Find the unit vector in the direction of (-5i + 12j).
\(\frac{1}{13}(-5i - 12j)\)
\(\frac{1}{13}(5i - 12j)\)
\(\frac{1}{13}(-5i + 12j)\)
\(\frac{1}{13}(5i + 12j)\)
Correct answer is C
The unit vector \(\hat{n} = \frac{\overrightarrow{r}}{|r|}\)
\(\hat{n} = \frac{-5i + 12j}{\sqrt{(-5)^{2} + (12)^{2}} \)
= \(\frac{-5i + 12j}{13} \)
-8
-6
-4
-3
Correct answer is C
\(f : x \to x^{2} - x - 6\); \(g : x \to x - 1\)
\(g(3) = 3 - 1 = 2\)
\(f(g(3)) = f(2) = 2^{2} - 2 - 6 = 4 - 2 - 6 = -4\)
Given that \(q = 9i + 6j\) and \(r = 4i - 6j\), which of the following statements is true?
r and q are collinear
r and q are perpendicular
The magnitude of r is 52 units
The projection of r on q is \(\sqrt{117}\) units.
Correct answer is B
The dot product of two perpendicular forces = 0
\((9i + 6j).(4i - 6j) = 36 - 36 = 0\)
Hence, r and q are perpendicular.
\(4 ms^{-2}\)
\(6 ms^{-2}\)
\(8 ms^{-2}\)
\(10 ms^{-2}\)
Correct answer is D
\(v(t) = (3t^{2} - 2t) ms^{-1}\)
\(a(t) = \frac{\mathrm d v}{\mathrm d t} = (6t - 2) ms^{-2}\)
\(a(2) = 6(2) - 2 = 12 - 2 = 10 ms^{-2}\)
2m
4m
6m
8m
Correct answer is B
\(v(t) = (3t^{2} - 2t) ms^{-1}\)
\(s(t) = \int v(t) \mathrm {d} t\)
= \(\int (3t^{2} - 2t) \mathrm {d} t = t^{3} - t^{2}\)
\(s(2) = 2^{3} - 2^{2} = 8 - 4 = 4m\)