A girl walks \(4km\) towards west,then she walk \(3km\) in a direction \(30°\)east of north and stops.Determine the girls displacement from her initial point to departure.
Let O and B be the initial and final positions of the girl respectively.
Then,the girl's position can be shown as:
Now,we have
\(\overrightarrow{OA}=-4\hat{i}\)
\(\overrightarrow{AB}=\hat{i}|\overrightarrow{AB}|cos60°+\hat{j}|\overrightarrow{AB}|sin 60°\)
\(=\hat{i}3\times\frac{1}{2}+j^{3}\times\frac{\sqrt{3}}{2}\)
\(=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
By the triangle law of vector addition,we have:
\(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\)
\(=(-4\hat{i})+(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{3}\hat{j})\)
\(=(-4+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
\(=(-8+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
\(=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
Hence,the girl's displacement from her initial point of departure is
\(\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
If \( X \) is a random variable such that \( P(X = -2) = P(X = -1) = P(X = 2) = P(X = 1) = \frac{1}{6} \), and \( P(X = 0) = \frac{1}{3} \), then the mean of \( X \) is
List-I | List-II |
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(A) 4î − 2ĵ − 4k̂ | (I) A vector perpendicular to both î + 2ĵ + k̂ and 2î + 2ĵ + 3k̂ |
(B) 4î − 4ĵ + 2k̂ | (II) Direction ratios are −2, 1, 2 |
(C) 2î − 4ĵ + 4k̂ | (III) Angle with the vector î − 2ĵ − k̂ is cos⁻¹(1/√6) |
(D) 4î − ĵ − 2k̂ | (IV) Dot product with −2î + ĵ + 3k̂ is 10 |