If the vector (hat{a} + 2hat{b}) is perpendic | vedclass

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

Question

(C) Given that the vectors $(\hat{a} + 2\hat{b})$ and $(5\hat{a} - 4\hat{b})$ are perpendicular,their dot product must be zero.$(\hat{a} + 2\hat{b}) \

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Given that the vectors $(\hat{a} + 2\hat{b})$ and $(5\hat{a} - 4\hat{b})$ are perpendicular,their dot product must be zero.$(\hat{a} + 2\hat{b}) \cdot (5\hat{a} - 4\hat{b}) = 0$Expanding the dot product:$5(\hat{a} \cdot \hat{a}) - 4(\hat{a} \cdot \hat{b}) + 10(\hat{b} \cdot \hat{a}) - 8(\hat{b} \cdot \hat{b}) = 0$Since $\hat{a}$ and $\hat{b}$ are unit vectors,$\hat{a} \cdot \hat{a} = 1$ and $\hat{b} \cdot \hat{b} = 1$. Also,$\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{a}$.$5(1) + 6(\hat{a} \cdot \hat{b}) - 8(1) = 0$$6(\hat{a} \cdot \hat{b}) - 3 = 0$$6(\hat{a} \cdot \hat{b}) = 3$$\hat{a} \cdot \hat{b} = \frac{3}{6} = \frac{1}{2}$Since $\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}| \cos 	heta$ and $|\hat{a}| = |\hat{b}| = 1$,we have:$\cos 	heta = \frac{1}{2}$$	heta = 60^\circ$

If the vector $(\hat{a} + 2\hat{b})$ is perpendicular to the vector $(5\hat{a} - 4\hat{b})$,then the angle between $\hat{a}$ and $\hat{b}$ is ........ $^\circ$.

Similar Questions

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

Obtain the scalar product of two vectors in terms of their Cartesian components.

The dot product of two mutually perpendicular vectors is:

$P$ and $Q$ are two non-zero vectors inclined to each other at an angle $\theta$. The component of $Q$ in the direction of $P$ is:

Why is $\vec{v} \times \vec{p} = 0$ for a particle moving in a straight line,and how does this relate to the angular momentum of a rotating particle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module