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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry3 Marks
Question
Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})=9$
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Answer

We know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
$\vec{\text{n}}_2=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6{\hat{\text{k}}\big|}\big|\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{2-16-12}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{-16}{(7)(3)}$
$=\frac{-16}{21}$
$\theta=\cos^{-1}\Big(\frac{-16}{21}\Big)$

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