COMEDK · Maths · 33. Vector Algebra
If \(|\vec{a}|=2 \sqrt{2}\) and \(|\vec{b}|=3\) and angle between \(\vec{a}\) and \(\vec{b}\) is \(\dfrac{\pi}{4}\). If a parallelogram is constructed with adjacent sides \(\vec{p}=2 \vec{a}-3 \vec{b}\) and \(\vec{q}=\vec{a}+\vec{b}\) then the product of length of both the diagonals is
- A 6
- B \(60 \sqrt{2}\)
- C \(18 \sqrt{260}\)
- D \(12 \sqrt{26}\)
Answer & Solution
Correct Answer
(D) \(12 \sqrt{26}\)
Step-by-step Solution
Detailed explanation
Given \(|\vec{a}| = 2\sqrt{2}\), \(|\vec{b}| = 3\), and the angle \(\theta = \dfrac{\pi}{4}\) between \(\vec{a}\) and \(\vec{b}\). The dot product is…
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