If overrightarrow{a} = hat{i} + 2hat{k},overr | vedclass

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

Question

(A) Given $\overrightarrow{r} 	imes \overrightarrow{b} + \overrightarrow{b} 	imes \overrightarrow{c} = \overrightarrow{0}$.This can be written as $\

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(A) Given $\overrightarrow{r} 	imes \overrightarrow{b} + \overrightarrow{b} 	imes \overrightarrow{c} = \overrightarrow{0}$.This can be written as $\overrightarrow{r} 	imes \overrightarrow{b} - \overrightarrow{c} 	imes \overrightarrow{b} = \overrightarrow{0}$,which implies $(\overrightarrow{r} - \overrightarrow{c}) 	imes \overrightarrow{b} = \overrightarrow{0}$.This means $\overrightarrow{r} - \overrightarrow{c} = \lambda \overrightarrow{b}$ for some scalar $\lambda$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda \overrightarrow{b}$.Given $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,we substitute $\overrightarrow{r}$:$(\overrightarrow{c} + \lambda \overrightarrow{b}) \cdot \overrightarrow{a} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{a} + \lambda (\overrightarrow{b} \cdot \overrightarrow{a}) = 0$.Calculating the dot products:$\overrightarrow{c} \cdot \overrightarrow{a} = (7)(1) + (-3)(0) + (4)(2) = 7 + 0 + 8 = 15$.$\overrightarrow{b} \cdot \overrightarrow{a} = (1)(1) + (1)(0) + (1)(2) = 1 + 0 + 2 = 3$.Thus,$15 + \lambda(3) = 0 \Rightarrow \lambda = -5$.Now,$\overrightarrow{r} = \overrightarrow{c} - 5\overrightarrow{b} = (7\hat{i} - 3\hat{j} + 4\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} - 8\hat{j} - \hat{k}$.Finally,$\overrightarrow{r} \cdot \overrightarrow{c} = (2\hat{i} - 8\hat{j} - \hat{k}) \cdot (7\hat{i} - 3\hat{j} + 4\hat{k}) = (2)(7) + (-8)(-3) + (-1)(4) = 14 + 24 - 4 = 34$.

Similar Questions

If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$,find $\lambda$ such that $\vec{a}$ is perpendicular to $\lambda\vec{b} + \vec{c}$.

$A$ particle acted on by constant forces $4i + j - 3k$ and $3i + j - k$ is displaced from the point $i + 2j + 3k$ to the point $5i + 4j + k$. The total work done by the force is ............... $unit$.

If $\vec{a} = -4 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and $\vec{b} = \sqrt{2} \hat{i} - \sqrt{2} \hat{j}$ are two vectors,then the angle between the vectors $2 \vec{a}$ and $\frac{\vec{b}}{2}$ is (in $^{\circ}$)

If $a = i + 2j - 3k$ and $b = 3i - j + 2k,$ then the angle between the vectors $a + b$ and $a - b$ is ............... $^o$

The vectors $a, b$ and $c$ are of the same length and taken pairwise,they form equal angles. If $a = i + j$ and $b = j + k,$ then the co-ordinates of $c$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module