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  • Find the angle between the planes whose vector equations are r vector dot 2 i caret +2 j caret -3 k caret = 5 and r vector dot 3 i caret - 3 j caret + 5 k caret= 3

12. Find the angle between the planes whose vector equations are \overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5 and \overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3.

          

Answers (1)

best_answer

Given two vector equations of plane

\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5 and \overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3.

Here, \vec{n_{1}} = 2\widehat{i}+2\widehat{j}-3\widehat{k}  and  \vec{n_{2}} = 3\widehat{i}-3\widehat{j}+5\widehat{k}

The formula for finding the angle between two planes,

\cos A = \left | \frac{\vec{n_{1}}.\vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|} \right |                                         .............................(1)

\vec{n_{1}}.\vec{n_{2}} = (2\widehat{i}+2\widehat{j}-3\widehat{k})(3\widehat{i}-3\widehat{j}+5\widehat{k}) = 2(3)+2(-3)-3(5) = -15

|\vec{n_{1}}| =\sqrt{(2)^2+(2)^2+(-3)^2} =\sqrt{17}

and       |\vec{n_{2}}| =\sqrt{(3)^2+(-3)^2+(5)^2} =\sqrt{43}

Now, we can substitute the values in the angle formula (1) to get,

\cos A = \left | \frac{-15}{\sqrt{17}\sqrt{43}} \right |

or  \cos A =\frac{15}{\sqrt{731}}

or  A = \cos^{-1}\left ( \frac{15}{\sqrt{731}} \right )

Posted by

Divya Prakash Singh

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