Find the angle between the planes whose vector equations are: r . (2 i +2 j -3 k )=5 and r . (3 i -3 j +5 k )=3.

The equations of the given planes are r . (2 i +2 j -3 k )=5 and r . (3 i -3 j +5 k )=3 It is known that if n _1 and n _2 are normal to the planes, r . n _1= d _1 and r . n _2= d _2, then the angle between them, Q is given by, = | n _1. n _2 | n _1|| n _2| | ...(1) Here, n _1=2 i +2 j -3 k and n _2=3 i -3 j +5 k n _1. n _2= (2 i +2 j -3 k ) ( 3 i -3 j +5 k ) = 2.3 + 2.(-3) + (-3).5 = -15 | n _1|=(2)^2+(2)^2+(-3)^2=17 | n _2|=(3)^2+(-3)^2+(5)^2=43 = |-15 17.43 | =15 731 Q= ^ -1 (15 731 ).

Find the angle between the planes whose vector equations are: r .…

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Question

Find the angle between the planes whose vector equations are:
$\vec{\text{r}}.\Big(2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)=5\ \text{and}\ \vec{\text{r}}.\Big(3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\Big)=3.$

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Answer

The equations of the given planes are
$\vec{\text{r}}.\Big(2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)=5\ \text{and}\ \vec{\text{r}}.\Big(3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\Big)=3$
It is known that if $\vec{\text{n}}_1\ \text{and}\ \vec{\text{n}}_2$ are normal to the planes, $\vec{\text{r}}.\vec{\text{n}}_1=\vec{\text{d}}_1\ \text{and}\ \vec{\text{r}}.\vec{\text{n}}_2=\vec{\text{d}}_2,$ then the angle between them, Q is given by,
$\cos\text{Q}=\Bigg|\frac{\vec{\text{n}}_1.\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}\Bigg|\ \ ...(1)$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\ \text{and }\vec{\text{n}}_2=3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}$
$\therefore\ \ \vec{\text{n}}_1.\vec{\text{n}}_2=\Big(2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)\Big( 3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\Big)$
= 2.3 + 2.(-3) + (-3).5 = -15
$|\vec{\text{n}}_1|=\sqrt{(2)^2+(2)^2+(-3)^2}=\sqrt{17}$
$|\vec{\text{n}}_2|=\sqrt{(3)^2+(-3)^2+(5)^2}=\sqrt{43}$
$\cos\text{Q}=\Big|\frac{-15}{\sqrt{17}.\sqrt{43}}\Big|$
$\Rightarrow\ \cos\text{Q}=\frac{15}{\sqrt{731}}$
$\Rightarrow\ \text{Q}=\cos^{-1}\Big(\frac{15}{\sqrt{731}}\Big).$

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