Vectors (Lines)

00:00 7 marks

A line L has vector equation \(\mathbf{r} = \begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix} + \lambda \begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix}.\) Point Q has coordinates (5, 3, -4). Point P is the point on L that is closest to Q.

(a) Find the coordinates of P. [4]
(b) Find the vector product \(\overrightarrow{QP} \times \begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix}.\) [3]

Part a

6 minutes 4 marks

Question

Find the coordinates of P. [4]

Hint Hint

Hint

Parameterise point P on L as \(P=(1+2\lambda,\lambda,2-\lambda)\), form QP and set its dot product with the direction vector to zero to solve for \(\lambda\), then substitute back to find P.

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Complete Answer

P has coordinates \(\left(\frac{20}{3},\,\frac{17}{6},\,-\frac{5}{6}\right)\).

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Part b

4 minutes 3 marks

Question

Find the vector product \(\overrightarrow{QP} \times \begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix}.\) [3]

Hint Hint

Hint

Use the components of QP found in part (a) and apply the cross product determinant formula to compute each component.

Answer Answer
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Explanation Explanation

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Complete Answer

\(\overrightarrow{QP} \times \begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix} = \begin{pmatrix} -3 \\ 8 \\ 2 \end{pmatrix}.\)

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Vectors (Lines)

Part a

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Part b

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Vectors (Lines)

Part a

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Part b

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