Find the angle between the planes \vec{r}.(4\hat{i}+\hat{j}-2\hat{k})=6 and \vec{r}.(5\hat{i}-6\hat{j}+\hat{k})=7?

Category: QuestionsFind the angle between the planes \vec{r}.(4\hat{i}+\hat{j}-2\hat{k})=6 and \vec{r}.(5\hat{i}-6\hat{j}+\hat{k})=7?
Editor">Editor Staff asked 11 months ago

Find the angle between the planes \vec{r}.(4\hat{i}+\hat{j}-2\hat{k})=6 and \vec{r}.(5\hat{i}-6\hat{j}+\hat{k})=7?
 
(a) cos^{-1}⁡\frac{12}{\sqrt{1302}}
 
(b) cos^{-1}⁡\frac{1}{\sqrt{1392}}
 
(c) cos^{-1}\frac{⁡23}{\sqrt{102}}
 
(d) cos^{-1}⁡\frac{15}{\sqrt{134}}
 
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I need to ask this question from Three Dimensional Geometry in division Three Dimensional Geometry of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

Right answer is (a) cos^{-1}⁡\frac{12}{\sqrt{1302}}
 
To explain: Given that, the normal to the planes are \vec{n_1}=4\hat{i}+\hat{j}-2\hat{k} \,and \,\vec{n_2}=5\hat{i}-6\hat{j}+\hat{k}
 
The angle between two planes of the form \vec{r}.\vec{n_1}=d_1 \,and \,\vec{r}.\vec{n_2}=d_2 is given by
 
cos⁡θ=\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |
 
|\vec{n_1}|=\sqrt{4^2+1^2+(-2)^2}=\sqrt{21}
 
|\vec{n_2}|=\sqrt{5^2+(-6)^2+1^2}=\sqrt{62}
 
\vec{n_1}.\vec{n_2}=4(5)+1(-6)-2(1)=20-6-2=12
 
cos⁡θ=\frac{12}{\sqrt{21}.\sqrt{62}}=\frac{12}{\sqrt{1302}}
 
∴θ=cos^{-1}⁡\frac{12}{\sqrt{1302}}.