Direction Of A Vector Formula With Solved Examples - Byju's
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\[\LARGE \theta =\tan^{-1}\frac{y}{x}\]
If (\(\begin{array}{l}x_{1}\end{array} \),\(\begin{array}{l}y_{1}\end{array} \) ) is the starting point and the ending point is (\(\begin{array}{l}x_{2}\end{array} \),\(\begin{array}{l}y_{2}\end{array} \) ), then the formula for direction is:
\(\begin{array}{l}\LARGE \theta =\tan^{-1}\frac{(y_{2}-y_{1})}{(x_{2}-x_{1})}\end{array} \)Solved Example
Question:
Find the direction of the vector \(\begin{array}{l}\overrightarrow{PQ}\end{array} \) whose initial point P is at (5, 2) and the endpoint is at Q is at (4, 3).
Solution:
Given \(\begin{array}{l}(x_{1}\end{array} \), \(\begin{array}{l}y_{1})\end{array} \) = (5, 2)
\(\begin{array}{l}(x_{2}\end{array} \), \(\begin{array}{l}y_{2})\end{array} \) = (4, 3)
According to the formula we have,
\(\begin{array}{l}\theta\end{array} \) = \(\begin{array}{l}tan^{-1}\end{array} \) \(\begin{array}{l}\frac{(y_{2} – y_{1})}{(x_{2} – x_{1})}\end{array} \) \(\begin{array}{l}\theta\end{array} \) = \(\begin{array}{l}tan^{-1}\end{array} \) \(\begin{array}{l}\frac{(3-2)}{(4-5)}\end{array} \)θ = tan-1(-1)
θ = -45° or θ = 135°
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