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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 48 of 78
Marks: +1, -0
Find the angle between the planes whose vector equations are r⃗⋅(2i^+2j^−3k^)=5\vec{r}\cdot(2\hat{i}+2\hat{j}-3\hat{k})=5 and r⃗⋅(3i^−3j^+5k^)=3.\vec{r}\cdot(3\hat{i}-3\hat{j}+5\hat{k})=3.
Solution:  
If "θ’ be the angle between the planes r⃗⋅n⃗1=d1\vec{r}\cdot\vec{n}_1=d_1 and r⃗⋅n⃗2=d2\vec{r}\cdot\vec{n}_2=d_2 ,
then cos⁡θ=∣n⃗1⋅n⃗2∣n⃗1∣∣n⃗2∣∣\cos\theta=\left|\frac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}\right|
Here n⃗1=2i^+2j^−3k^\vec{n}_1=2\hat{i}+2\hat{j}-3\hat{k} and n⃗2=3i^−3j^+5k^\vec{n}_2=3\hat{i}-3\hat{j}+5\hat{k}
cos⁡θ=∣(2i^+2j^−3k^)⋅(3i^−3j^+5k^)4+4+99+9+25∣\cos\theta=\left|\frac{(2\hat{i}+2\hat{j}-3\hat{k})\cdot(3\hat{i}-3\hat{j}+5\hat{k})}{\sqrt{4+4+9}\sqrt{9+9+25}}\right|
=∣−151743∣=\left|\frac{-15}{\sqrt{17}\sqrt{43}}\right|
Hence θ=cos⁡−1∣−151743∣\theta=\cos^{-1}\left|\frac{-15}{\sqrt{17}\sqrt{43}}\right| =cos⁡−1(15731)=\cos^{-1}\left(\frac{15}{\sqrt{731}}\right)
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