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

Let $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}$,$\vec{c}=\hat{i}-3 \hat{j}-7 \hat{k}$. If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$,$\vec{r} \cdot \vec{a}=9$,$\vec{r} \cdot \vec{b}=7$,$\vec{r} \cdot \vec{c}=6$,then $(x, y, z) = $

Let the unit vectors $a$ and $b$ be perpendicular and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c = \alpha a + \beta b + \gamma (a \times b)$,then

If $a, b, c$ are three mutually perpendicular vectors such that the magnitudes of $b$ and $c$ are $1/2$ times and $\sqrt{3}/2$ times that of $a$,respectively,then the angle between the vectors $a+b+c$ and $b$ is

$A(\vec{a}), B(\vec{b}), C(\vec{c}), D(\vec{d})$ are four concyclic points,such that $x \vec{a}+y \vec{b}+z \vec{c}+t \vec{d}=\vec{0}$ and $x+y+z+t=0$,where $x, y, z, t$ are constants not all zero. If the chords $AB$ and $CD$ intersect at $P$,then:

Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module