Four vectors \(r = \alpha i + \beta j\), where \(\alpha \text{ and } \beta\)
Four vectors \(r = \alpha i + \beta j\), where \(\alpha \text{ and } \beta\) are constants, \(s = 2i -j, m = 3i + 2j\) and \(n = i + j\) are such that the magnitude of r is three times as s and is parallel to the vactor (m - n).
(a) Find the values of \(\alpha\) and \(\beta\).
(b) Calculate the magnitude and direction of (r - s).
Explanation
(a) \(r = \alpha i + \beta j ; s = 2i - j\)
\(|r| = \sqrt{\alpha^{2} + \beta^{2}} ; |s| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}\)
\(|r| = 3|s| = 3\sqrt{5}\)
\(\implies \alpha^{2} + \beta^{2} = (3\sqrt{5})^{2} = 45 .... (1)\)
\(m = 3i + 2j ; n = i + j\)
\((m - n) = (3i + 2j) - (i + j) = 2i + j\)
\(r\) is parallel to \((m - n)\), hence, they both have the same gradient.
\(\frac{\beta}{\alpha} = \frac{1}{2} \implies \alpha = 2\beta\)
Put \(\alpha = 2\beta\) in (1),
\((2\beta)^{2} + \beta^{2} = 45 \implies 5\beta^{2} = 45\)
\(\beta^{2} = 9 \implies \beta = 3\)
\(\alpha = 2\beta = 2(3) = 6\)
\(r = 6i + 3j\).
(b) \(r = 6i + 3j ; s = 2i - j\)
\((r - s) = (6i + 3j) - (2i - j) = 4i + 4j\)
\(|r - s| = \sqrt{4^{2} + 4^{2}} = \sqrt{32} = 4\sqrt{2}\)
\(Direction : \tan \theta = \frac{4}{4} = 1\)
\(\theta = 45°\)
(r - s) makes an angle of 45° with the x- axis.