2015 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

By KS Teng

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
\theta = \mathrm{cos^{-1}} |\frac{\begin{pmatrix}2\\3\\{-6}\end{pmatrix}\cdot \begin{pmatrix}1\\0\\{0}\end{pmatrix}}{\sqrt{49} \cdot 1}| = 73.4^{\circ}

(ii)
Let \vec{ON} be point on L that makes \sqrt{33} from P.

\vec{ON} = \begin{pmatrix}1\\{-2}\\{-4}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix} for some \lambda

\vec{PN} = \begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix}

|\vec{PN}| = \sqrt{(-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2}

33 = 49 \lambda^2 - 70 \lambda + 54

\lambda = 1 \mathrm{~or~} \frac{3}{7}

\vec{ON} = \begin{pmatrix}3\\{1}\\{-10}\end{pmatrix} \mathrm{~or~} \frac{1}{7}\begin{pmatrix}13\\{-5}\\{-46}\end{pmatrix}

L = |\vec{PN}|^2 = (-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2 = 49 \lambda^2 - 70 \lambda +54

\frac{dL}{d\lambda} = 98 \lambda - 70

\frac{d^2L}{d\lambda ^2} = 98 > 0

So when \lambda = \frac{70}{98} = \frac{5}{7}, L is minimum.

\vec{ON} = \frac{1}{7} \begin{pmatrix}{17}\\{1}\\{-58}\end{pmatrix}

(iii)
\begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} \times \begin{pmatrix}{2}\\{3}\\{-6}\end{pmatrix} = \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix}

\begin{pmatrix}{1}\\{-2}\\{-4}\end{pmatrix} \bullet \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix} = -4

\therefore, \pi : 36x - 2y +11z=-4

KS Comments:

Leave a Comment

Contact Us

CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

Not readable? Change text. captcha txt
0

Start typing and press Enter to search

2015 A-level H2 Mathematics (9740) Paper 2 Question 1 Suggested SolutionsJC Mathematics
2015 A-level H2 Mathematics (9740) Paper 2 Question 3 Suggested SolutionsJC Mathematics