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  • Find the angle between the line and plane  where equation of line is <img alt="r=\widehat{i}+\widehat{j}+\widehat{k}+\lambda (\widehat{i}-\widehat{j}+\widehat{k})" src="https://learn.care

Find the angle between the line and plane  where equation of line is r=\widehat{i}+\widehat{j}+\widehat{k}+\lambda (\widehat{i}-\widehat{j}+\widehat{k}) and the equation of plane r.(\widehat{i}-\widehat{j}+\widehat{k})=2

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The line - \vec{r}=\hat{i}+\hat{j}+\hat{k}+\lambda(\hat{i}-\hat{j}+\hat{k})

Plane - \vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=2

Normal vector - \vec{m}=\hat{i}-\hat{j}+\hat{k}

Formula to use - \sin \theta=\frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| \cdot|\vec{n}|}

The angle between the line and the plane is 90 degrees

\begin{aligned} & \vec{d} \cdot \vec{n}=(1)(1)+(-1)(-1)+(1)(1)=1+1+1=3 \\ & |\vec{d}|=\sqrt{1^2+(-1)^2+1^2}=\sqrt{3} \\ & |\vec{n}|=\sqrt{1^2+(-1)^2+1^2}=\sqrt{3} \\ & \qquad \cos \phi=\frac{3}{\sqrt{3} \cdot \sqrt{3}}=\frac{3}{3}=1 \Rightarrow \phi=\cos ^{-1}(1)=0^{\circ} \end{aligned}

line is parallel to the normal

Final answer would be 90 degrees

Posted by

Divya Sharma

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