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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.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})=1$ and $\vec{\text{r}}\cdot(-\hat{\text{i}}+\hat{\text{j}})=4$
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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, $\vec{\text{n}}_1=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{n}}_2=-\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})\cdot(-\hat{\text{i}}+\vec{\text{j}}+0\hat{\text{k}})}{\big|2\hat{\text{i}}-3\hat{\text{j}}+4{\hat{\text{k}}\big|}\big|\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big|}$
$=\frac{-2-3}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}$
$=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$ 

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