Let overrightarrow{a} = 2hat{i} - 7hat{j} + 5 | vedclass

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

Question

(A) Given $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j}

Vedclass · Accepted Answer

$\frac{11}{7} \sqrt{2}$

Vedclass · Answer

(A) Given $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j} - 3\hat{k}$.From the condition $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{c} 	imes \overrightarrow{a}$,we have $(\overrightarrow{r} - \overrightarrow{c}) 	imes \overrightarrow{a} = 0$.This implies that $\overrightarrow{r} - \overrightarrow{c}$ is parallel to $\overrightarrow{a}$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda\overrightarrow{a}$ for some scalar $\lambda$.Given $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,we substitute $\overrightarrow{r}$:$(\overrightarrow{c} + \lambda\overrightarrow{a}) \cdot \overrightarrow{b} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{b} + \lambda(\overrightarrow{a} \cdot \overrightarrow{b}) = 0$.Calculate the dot products:$\overrightarrow{c} \cdot \overrightarrow{b} = (1)(1) + (2)(0) + (-3)(1) = 1 - 3 = -2$.$\overrightarrow{a} \cdot \overrightarrow{b} = (2)(1) + (-7)(0) + (5)(1) = 2 + 5 = 7$.Substituting these values: $-2 + 7\lambda = 0 \Rightarrow \lambda = \frac{2}{7}$.Now,$\overrightarrow{r} = \overrightarrow{c} + \frac{2}{7}\overrightarrow{a} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \frac{2}{7}(2\hat{i} - 7\hat{j} + 5\hat{k}) = (1 + \frac{4}{7})\hat{i} + (2 - 2)\hat{j} + (-3 + \frac{10}{7})\hat{k} = \frac{11}{7}\hat{i} - \frac{11}{7}\hat{k}$.Finally,$|\overrightarrow{r}| = \sqrt{(\frac{11}{7})^2 + 0^2 + (-\frac{11}{7})^2} = \sqrt{2 	imes (\frac{11}{7})^2} = \frac{11}{7}\sqrt{2}$.

Similar Questions

If $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$,then the value of $|\vec{b}|$ is

The equation of the perpendicular bisector of the line segment joining the points whose position vectors are $a$ and $b$ respectively is

Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:

If vectors $\vec{a}$ and $\vec{b}$ have the same magnitude, the angle between them is $60^\circ$, and their scalar product is $1/2$, then $|\vec{a}|$ is:

Let $a, b$ and $c$ be unit vectors such that $a$ is perpendicular to the plane containing $b$ and $c$ and the angle between $b$ and $c$ is $\frac{\pi}{3}$. Then,$|a+b+c|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module