Moderate -0.8 Part (a) requires straightforward vector subtraction and magnitude calculation to solve for a constant—a routine exercise. Part (b) involves applying the parallelogram property (opposite sides equal) using basic vector addition. Both parts are standard textbook exercises requiring only direct application of fundamental vector operations with no problem-solving insight needed.
2. Relative to a fixed origin \(O\),
the point \(A\) has position vector \(( 3 \mathbf { i } - \mathbf { j } + 2 \mathbf { k } )\)
the point \(B\) has position vector ( \(\mathbf { i } + 2 \mathbf { j } - 4 \mathbf { k }\) )
and the point \(C\) has position vector \(( - \mathbf { i } + \mathbf { j } + a \mathbf { k } )\), where \(a\) is a constant and \(a > 0\).
Given that \(| \overrightarrow { B C } | = \sqrt { 41 }\)
a. show that \(a = 2\).
\(D\) is the point such that \(A B C D\) forms a parallelogram.
b. Find the position vector of \(D\).
2. Relative to a fixed origin $O$,\\
the point $A$ has position vector $( 3 \mathbf { i } - \mathbf { j } + 2 \mathbf { k } )$\\
the point $B$ has position vector ( $\mathbf { i } + 2 \mathbf { j } - 4 \mathbf { k }$ )\\
and the point $C$ has position vector $( - \mathbf { i } + \mathbf { j } + a \mathbf { k } )$, where $a$ is a constant and $a > 0$.\\
Given that $| \overrightarrow { B C } | = \sqrt { 41 }$\\
a. show that $a = 2$.\\
$D$ is the point such that $A B C D$ forms a parallelogram.\\
b. Find the position vector of $D$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q2 [5]}}