Question

If A→=i^+j^+k^ and B→=−i^−j^−k^. Then angle made by (A→−B→) with A→ is

Answer 1

SOLUTION

GIVEN

$\vec{A} = \hat{i} + \hat{j} + \hat{k} \:$

$\vec{B} = \hat{i} - \hat{j} - \hat{k} \:$

TO DETERMINE

The angle made by $\vec{A} - \vec{B}$ with $\vec{A}$

EVALUATION

Here the given vectors are

$\vec{A} = \hat{i} + \hat{j} + \hat{k} \:$

$\vec{B} = \hat{i} - \hat{j} - \hat{k} \:$

Now

$\vec{A} - \vec{B}$

$= ( \hat{i} + \hat{j} + \hat{k} ) - ( \hat{i} - \hat{j} - \hat{k} \: )$

$= 2\hat{j} + 2 \hat{k}$

Now

$( \vec{A} - \vec{B} ). \vec{A}$

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

= 0 + 2 + 2

= 4

Let θ be the required angle

So we get

$\displaystyle \cos \theta = \frac{( \vec{A} - \vec{B} ). \vec{A} }{ | ( \vec{A} - \vec{B} ) || \vec{A} | }$

$\displaystyle \implies \: \cos \theta = \frac{4 }{ \sqrt{4 + 4} \times \sqrt{1 + 1 + 1} }$

$\displaystyle \implies \: \cos \theta = \frac{4 }{ \sqrt{8} \times \sqrt{3} }$

$\displaystyle \implies \: \cos \theta = \frac{4 }{ 2 \sqrt{2} \times \sqrt{3} }$

$\displaystyle \implies \: \cos \theta = \frac{2 }{ \sqrt{6} }$

$\displaystyle \implies \: \theta = { \cos}^{ - 1} \bigg( \frac{2}{ \sqrt{6} } \bigg)$

FINAL ANSWER

Hence the required angle $\displaystyle = { \cos}^{ - 1} \bigg( \frac{2}{ \sqrt{6} } \bigg)$

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