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Relative to a fixed origin O, the point A has position vector i − 3j + 2k and the point B has position vector −2i + 2j − k - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 6
Question 6
Relative to a fixed origin O, the point A has position vector i − 3j + 2k and the point B has position vector −2i + 2j − k. The points A and B lie on a straight line... show full transcript
Worked Solution & Example Answer:Relative to a fixed origin O, the point A has position vector i − 3j + 2k and the point B has position vector −2i + 2j − k - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 6
Step 1
Find AB.
Answer
To find the vector ( \overrightarrow{AB} ), we will subtract the position vector of A from that of B:
[ \overrightarrow{AB} = \text{Position Vector of B} - \text{Position Vector of A} = (-2i + 2j - k) - (i - 3j + 2k) ]
Calculating this gives: [ \overrightarrow{AB} = -2i + 2j - k - i + 3j - 2k = -3i + 5j - 3k ]
Thus, ( \overrightarrow{AB} = -3i + 5j - 3k ).
Step 2
Find a vector equation of l.
Answer
The vector equation of the line l can be stated in terms of a position vector and a direction vector. The direction vector can be taken as ( \overrightarrow{AB} ), which we found earlier:
[ \overrightarrow{AB} = -3i + 5j - 3k ]
Using point A as the position vector, the vector equation can be written as:
[ \text{r} = (i - 3j + 2k) + t(-3i + 5j - 3k) ] where ( t ) is a scalar parameter.
Step 3
the value of p.
Answer
Since vector ( \overrightarrow{AC} ) must be perpendicular to the direction vector ( \overrightarrow{AB} ), we can use the dot product:
The position vector of C is given as ( 2i + pj - 4k ). Thus,
[ \overrightarrow{AC} = (2i + pj - 4k) - (i - 3j + 2k) = (2 - 1)i + (p + 3)j + (-4 - 2)k = i + (p + 3)j - 6k ]
To satisfy the condition of perpendicularity:
[ \overrightarrow{AB} \cdot \overrightarrow{AC} = 0 ]
[ (-3i + 5j - 3k) \cdot (i + (p + 3)j - 6k) = -3(1) + 5(p + 3) - 3(-6) = 0 ]
Simplifying gives:
[ -3 + 5p + 15 + 18 = 0 \Rightarrow 5p + 30 = 0 \Rightarrow p = -6 ]
Step 4
the distance AC.
Answer
To find the distance ( AC ), we will first find the magnitude of ( \overrightarrow{AC} ):
Using ( p = -6 ), the position vector of C becomes:
[ C = 2i - 6j - 4k ]
Then:
[ \overrightarrow{AC} = (2i - 6j - 4k) - (i - 3j + 2k) ]
Calculating this:
[ \overrightarrow{AC} = (2 - 1)i + (-6 + 3)j + (-4 - 2)k = i - 3j - 6k ]
Now, finding the magnitude:
[ |\overrightarrow{AC}| = \sqrt{(1)^2 + (-3)^2 + (-6)^2} = \sqrt{1 + 9 + 36} = \sqrt{46}]