Find the angle between two planes \(\vec{r}.(2\hat{i}-\hat{j}+\hat{k})=3\) and \(\vec{r}.(3\hat{i}+2\hat{j}-3\hat{k})\)=5.

Category: QuestionsFind the angle between two planes \(\vec{r}.(2\hat{i}-\hat{j}+\hat{k})=3\) and \(\vec{r}.(3\hat{i}+2\hat{j}-3\hat{k})\)=5.
Editor">Editor Staff asked 11 months ago

Find the angle between two planes \(\vec{r}.(2\hat{i}-\hat{j}+\hat{k})=3\) and \(\vec{r}.(3\hat{i}+2\hat{j}-3\hat{k})\)=5.
 
(a) \(cos^{-1}⁡\frac{1}{\sqrt{22}}\)
 
(b) \(cos^{-1}⁡\frac{1}{\sqrt{6}}\)
 
(c) \(cos^{-1}⁡\frac{1}{\sqrt{132}}\)
 
(d) \(cos^{-1}⁡\frac{1}{\sqrt{13}}\)
 
This question was posed to me by my school principal while I was bunking the class.
 
This intriguing question comes from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

The correct answer is (c) \(cos^{-1}⁡\frac{1}{\sqrt{132}}\)
 
The explanation: Given that, the normal to the planes are \(\vec{n_1}=2\hat{i}-\hat{j}+\hat{k}\) and \(\vec{n_2}=3\hat{i}+2\hat{j}-3\hat{k}\)
 
cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
 
\(|\vec{n_1}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}\)
 
\(|\vec{n_2}|=\sqrt{3^2+2^2+(-3)^2}=\sqrt{22}\)
 
\(\vec{n_1}.\vec{n_2}\)=2(3)-1(2)+1(-3)=6-2-3=1
 
cos⁡θ=\(\frac{1}{\sqrt{22}.\sqrt{6}}=\frac{1}{\sqrt{132}}\)
 
∴θ=\(cos^{-1}⁡\frac{1}{\sqrt{132}}\).