What is the angle between \(a = (3i - 4j)\) and \(b = (6i + 4j)\)?
13°
87°
100°
110°
Correct answer is B
\(a . b = |a||b| \cos \theta\)
\(a = 3i - 4j; b = 6i + 4j\)
\(18 - 16 = (\sqrt{3^{2} + (-4)^{2}})(\sqrt{6^{2} + 4^{2}}) \cos \theta\)
\(2 = 5\sqrt{52} \cos \theta\)
\(\cos \theta = \frac{2}{5\sqrt{52}} = 0.0555\)
\(\theta = 86.8° \approxeq 87°\)
Simplify \((1 + 2\sqrt{3})^{2} - (1 - 2\sqrt{3})^{2}\)
0
\(8\sqrt{3}\)
13
\(2 - 4\sqrt{3}\)
Correct answer is B
\((1 + 2\sqrt{3})^{2} = 1 + 4\sqrt{3} + 12 = 13 + 4\sqrt{3}\)
\((1 - 2\sqrt{3})^{2} = 1 - 4\sqrt{3} + 12 = 13 - 4\sqrt{3}\)
\(13 + 4\sqrt{3} - (13 - 4\sqrt{3}) = 13 + 4\sqrt{3} - 13 + 4\sqrt{3}\)
= \(8\sqrt{3}\)
Find the maximum value of \(2 + \sin (\theta + 25)\).
1
2
3
4
Correct answer is C
\(\sin \theta \leq 1\)
i.e Maximum value of \(\sin \theta \forall \theta = 1\).
Therefore, \(2 + \sin \theta \leq 2 + 1 = 3\)
\(\begin{pmatrix} -14 \\ 15 \end{pmatrix} ms^{-1}\)
\(\begin{pmatrix} -2 \\ 1 \end{pmatrix} ms^{-1}\)
\(\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\)
\(\begin{pmatrix} 14 \\ -9 \end{pmatrix} ms^{-1}\)
Correct answer is C
\(u = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ms^{-1}\)
\(a = \begin{pmatrix} 3 \\ -4 \end{pmatrix} ms^{-2}; t = 3 secs\)
\(v = u + at \implies v = \begin{pmatrix} -5 \\ 3 \end{pmatrix} + \begin{pmatrix} 3 \\ -4 \end{pmatrix} \times 3\)
= \(\begin{pmatrix} -5 \\ 3 \end{pmatrix} + \begin{pmatrix} 9 \\ -12 \end{pmatrix} = \begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\)\)
5
7
8
10
Correct answer is A
No explanation has been provided for this answer.