Question

If \(|\overset{⃗}{a}\cdot \overset{⃗}{b}| = |\overset{⃗}{a}\times \overset{⃗}{b}|\), \(\overset{⃗}{a}\cdot \overset{⃗}{b} < 0\) and \(θ\) is the angle between \(\overset{⃗}{a}\) and \(\overset{⃗}{b}\), then the value of \(sinθ+tanθ\) is...

• A. \(\frac{\sqrt{2}-2}{2}\)
• B. \(\frac{\sqrt{2}+2}{2}\)
• C. \(\frac{1+\sqrt{2}}{2}\)
• D. \(\frac{2-\sqrt{2}}{2}\)

Hint:

Area of a quadrilateral is half the magnitude of the cross product of its diagonals.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a quadrilateral ABCD, area \(= \frac12|\overrightarrow{AC}\times\overrightarrow{BD}|\).

Step 2: Find BD:
\(\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} = \vec b - \vec a\).

Step 3: Cross product:
\[ \overrightarrow{AC}\times\overrightarrow{BD} = (3\vec a + 2\vec b)\times(\vec b - \vec a) = 3\vec a\times\vec b - 2\vec b\times\vec a = 3\vec a\times\vec b + 2\vec a\times\vec b = 5\,\vec a\times\vec b \]
Area of ABCD = \(\frac52|\vec a\times\vec b|\). The parallelogram on AB and AD has area \(|\vec a\times\vec b|\). So \(\alpha = \frac52\).

Final Answer:
\(\alpha = \frac52\), option (B). \[ \boxed{\frac{5}{2}} \]