Question

If A= 2i+3j and B= 2j+3k the component of B along A is

Answer 1

Step-by-step explanation:

hope it will help thank you for

Answer 2

The component of $\vec{B}$ along $\vec{A}$ is $\displaystyle \sf{ \frac{6}{ \sqrt{13} } }$

Given :

$\vec{A} = 2 \hat{ i} + 3 \hat{j}$

$\vec{B} = 2 \hat{ j} + 3\hat{ k}$

To find :

The component of $\vec{B}$ along $\vec{A}$

Concept :

The component of $\vec{B}$ along $\vec{A}$

$= \dfrac{\vec{A} .\vec{B}}{ |\vec{A}| }$

Solution :

Step 1 of 3 :

Write down the given vectors

Here the given vectors are

$\vec{A} = 2 \hat{ i} + 3 \hat{ j}$

$\vec{B} = 2 \hat{ j} + 3\hat{ k}$

Step 2 of 3 :

Calculate dot product of the vectors

In Vector Dot product

$(\hat{ \imath}.\hat{ \imath}) = (\hat{ \jmath}.\hat{ \jmath}) = (\hat{ k}.\hat{ k}) = 1$

$(\hat{ \imath}.\hat{ \jmath}) = (\hat{ \jmath}.\hat{ \imath}) = 0$

$(\hat{ \imath}.\hat{k}) = (\hat{k}.\hat{ \imath}) = 0$

$(\hat{ \jmath}.\hat{ k}) = (\hat{k}.\hat{ \jmath}) = 0$

Thus we get

$\vec{A} .\vec{B}$

$=(2 \hat{ i} + 3 \hat{ j}).(2 \hat{ j} + 3\hat{ k})$

$\displaystyle \sf{ = 6 }$

Also

$| \vec{A}|$

$= | 2 \hat{ i} + 3\hat{ j} |$

$= \sqrt{ {2}^{2} + {3}^{2} + {0}^{2} }$

$= \sqrt{13}$

Step 3 of 3 :

Find component of $\vec{B}$ along $\vec{A}$

The component of $\vec{B}$ along $\vec{A}$

$= \dfrac{\vec{A} .\vec{B}}{ |\vec{A}| }$

$\displaystyle \sf{ = \frac{6}{ \sqrt{13} } }$

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