If vec{r} is a unit vector satisfying vec{r} | vedclass

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

Question

(D) Given, $\vec{r} 	imes \vec{a}=\vec{b}$.Taking the cross product with $\vec{a}$ on both sides: $(\vec{r} 	imes \vec{a}) 	imes \vec{a}=\vec{b}

Vedclass · Accepted Answer

$\frac{1}{4}[\vec{a}-(\vec{b} 	imes \vec{a})]$

Vedclass · Answer

(D) Given, $\vec{r} 	imes \vec{a}=\vec{b}$.Taking the cross product with $\vec{a}$ on both sides: $(\vec{r} 	imes \vec{a}) 	imes \vec{a}=\vec{b} 	imes \vec{a}$.Using the vector triple product formula $(\vec{A} 	imes \vec{B}) 	imes \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$, we get:$(\vec{r} \cdot \vec{a})\vec{a} - (\vec{a} \cdot \vec{a})\vec{r} = \vec{b} 	imes \vec{a}$.Since $|\vec{a}|=2$, we have $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 4$.Thus, $4\vec{r} = (\vec{r} \cdot \vec{a})\vec{a} - (\vec{b} 	imes \vec{a})$.Now, $|\vec{r} 	imes \vec{a}| = |\vec{b}| \Rightarrow |\vec{r}||\vec{a}|\sin 	heta = \sqrt{3}$.$1 	imes 2 \sin 	heta = \sqrt{3} \Rightarrow \sin 	heta = \frac{\sqrt{3}}{2} \Rightarrow 	heta = \frac{\pi}{3}$.Then, $\vec{r} \cdot \vec{a} = |\vec{r}||\vec{a}|\cos 	heta = 1 	imes 2 	imes \cos \frac{\pi}{3} = 2 	imes \frac{1}{2} = 1$.Substituting $\vec{r} \cdot \vec{a} = 1$ into the equation $4\vec{r} = (\vec{r} \cdot \vec{a})\vec{a} - (\vec{b} 	imes \vec{a})$:$4\vec{r} = \vec{a} - (\vec{b} 	imes \vec{a}) \Rightarrow \vec{r} = \frac{1}{4}[\vec{a} - (\vec{b} 	imes \vec{a})]$.

If $\vec{r}$ is a unit vector satisfying $\vec{r} \times \vec{a}=\vec{b}$, $|\vec{a}|=2$ and $|\vec{b}|=\sqrt{3}$, then one such $\vec{r}=$

Similar Questions

If $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$,then the value of $(2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} \times 2\vec{b})]$ is:

For a triangle $\Delta ABC$,if $\vec{BC} = \vec{a}$,$\vec{CA} = \vec{b}$,and $\vec{AB} = \vec{c}$,then which of the following is true?

If $\overrightarrow{OA} = 3i + 2j - k$ and $\overrightarrow{OB} = i + 3j + k$,then the area of the triangle $OAB$ is

Find a unit vector perpendicular to each of the vectors $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b}),$ where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}.$

If the position vectors of three points $A, B$ and $C$ are respectively $i + j + k, 2i + 3j - 4k$ and $7i + 4j + 9k$,then the unit vector perpendicular to the plane containing the triangle $ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module