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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 30 of 78
Marks: +1, -0
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3i^+5j^−6k^.\hat{3i}+\hat{5j}-\hat{6k}.
Solution:  
Let n⃗=3i^+5j^−6k^\vec{n}=\hat{3i}+\hat{5j}-\hat{6k}
Then ∣n⃗∣=9+25+36=70|\vec{n}|=\sqrt{9+25+36}=\sqrt{70}
n^=n⃗∣n⃗∣=170(3i^+5j^−6k^)\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{1}{\sqrt{70}}\left(\hat{3i}+\hat{5j}-\hat{6k}\right)
r⃗⋅170(3i^+5j^−6k^)=7\vec{r}\cdot \frac{1}{\sqrt{70}}\left(\hat{3i}+\hat{5j}-\hat{6k}\right)=7
⇒r⃗(3i^+5j^−6k^70)=7\Rightarrow \vec{r}\left( \frac{\hat{3i}+\hat{5j}-\hat{6k}}{\sqrt{70}} \right)=7
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