Question

If \(A(-2,1)\), \(B(2,3)\) and \(C(-2,-4)\) are three points. Then, the angle between \(BA\) and \(BC\) is

• A. \(\tan^{-1}\left(\frac{2}{3}\right)\)
• B. \(\tan^{-1}\left(\frac{3}{2}\right)\)
• C. \(\tan^{-1}\left(\frac{1}{3}\right)\)
• D. \(\tan^{-1}\left(\frac{1}{2}\right)\)

Hint:

Angle between vectors: \(\tan\theta=\dfrac{|x_1y_2-y_1x_2|}{x_1x_2+y_1y_2}\). This avoids calculating magnitudes and \(\cos\theta\).

Answer 1

The Correct Option is A

Step 1: Find vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\).
\[ \overrightarrow{BA}=A-B=(-2-2,\ 1-3)=(-4,-2) \] \[ \overrightarrow{BC}=C-B=(-2-2,\ -4-3)=(-4,-7) \] Step 2: Use angle between two vectors formula.
\[ \tan\theta=\left|\frac{\overrightarrow{BA}\times \overrightarrow{BC}}{\overrightarrow{BA}\cdot \overrightarrow{BC}}\right| \] Step 3: Compute dot product.
\[ \overrightarrow{BA}\cdot \overrightarrow{BC}=(-4)(-4)+(-2)(-7)=16+14=30 \] Step 4: Compute cross product magnitude in 2D.
\[ |\overrightarrow{BA}\times \overrightarrow{BC}|=|x_1y_2-y_1x_2| \] \[ =|(-4)(-7)-(-2)(-4)| =|28-8|=20 \] Step 5: Compute \(\tan\theta\).
\[ \tan\theta=\frac{20}{30}=\frac{2}{3} \] Thus:
\[ \theta=\tan^{-1}\left(\frac{2}{3}\right) \] Final Answer: \[ \boxed{\tan^{-1}\left(\frac{2}{3}\right)} \]