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Relative to a fixed origin O, the point A has position vector (10i + 2j + 3k), and the point B has position vector (8i + 3j + 4k) - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 8
Question 2
Relative to a fixed origin O, the point A has position vector (10i + 2j + 3k), and the point B has position vector (8i + 3j + 4k). The line l passes through the poi... show full transcript
Worked Solution & Example Answer:Relative to a fixed origin O, the point A has position vector (10i + 2j + 3k), and the point B has position vector (8i + 3j + 4k) - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 8
Step 1
Find the vector AB.
Answer
To find the vector (ar{AB}), we calculate the difference between the position vectors of points B and A.
Let:
- Position Vector of A: (\bar{A} = (10, 2, 3))
- Position Vector of B: (\bar{B} = (8, 3, 4))
The calculation is: [ \bar{AB} = \bar{B} - \bar{A} = (8, 3, 4) - (10, 2, 3) = (-2, 1, 1) ]
Thus, the vector (\bar{AB} = -2i + 1j + 1k.
Step 2
Find a vector equation for the line l.
Answer
The line l passing through points A and B can be expressed in vector form as: [ r(t) = \bar{A} + t \bar{AB} ] where (t) is a scalar parameter, and (\bar{AB}) is the direction vector.
Substituting in the known values: [ r(t) = (10, 2, 3) + t(-2, 1, 1) ] Simplifying this gives: [ r(t) = (10 - 2t, 2 + t, 3 + t) ]
Step 3
Find the position vector of the point P.
Answer
The vector (\bar{CP}) can be expressed as: [ \bar{CP} = \bar{P} - \bar{C} ] Given (\bar{C} = (3, 12, 3)) and knowing (\bar{P}) is given by the parameterization of line l: [ \bar{P} = (10 - 2t, 2 + t, 3 + t) ]
Since (\bar{CP}) is perpendicular to (l), the dot product (\bar{CP} \cdot \bar{AB} = 0):
Substituting:
[
\bar{CP} = [(10 - 2t - 3), (2 + t - 12), (3 + t - 3)]
]
This leads to:
[
(7 - 2t)(-2) + (-10 + t)(1) + (t)(1) = 0
]
Simplifying and solving:
[
-14 + 4t - 10 + t + t = 0
= -10 + 6t = 0
= t = \frac{5}{3}
]
Finally, substituting (t = \frac{5}{3}) back into the equation for (\bar{P}):
[
\bar{P} = \left(10 - 2\left(\frac{5}{3}\right), 2 + \frac{5}{3}, 3 + \frac{5}{3}\right) = \left(\frac{20}{3}, \frac{11}{3}, \frac{14}{3}\right)
]
So the position vector of point P is (\left(\frac{20}{3}i + \frac{11}{3}j + \frac{14}{3}k\right).