(a) Given that (x = 3i – j, y = 2i + kj) and the cosine of the angle between x and y is (frac{sqrt{5}}{5}), find the values of the constant k.
(b) In the quadrilateral ABCD,
(overrightarrow{AB} = begin{pmatrix} -5 \ -1 end{pmatrix}, overrightarrow{AC} = begin{pmatrix} -6 \ -9 end{pmatrix})
and (overrightarrow{BD} = begin{pmatrix} 4 \ -7 end{pmatrix}). Show whether or not ABCD is a parallelogram.
Explanation
(a) (x = 3i – j ; y = 2i + kj)
(x cdot y = |x||y| cos theta)
((3i – j) cdot (2i + kj) = (sqrt{3^{2} + (-1)^{2}})(sqrt{2^{2} + k^{2}}) cos theta)
(6 – k = (sqrt{10})(sqrt{4 + k^{2}})(frac{sqrt{5}}{5}))
(5(6 – k) = sqrt{10(4 + k^{2})(5)})
(30 – 5k = sqrt{200 + 50k^{2}})
((30 – 5k)^{2} = 200 + 50k^{2})
(900 – 300k + 25k^{2} = 200 + 50k^{2})
(900 – 200 – 300k + 25k^{2} – 50k^{2} = 0)
(25k^{2} + 300k – 700 = 0 implies k^{2} + 12k – 28 = 0)
(k^{2} – 2k + 14k – 28 = 0 implies k(k – 2) + 14(k – 2) = 0)
(text{k = 2 or -14}).
(b)
ABCD is a parallelogram if AD = BC.
(overrightarrow{AD} = overrightarrow{AB} + overrightarrow{BD})
= (begin{pmatrix} -5 \ -1 end{pmatrix} + begin{pmatrix} 4 \ -7 end{pmatrix})
= (begin{pmatrix} -1 \ -8 end{pmatrix})
(overrightarrow{BC} = overrightarrow{BA} + overrightarrow{AC})
= (begin{pmatrix} 5 \ 1 end{pmatrix} + begin{pmatrix} -6 \ -9 end{pmatrix})
= (begin{pmatrix} -1 \ -8 end{pmatrix}).
(overrightarrow{AD} = overrightarrow{BC}). ABCD is a parallelogram.