The vector equation of line $L$ is given by r $=\left(\begin{array}{c}-1\\ 3\\ 8\end{array}\right)+t\left(\begin{array}{c}4\\ 5\\ -1\end{array}\right)$.

Point P is the point on $L$ that is closest to the origin. Find the coordinates of P.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 (Distance between the origin and P)

correct position vector for OP       (A1)

eg  $\overrightarrow{\text{OP}}=\left(\begin{array}{c}-1+4t\\ 3+5t\\ 8-t\end{array}\right)$ ,  $\text{P}=\left(-1+4t\text{,}3+5t\text{,}8-t\right)$

correct expression for OP or OP² (seen anywhere)       A1

eg   $\sqrt{{\left(-1+4t\right)}^2+{\left(3+5t\right)}^2+{\left(8-t\right)}^2}$ ,  ${\left(-1+4x\right)}^2+{\left(3+5x\right)}^2+{\left(8-x\right)}^2$

valid attempt to find the minimum of OP       (M1)

eg   $d^{\prime }=0$, root on sketch of $d^{\prime }$,  min indicated on sketch of $d$

$t=-\frac{1}{14}\text{,}-0.0714285$      (A1)

substitute their value of $t$ into $L$ (only award if there is working to find $t$)       (M1)

eg   one correct coordinate,  $-1+4\left(-\frac{1}{14}\right)$

$\left(-1.28571\text{,}2.64285\text{,}8.07142\right)$

$\left(-\frac{9}{7}\text{,}\frac{37}{14}\text{,}\frac{113}{14}\right)=\left(-1.29\text{,}2.64\text{,}8.07\right)$       A1  N2

METHOD 2 (Perpendicular vectors)

recognizing that closest implies perpendicular     (M1)

eg  $\overrightarrow{\text{OP}}\mathrm{\perp }L$  (may be seen on sketch), $a\bullet b=0$

valid approach involving $\overrightarrow{\text{OP}}$       (M1)

eg   $\overrightarrow{\text{OP}}=\left(\begin{array}{c}-1+4t\\ 3+5t\\ 8-t\end{array}\right)\text{,}\left(\begin{array}{c}4\\ 5\\ -1\end{array}\right)\bullet \overrightarrow{\text{OP}}\text{,}\left(\begin{array}{c}4\\ 5\\ -1\end{array}\right)\mathrm{\perp }\overrightarrow{\text{OP}}$

correct scalar product        A1

eg   $4\left(-1+4t\right)+5\left(3+5t\right)-1\left(8-t\right)$ ,  $-4+16t+15+25t-8+t=0$,  $42t+3$

$t=-\frac{1}{14}\text{,}-0.0714285$      (A1)

substitute their value of $t$ into $L$ or $\overrightarrow{\text{OP}}$ (only award if scalar product used to find $t$)      (M1)

eg   one correct coordinate,  $-1+4\left(-\frac{1}{14}\right)$

$\left(-1.28571\text{,}2.64285\text{,}8.07142\right)$

$\left(-\frac{9}{7}\text{,}\frac{37}{14}\text{,}\frac{113}{14}\right)=\left(-1.29\text{,}2.64\text{,}8.07\right)$       A1  N2

[6 marks]