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Vector m = \( \begin{pmatrix} 2 \\ k \end{pmatrix} \) and vector n = \( \begin{pmatrix} 3 \\ 11 \end{pmatrix} \) - OCR - GCSE Maths - Question 20 - 2020 - Paper 6
Question 20
Vector m = \( \begin{pmatrix} 2 \\ k \end{pmatrix} \) and vector n = \( \begin{pmatrix} 3 \\ 11 \end{pmatrix} \). Vector 2m + n is parallel to \( \begin{pmatrix} 1 ... show full transcript
Worked Solution & Example Answer:Vector m = \( \begin{pmatrix} 2 \\ k \end{pmatrix} \) and vector n = \( \begin{pmatrix} 3 \\ 11 \end{pmatrix} \) - OCR - GCSE Maths - Question 20 - 2020 - Paper 6
Step 1
Find 2m + n
Answer
First, calculate the vector ( 2m ):
[ 2m = 2 \times \begin{pmatrix} 2 \ k \end{pmatrix} = \begin{pmatrix} 4 \ 2k \end{pmatrix} ]
Next, add vector n:
[ 2m + n = \begin{pmatrix} 4 \ 2k \end{pmatrix} + \begin{pmatrix} 3 \ 11 \end{pmatrix} = \begin{pmatrix} 4 + 3 \ 2k + 11 \end{pmatrix} = \begin{pmatrix} 7 \ 2k + 11 \end{pmatrix} ]
Step 2
Set up the parallel condition
Answer
Since vector ( 2m + n ) is parallel to ( \begin{pmatrix} 1 \ -1 \end{pmatrix} ), there exists a scalar ( k ) such that:
[ \begin{pmatrix} 7 \ 2k + 11 \end{pmatrix} = k \begin{pmatrix} 1 \ -1 \end{pmatrix} ]
This gives us two equations:
- ( 7 = k )
- ( 2k + 11 = -k )
Step 3
Solve for k
Answer
From the first equation we have:
[ k = 7 ]
Substituting ( k = 7 ) into the second equation:
[ 2(7) + 11 = -7 ] [ 14 + 11 = -7 ] [ 25 = -7 ] \text{ (this is incorrect).}
Thus, we return to check the second equation: [ 2k + 11 = -k ] [ 3k + 11 = 0 ] [ 3k = -11 ] [ k = -\frac{11}{3} \text{ (invalid due to inconsistency).} } ]
Step 4