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The components of $\vec a = 2\hat i + 3\hat j$ along the direction of vector $\left( {\hat i + \hat j} \right)$ is
$\left( {\hat i + \hat j} \right)$
$\frac{1}{{2\,}}\,\left( {\hat i + \hat j} \right)$
$\frac{5}{\sqrt{2}}\,\left( {\hat i + \hat j} \right)$
$\frac{5}{\sqrt{2}}\,\left( {\hat i - \hat j} \right)$
Solution
The vector component of $\mathrm{A}$ in the direction of $\mathrm{B}$ is $A \cos \theta=\frac{5}{\sqrt{2}}$
Vector $\mathrm{A}=2 \hat{i}+3 \hat{j}$
Vector $\mathrm{B}=\hat{i}+\hat{j}$
We know that,
The vector component of $A$ in the direction of $B$ is
$\vec{A} \cdot \vec{B}=|A||B| \cos \theta$
$A \cos \theta=\frac{\vec{A} \cdot \vec{B}}{|B|}$
$A \cos \theta=\frac{2 \hat{\imath}+3 \hat{\jmath} \cdot \hat{\imath}+\hat{\jmath}}{\sqrt{2}}$
$A \cos \theta=\frac{5}{\sqrt{2}}$
Hence, The vector component of $A$ in the direction of $B$ is $A \cos \theta=\frac{5}{\sqrt{2}}$
Similar Questions
If $\left| {\vec A } \right|\, = \,2$ and $\left| {\vec B } \right|\, = \,4$ then match the relation in Column $-I$ with the angle $\theta $ between $\vec A$ and $\vec B$ in Column $-II$.
| Column $-I$ | Column $-II$ |
| $(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$ | $(i)$ $\theta = \,{0^o}$ |
| $(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$ | $(ii)$ $\theta = \,{90^o}$ |
| $(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$ | $(iii)$ $\theta = \,{180^o}$ |
| $(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$ | $(iv)$ $\theta = \,{60^o}$ |
