If overrightarrow{a} cdot hat{i} = overrighta | vedclass

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

Question

(D) Let $\overrightarrow{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Given $\overrightarrow{a} \cdot \hat{i} = 1$,we have $(a_1 \hat{i} + a_2 \hat{j

Vedclass · Accepted Answer

$\frac{1}{3}(3 \hat{i} - 3 \hat{j} + \hat{k})$

Vedclass · Answer

(D) Let $\overrightarrow{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Given $\overrightarrow{a} \cdot \hat{i} = 1$,we have $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot \hat{i} = 1$,which implies $a_1 = 1$.Next,$\overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = 1$,so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (2 \hat{i} + \hat{j}) = 1$.This gives $2a_1 + a_2 = 1$. Substituting $a_1 = 1$,we get $2(1) + a_2 = 1$,so $a_2 = -1$.Finally,$\overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$.This gives $a_1 + a_2 + 3a_3 = 1$. Substituting $a_1 = 1$ and $a_2 = -1$,we get $1 - 1 + 3a_3 = 1$,so $3a_3 = 1$,which means $a_3 = \frac{1}{3}$.Thus,$\overrightarrow{a} = \hat{i} - \hat{j} + \frac{1}{3} \hat{k} = \frac{1}{3}(3 \hat{i} - 3 \hat{j} + \hat{k})$.

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to :

Similar Questions

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is:

If the force $\overrightarrow{F} = \hat{i} + 2\hat{j} + 3\hat{k}$ moves a particle from position $\vec{r_1} = \hat{i} + \hat{j} - \hat{k}$ to $\vec{r_2} = 2\hat{i} - \hat{j} + \hat{k},$ then the work done is:

The component of $\vec{i} + \vec{j}$ along $\vec{j} + \vec{k}$ is:

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 $\lambda\vec{b} + \vec{c}$.

Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $7 \hat{i}-\hat{j}+8 \hat{k}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module