1992 - JAMB Mathematics Past Questions and Answers - page 3

p = \(\begin{vmatrix} 0 & 3 & 0 \ 2 & 1 & 3\ 4 & 2 & 2 \end{vmatrix}\)

P^T = \(\begin{vmatrix}0 & 2 & 4 \ 2 & 1 & 3\ 0 & 3 & 2 \end{vmatrix}\)

/p^T/ = \(\begin{vmatrix}0 & 2 & 4 \ 3 & 1 & 3\ 0 & 3 & 2 \end{vmatrix}\)

= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]

= 0 - 12 + 36 = 24

p = \(\begin{vmatrix} 0 & 3 & 0 \ 2 & 1 & 3\ 4 & 2 & 2 \end{vmatrix}\)

Q = \(\begin{vmatrix} 0 & 2 & 4 \ 3 & 1 & 2\ 0 & 3 & 2 \end{vmatrix}\) = p^T

pq = pp^T

2x - 4y + 3 = 0

Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\)

= \(\frac{4 - 12 + 3}{\sqrt{20}}\)

= \(\frac{-5}{-2\sqrt{5}}\}

= -\{\frac{\sqrt{5}{2}\)

Equation (5, 7) parallel to the line 7x + 5y = 12

5Y = -7x + 12

y = \(\frac{-7x}{5}\) + \(\frac{12}{5}\)

Gradient = \(\frac{-7}{5}\)

∴ Required equation = \(\frac{y - 7}{x - 5}\) = \(\frac{-7}{5}\) i.e. 5y - 35 = -7x + 35

5y + 7x = 70

sin \(\theta\) = \(\frac{m}{n}\) x x²

= n² - m²

x = n² - m² cos \(\theta\)

\(\frac{x}{n}\) = \(\frac{n^2 - m^2}{n}\)

cos \(\theta\) = \(\frac{n^2 - m^2}{n}\)

= \(\frac{(n + m)(n - m)}{n}\)