Sure, let’s determine the value of [tex]\( y \)[/tex] such that points [tex]\( P, Q, \)[/tex] and [tex]\( R \)[/tex] are collinear.
We start with the given vectors:
[tex]\[ OP = \begin{pmatrix} -2 \\ 7 \end{pmatrix} \][/tex]
[tex]\[ OQ = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \][/tex]
[tex]\[ OR = \begin{pmatrix} 5 \\ y \end{pmatrix} \][/tex]
First, we need to find vectors [tex]\( \overrightarrow{PQ} \)[/tex] and [tex]\( \overrightarrow{PR} \)[/tex]:
[tex]\[ \overrightarrow{PQ} = OQ - OP = \begin{pmatrix} 1 - (-2) \\ 1 - 7 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \end{pmatrix} \][/tex]
[tex]\[ \overrightarrow{PR} = OR - OP = \begin{pmatrix} 5 - (-2) \\ y - 7 \end{pmatrix} = \begin{pmatrix} 7 \\ y - 7 \end{pmatrix} \][/tex]
For the points [tex]\( P, Q, \)[/tex] and [tex]\( R \)[/tex] to be collinear, vectors [tex]\( \overrightarrow{PQ} \)[/tex] and [tex]\( \overrightarrow{PR} \)[/tex] must be linearly dependent. This can be checked by setting the cross product of these vectors to zero. In 2D vectors, this is equivalent to ensuring the determinant of the matrix formed by these vectors is zero.
Create the matrix:
[tex]\[
\begin{vmatrix}
3 & 7 \\
-6 & y - 7 \\
\end{vmatrix} = 0
\][/tex]
Calculate the determinant:
[tex]\[ \text{Determinant} = (3)(y - 7) - (7)(-6) = 0 \][/tex]
[tex]\[ 3(y - 7) + 42 = 0 \][/tex]
[tex]\[ 3y - 21 + 42 = 0 \][/tex]
[tex]\[ 3y + 21 = 0 \][/tex]
[tex]\[ 3y = -21 \][/tex]
[tex]\[ y = -7 \][/tex]
Therefore, the value of [tex]\( y \)[/tex] for which points [tex]\( P, Q, \)[/tex] and [tex]\( R \)[/tex] are collinear is [tex]\( \boxed{-7} \)[/tex].
