Let |A₁| = 3, |A₂| = 5 and |A₁ + A₂| = 5. The value of (2A₁ + 3A₂) · (3A₁ - 2A₂) is:- (a) –118.5 (b) –112.5 (c) –99.5 (d) –106.5

Answer: (a). |A₁ + A₂|² = 9 + 25 + 2A₁ · A₂ = 25 ⇒ A₁ · A₂ = -4.5 (2A₁ + 3A₂) · (3A₁ - 2A₂) = 6A₁² + 5A₁ · A₂ - 6A₂² = 54 - 22.5 - 150 = -118.5 So the answer is (a) -118.5.

Let |A₁| = 3, |A₂| = 5 and |A₁ + A₂| = 5. The value of (2A₁ + 3A₂) ·… | JEE Main 2019 (8 Apr, Shift 2)

Q23JEE Main 2019 · 8 APR · SHIFT 2Difficult

Scalar (Dot) Product

Let $\left|\vec{A}_1\right| = 3, \ \left|\vec{A}_2\right| = 5$ and $\left|\vec{A}_1 + \vec{A}_2\right| = 5$ . The value of $\left(2\vec{A}_1 + 3\vec{A}_2\right) \cdot \left(3\vec{A}_1 - 2\vec{A}_2\right)$ is :-

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Answer: (a)

$|\vec{A}_1 + \vec{A}_2|^2 = 9 + 25 + 2\vec{A}_1 \cdot \vec{A}_2 = 25 \Rightarrow \vec{A}_1 \cdot \vec{A}_2 = -4.5$
$(2\vec{A}_1 + 3\vec{A}_2) \cdot (3\vec{A}_1 - 2\vec{A}_2) = 6A_1^2 + 5\vec{A}_1 \cdot \vec{A}_2 - 6A_2^2$
$= 54 - 22.5 - 150 = -118.5$
So the answer is (a) $-118.5$ .

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Scalar (Dot) Product question from JEE Main 2019 (8 Apr, Shift 2)

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