The position vectors of points A, B and C with respect to the origin are (8i – 2j), (2i + 6j) and (-10i + 4j) respectively. If ABCN is a parallelogram, find :
(a) the position vector of N;
(b) AN and AB ;
(c) correct to two decimal place, the acute angle between AN and AB.
Explanation
(a)
Position vector of A is (8i – 10j)
Position vector of B is (2i + 6j)
Position vector of C is (-10i + 4j).
Let position vector of N be (ai + bj).
ABCN is a parallelogram.
(therefore overrightarrow{AB} = overrightarrow{NC}).
(begin{bmatrix} (2 – 8)i \ (6 + 10)j end{bmatrix} = begin{bmatrix} (-10 – a)i \ (4 – b)j end{bmatrix})
Equating components,
(2 – 8 = -10 – a implies a = -4)
(6 + 10 = 4 – b implies b = -12)
The position vector of N is (-4i – 12j).
(b) (overrightarrow{AN} cdot overrightarrow{AB})
(overrightarrow{AN} = bar{A} – bar{N} )
= ((8i – 10j) – (-4i – 12j))
= (12i + 2j)
(overrightarrow{AB} = bar{A} – bar{B})
= ((8i – 10j) – (2i + 6j))
= (6i – 16j)
(therefore overrightarrow{AN} cdot overrightarrow{AB} = (12i + 2j) cdot (6i – 16j))
= (72 – 32 = 40)
(c) Let (theta) be the acute angle between (overrightarrow{AN}) and (overrightarrow{AB}).
(overrightarrow{AN} cdot overrightarrow{AB} = |AN||AB| cos theta)
(|AN| = |12i + 2j| = sqrt{12^{2} + 2^{2}} = sqrt{144 + 4} = sqrt{148})
(|AB| = |6i – 16j| = sqrt{6^{2} + (-16)^{2}} = sqrt{36 + 256} = sqrt{292})
(therefore 40 = (sqrt{148})(sqrt{292}) cos theta)
(cos theta = frac{40}{(sqrt{148})(sqrt{292})})
(cos theta = frac{40}{207.88} = 0.1924)
(theta = cos^{-1} (0.1924) = 78.907°)
(approxeq 78.9°)