Obtain scalar product in terms of Cartesian component of vectors.

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3-1.Vectors

easy

Obtain scalar product in terms of Cartesian component of vectors.

Option A

Option B

Option C

Option D

Solution

$\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is written in Cartesian component as follow :

$\overrightarrow{\mathrm{A}}= \mathrm{A}_{x} \hat{i}+\mathrm{A}_{y} \hat{j}+\mathrm{A}_{z} \hat{k}$

$\overrightarrow{\mathrm{B}}=\mathrm{B}_{x} \hat{i}+\mathrm{B}_{y} \hat{j}+\mathrm{B}_{z} \hat{k}$

$\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\left(\mathrm{A}_{x} \hat{i}+\mathrm{A}_{y} \hat{j}+\mathrm{A}_{z} \hat{k}\right) \cdot\left(\mathrm{B}_{x} \hat{i}+\mathrm{B}_{y} \hat{j}+\mathrm{B}_{z} \hat{k}\right)$

$= \mathrm{A}_{x} \mathrm{~B}_{x}(\hat{i} \cdot \hat{i})+\mathrm{A}_{x} \mathrm{~B}_{y}(\hat{i} \cdot \hat{j})+\mathrm{A}_{z} \mathrm{~B}_{z}(\hat{i} \cdot \hat{k})$

$+\mathrm{A}_{y} \mathrm{~B}_{x}(\hat{j} \cdot \hat{i})+\mathrm{A}_{y} \mathrm{~B}_{y}(\hat{j} \cdot \hat{j})+\mathrm{A}_{y} \mathrm{~B}_{z}(\hat{j} \cdot \hat{k})$

$+\mathrm{A}_{z} \mathrm{~B}_{x}(\hat{k} \cdot \hat{i})+\mathrm{A}_{z} \mathrm{~B}_{y}(\hat{k} \cdot \hat{j})+\mathrm{A}_{z} \mathrm{~B}_{z}(\hat{k} \cdot \hat{k})$

In this equation $\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1$ and $\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{i}=0, \hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{j}=0$, and $\hat{k} \cdot \hat{i}=\hat{i} \cdot \hat{k}=0$

So, $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{A}_{x} \mathrm{~B}_{x}+\mathrm{A}_{y} \mathrm{~B}_{y}+\mathrm{A}_{z} \mathrm{~B}_{z}$

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(AIPMT-2004)

Obtain scalar product in terms of Cartesian component of vectors.

Solution

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