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})