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The line \( l_1 \) has vector equation \( \, r = 8i + 12j + 14k + \lambda(i + j - k) \), where \( \lambda \) is a parameter - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 7
Question 7
The line \( l_1 \) has vector equation \( \, r = 8i + 12j + 14k + \lambda(i + j - k) \), where \( \lambda \) is a parameter. The point A has coordinates (4, 8, ... show full transcript
Worked Solution & Example Answer:The line \( l_1 \) has vector equation \( \, r = 8i + 12j + 14k + \lambda(i + j - k) \), where \( \lambda \) is a parameter - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 7
Step 1
(a) Find the values of a and b.
Answer
To find the values of ( a ) and ( b ), we need to express the coordinates of points A and B in terms of ( \lambda ). The coordinates from the vector equation can be written as:
- For point A:
( 8 + \lambda ) for the x-coordinate,
( 12 + \lambda ) for the y-coordinate,
( 14 - \lambda ) for the z-coordinate. - For point A, we have ( A(4, 8, a) ):
This gives us the following equations: [ 8 + \lambda = 4 ]
[ 12 + \lambda = 8 ]
[ 14 - \lambda = a ]
Solving the first equation:
[ \lambda = 4 - 8 ]
[ \lambda = -4 ]
Solving the second equation:
[\lambda = 8 - 12 ]
[\lambda = -4 ]
Plugging ( \lambda = -4 ) into the equation for A to find ( a ):
[ a = 14 - (-4) = 14 + 4 = 18. ]
Next, substitute ( \lambda = -4 ) into the coordinate equation for B:
[ b = 8 + \lambda = 8 - 4 = 4, ]
[ b = 9 .]
Thus, ( a = 18 ) and ( b = 9. )
Step 2
(b) find the coordinates of P.
Answer
To find the coordinates of P, we need to determine at what value of ( \lambda ) the vector ( OP ) is perpendicular to the direction vector of line ( l_1 ).
The direction vector of ( l_1 ) can be identified as ( (1, 1, -1) ). Set the coordinates of point P as ( P(8 + \lambda, 12 + \lambda, 14 - \lambda) ).
The vector ( OP ) is given by: [ OP = (8 + \lambda, 12 + \lambda, 14 - \lambda) - (0, 0, 0) = (8 + \lambda, 12 + \lambda, 14 - \lambda) ]
The dot product must equal zero for the vectors to be perpendicular:
[ (8 + \lambda, 12 + \lambda, 14 - \lambda) \cdot (1, 1, -1) = 0. ]
Expanding this gives:
[ (8 + \lambda) + (12 + \lambda) - (14 - \lambda) = 0 ]
[ 8 + \lambda + 12 + \lambda - 14 + \lambda = 0. ]
So,
[ 3\lambda + 6 = 0 ]
[ 3\lambda = -6 ]
[ \lambda = -2. ]
Now substituting ( \lambda = -2 ) back into the coordinates of P gives: [ P = (8 - 2, 12 - 2, 14 + 2) = (6, 10, 16).]
Step 3
(c) Hence find the distance OP, giving your answer as a simplified surd.
Answer
To find the distance ( OP ), we can use the distance formula:
[ OP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} ]
Substituting the coordinates of O (0, 0, 0) and P (6, 10, 16):
[ OP = \sqrt{(6 - 0)^2 + (10 - 0)^2 + (16 - 0)^2} ]
[ = \sqrt{6^2 + 10^2 + 16^2} ]
[ = \sqrt{36 + 100 + 256} ]
[ = \sqrt{392} ]
[ = 14\sqrt{2}.]