\(2\sqrt{61}\)
\(\sqrt{42}\)
\(2\sqrt{21}\)
\(2\sqrt{10}\)
Correct answer is C
Given lines \(OA\) and \(OB\) inclined at angle \(\theta\), the line \(AB\) is gotten using cosine rule.
\(|AB|^{2} = |OA|^{2} + |OB|^{2} - 2|OA||OB|\cos \theta\)
\(|AB|^{2} = 4^{2} + 5^{2} - 2(4)(5)\cos 60\)
= \(16 + 25 - 20\)
\(|AB|^{2} = 21 \implies |AB| = \sqrt{21}\)
\(\implies \text{The two particles are} \sqrt{21} m \text{apart in 1 sec}\)
In two seconds, the particles will be \(2\sqrt{21} m\) apart.
\(\frac{\sqrt{2}}{5}\)
\(5\sqrt{2}\)
\(2\sqrt{5}\)
\(\frac{5\sqrt{2}}{2}\)
Correct answer is A
\(F = F_{1} + F_{2}\)
\((2i - 5j) + (-3i + 4j) = (-i - j)\)
\(F = ma \implies (-1, -1) = 5a\)
\(a = (-\frac{1}{5}, -\frac{1}{5})\)
\(|a| = \sqrt{(\frac{-1}{5})^2 + (\frac{-1}{5})^2} = \sqrt{2}{25}\)
\(|a| = \frac{\sqrt{2}}{5} ms^{-2}\)
\(\frac{1}{24}\)
\(\frac{1}{6}\)
\(\frac{2}{13}\)
\(\frac{1}{4}\)
Correct answer is A
The number of arrangements for the 4 letters = \(^{4}P_{4} = \frac{4!}{(4 - 4)!}\)
\(4! = 24\)
Alphabetical order is just 1 of the arrangements for the letters
= \(\frac{1}{24}\)
Find the direction cosines of the vector \(4i - 3j\).
\(\frac{9}{10}, \frac{27}{10}\)
\(\frac{17}{27}, -\frac{17}{27}\)
\(\frac{4}{5}, -\frac{3}{5}\)
\(\frac{4}{7}, \frac{-3}{7}\)
Correct answer is C
Given \(V = xi +yj\), the direction cosines are \(\frac{x}{|V|}, \frac{y}{|V|}\).
\(|4i - 3j| = \sqrt{4^{2} + (-3)^{2}} = \sqrt{25} = 5\)
Direction cosines = \(\frac{4}{5}, \frac{-3}{5}\).
-2i - 10j
-2i + 2j
4i - j
4i + j
Correct answer is C
\(\overrightarrow{OA} = (3, 4)\)
\(\overrightarrow{OB} = (5, -6)\)
\(\overrightarrow{OM} = (\frac{3 + 5}{2}, \frac{4 + (-6)}{2})\)
= \((4, -1) = 4i - j\)