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20 (a) b is a vector - OCR - GCSE Maths - Question 20 - 2018 - Paper 6
Question 20
20 (a) b is a vector. Given that b + \( \begin{pmatrix} 5 \ 2 \end{pmatrix} \) is parallel to \( \begin{pmatrix} 2 \ 1 \end{pmatrix} \), find two possible answer... show full transcript
Worked Solution & Example Answer:20 (a) b is a vector - OCR - GCSE Maths - Question 20 - 2018 - Paper 6
Step 1
Given that b + \( \begin{pmatrix} 5 \ 2 \end{pmatrix} \) is parallel to \( \begin{pmatrix} 2 \ 1 \end{pmatrix} \)
Answer
To find vector b, we note that if two vectors are parallel, one can be expressed as a scalar multiple of the other. Therefore, we can set up the equation:
( b + \begin{pmatrix} 5 \ 2 \end{pmatrix} = k \begin{pmatrix} 2 \ 1 \end{pmatrix} ) for some scalar k.
Rearranging gives:
( b = k \begin{pmatrix} 2 \ 1 \end{pmatrix} - \begin{pmatrix} 5 \ 2 \end{pmatrix} )
Calculating for k = 1 and k = -1 gives us two possible solutions for b.
- When k = 1:
( b = \begin{pmatrix} 2 \ 1 \end{pmatrix} - \begin{pmatrix} 5 \ 2 \end{pmatrix} = \begin{pmatrix} -3 \ -1 \end{pmatrix} ) - When k = -1:
( b = -\begin{pmatrix} 2 \ 1 \end{pmatrix} - \begin{pmatrix} 5 \ 2 \end{pmatrix} = \begin{pmatrix} -7 \ -3 \end{pmatrix} )
Step 2
Given that \( m \begin{pmatrix} 4 \ 1 \end{pmatrix} + n \begin{pmatrix} 5 \ 2 \end{pmatrix} = \begin{pmatrix} 12 \ 6 \end{pmatrix} \)
Answer
We can write two equations based on the components of the above vector equation:
- ( 4m + 5n = 12 )
- ( m + 2n = 6 )
We can solve these equations simultaneously. From the second equation:
( m = 6 - 2n )
Substituting into the first equation gives:
( 4(6 - 2n) + 5n = 12 )
This simplifies to:
( 24 - 8n + 5n = 12 )
( 24 - 3n = 12 )
( -3n = -12 )
So, ( n = 4 ). Substituting back to find m:
( m = 6 - 2(4) = 2 ).
Thus, we find:
m = 2 and n = 4.