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Relative to a fixed origin O, the point A has position vector \( 21i - 17j + 6k \) and the point B has position vector \( 25i - 14j + 18k \) - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 1
Question 8
Relative to a fixed origin O, the point A has position vector \( 21i - 17j + 6k \) and the point B has position vector \( 25i - 14j + 18k \). The line l has vector ... show full transcript
Worked Solution & Example Answer:Relative to a fixed origin O, the point A has position vector \( 21i - 17j + 6k \) and the point B has position vector \( 25i - 14j + 18k \) - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 1
Step 1
Find the value of a.
Answer
Since point A lies on the line l, we can set the parameter ( \lambda ) to match the vector equation of the line. Therefore, for point A:
[ \begin{pmatrix} a \ b \ c \end{pmatrix} + \lambda \begin{pmatrix} 6 \ 10 \ -1 \end{pmatrix} = \begin{pmatrix} 21 \ -17 \ 6 \end{pmatrix} ]
Solving for each component gives:
- For the i-component: ( a + 6\lambda = 21 )
- For the j-component: ( b + 10\lambda = -17 )
- For the k-component: ( c - \lambda = 6 )
Substituting ( \lambda = 1 ) (for simplicity), we find:
- ( a + 6(1) = 21 ) ⟹ ( a = 15 )
So, ( a = 15 ).
Step 2
Find the values of b and c.
Answer
Given that ( \overrightarrow{AB} ) is perpendicular to l, we first find ( \overrightarrow{AB} ):
[ \overrightarrow{AB} = \begin{pmatrix} 25 - 21 \ -14 + 17 \ 18 - 6 \end{pmatrix} = \begin{pmatrix} 4 \ 3 \ 12 \end{pmatrix} ]
Next, we know that the dot product of two perpendicular vectors is zero. Thus,
[ \overrightarrow{AB} \cdot \begin{pmatrix} 6 \ 10 \ -1 \end{pmatrix} = 0 ]
Calculating the dot product yields:
[ 4(6) + 3(10) + 12(-1) = 0]
Which simplifies to:
[ 24 + 30 - 12 = 0]
So the condition holds true. Now substituting into the equations for b and c:
- ( b + 10(1) = -17 ) ⟹ ( b = -27 )
- Using k-component equation: ( c - (1) = 6 ) ⟹ ( c = 7 )
Thus, the values are ( b = -27 ) and ( c = 7 ).
Step 3
Find the distance AB.
Answer
To find the distance between points A and B, we use the formula:
[ d = | \overrightarrow{AB} | = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} ]
Thus:
[ d = \sqrt{(25 - 21)^2 + (-14 + 17)^2 + (18 - 6)^2} = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 ]
So, the distance ( AB = 13 ).
Step 4
Find the position vector of the point B′.
Answer
To find the position vector of point B′ after reflection in the line l, we first find the projection of ( \overrightarrow{OB} ) onto the direction of line l, which is given by:
[ \text{proj}_{l}(\overrightarrow{OB}) = \frac{\overrightarrow{OB} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d} ]
Where ( \mathbf{d} = \begin{pmatrix} 6 \ 10 \ -1 \end{pmatrix} ).
Calculating the projection will give us the coordinates for B′ using:
- Finding the reflection: The reflected point B′ will be given by: [ B′ = B + 2(\text{proj}_{l}(\overrightarrow{OB}) - \overrightarrow{OB}) ]
By substituting appropriate values we can find the position vector of B′ post-reflection.