gives: [ 10 + 4\lambda - 12 + 16\lambda + 4\lambda = 0 ] [ 24\lambda - 2 = 0 \implies 24\lambda = 2 \implies \lambda = \frac{1}{12} ]

Now substituting ( \lambda ) back into the expression for P: [ P = (9 + 2(\frac{1}{12}))i + (13 + 4(\frac{1}{12}))j + (-3 - 2(\frac{1}{12}))k ] [ = (9 + \frac{1}{6})i + (13 + \frac{1}{3})j + (-3 - \frac{1}{6})k ] [ = \left(\frac{54}{6} + \frac{1}{6}\right)i + \left(\frac{39}{3} + \frac{1}{3}\right)j + \left(-\frac{18}{6} - \frac{1}{6}\right)k ] [ = \frac{55}{6}i + \frac{40}{3}j - \frac{19}{6}k ]

Thus, the exact coordinates of P are ( P = \left(\frac{55}{6}, \frac{40}{3}, -\frac{19}{6}\right) ).

Given that \( l_1 \) and \( l_2 \) meet, find the position vector of their point of intersection.

Find the acute angle between \( l_1 \) and \( l_2 \) 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 \( l_1 \) such that AP is perpendicular to \( l_1 \), find the exact coordinates of P.

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