Question
There are two vectors 〈1, -4〉 and 〈2, k〉 where k is an unknown quantity.
(A) Find a value of k such that the vectors are orthogonal.
(B) Find a value of k such that the vectors are parallel.
(C) If k = 6, find the angle between the two vectors. Round to the nearest tenth of a degree.
Answer/Explanation
(A) If vectors are orthogonal, the angle between them is \(\frac{\pi }{2}\) so that their dot product is defined by \(r_{1}r_{2}\cos \Theta =0\) Then \(r_{1}r_{2}=x_{1}x_{2}+y_{1}y_{2}=\left \langle 1,-4 \right \rangle.\left \langle 2,k \right \rangle=1(2)+(-4)(k)=0\).Then \(-4k=-2\Rightarrow k=\frac{1}{2}\)
(B)If vectors are parallel, the angle between them is 0, where cos(0) = 1. You know that, in the case of parallel vectors,\(r_2=Cr_1\) , where C is a constant. In this case, the horizontal component of \(r_{2}\) differs from \(r_{1}\) by a factor of 2. Multiplying the y-component by the same factor, you get k = -8. You can get the same result by applying the equation \(\cos \Theta =\frac{r_{1}.r_{2}}{\left \| r_{1} \right \|\left \| r_{2} \right \|}=-1\). You find that \(\frac{2-4k}{(\sqrt{17})\sqrt{4+k^{2}}}=-1\). If you multiply through by the denominator and then square both sides, you find that
\(4-16k+16k^{2}=17(4+k^{2})\) and, combining terms, you get \(k^{2}+16k+64=0\). You can then factor to find that (k + 8)(k + 8) = 0 or k = -8.
(C) Find the angle between two vectors by applying the equation \(\Theta =\cos ^{-1}\frac{r_1.r_2}{\left \| r_{1} \right \|\left \| r_{2} \right \|}.\)Here,\(r_{1}.r_{2}=1(2)-4(6)=-22,\left \| r_{1} \right \|=\sqrt{1^{2}+4^{2}}=\sqrt{17}\) and \( \left \| r_{2} \right \|=\sqrt{2^{2}+6^{2}}=\sqrt{40}=2\sqrt{10}.\)Substituting, we find that \(\Theta =-\frac{22}{2\sqrt{170}}\approx 147.5^{\circ}\)