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 $\vec{A} \times \vec{B} = \vec{0}$ and $\vec{B} \times \vec{C} = \vec{0}$,what is the angle between $\vec{A}$ and $\vec{C}$?

The angle between the two vectors $\overrightarrow{A} = 5\hat{i} + 5\hat{j}$ and $\overrightarrow{B} = 5\hat{i} - 5\hat{j}$ will be ....... $^\circ$.

The area of the parallelogram having diagonals $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$ is:

Given two vectors $\vec{A} = 3\hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + 2\hat{k}$. Find the unit vector perpendicular to both $\vec{A}$ and $\vec{B}$.

The vector component of $\vec{a} = 2\hat{i} + 3\hat{j}$ along the direction of vector $\vec{b} = (\hat{i} + \hat{j})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module