# 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
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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=$$cos^{-1}⁡\frac{1}{\sqrt{132}}$$.