Let |vec{A}_1| = 3,|vec{A}_2| = 5,and |vec{A} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: $|\vec{A}_1| = 3$,$|\vec{A}_2| = 5$,and $|\vec{A}_1 + \vec{A}_2| = 5$.Using the property $|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2

Vedclass · Accepted Answer

$-118$

Vedclass · Answer

(C) Given: $|\vec{A}_1| = 3$,$|\vec{A}_2| = 5$,and $|\vec{A}_1 + \vec{A}_2| = 5$.Using the property $|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta$:$5^2 = 3^2 + 5^2 + 2(3)(5) \cos 	heta$$25 = 9 + 25 + 30 \cos 	heta$$30 \cos 	heta = -9 \implies \vec{A}_1 \cdot \vec{A}_2 = |\vec{A}_1||\vec{A}_2| \cos 	heta = 3 	imes 5 	imes (-9/30) = -4.5$.Now,calculate the dot product:$(2\vec{A}_1 + 3\vec{A}_2) \cdot (3\vec{A}_1 - 2\vec{A}_2) = 6|\vec{A}_1|^2 - 4\vec{A}_1 \cdot \vec{A}_2 + 9\vec{A}_1 \cdot \vec{A}_2 - 6|\vec{A}_2|^2$$= 6|\vec{A}_1|^2 + 5(\vec{A}_1 \cdot \vec{A}_2) - 6|\vec{A}_2|^2$$= 6(3^2) + 5(-4.5) - 6(5^2)$$= 6(9) - 22.5 - 6(25)$$= 54 - 22.5 - 150 = -118.5$.

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

Similar Questions

Let $\vec{A} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = 4 \hat{i} + \hat{j} + 2 \hat{k}$,then $|\vec{A} \times \vec{B}|$ is equal to:

The unit vector perpendicular to the two vectors $(2\hat{i} + 3\hat{j} + \hat{k})$ and $(\hat{i} - \hat{j} + 2\hat{k})$ is:

If $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{C} + \overrightarrow{D},$ then select the correct alternative-

If $\vec{A}$ and $\vec{B}$ are vectors,which of the following statements is incorrect?

The two vectors have magnitudes $3$ and $5$. If the angle between them is $60^o$,then the dot product of the two vectors will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module