Angle between a Line and a Plane

The angle between the line $\overrightarrow{r}=(\hat{i}+2\hat{ȷ}-\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and the plane $\overrightarrow{r}·(2\hat{i}-\hat{ȷ}+\hat{k})=5$ is 

The angle between the plane $\overrightarrow{r}.\left(\hat{i}-2\hat{j}+3\hat{k}\right)=5$ and the line $\overrightarrow{r}=\left(\hat{i}+\hat{j}-\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)$ is 

The angle between the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z+2}{4}$ and the plane $2x+y-3z+4=0$ is 

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$ is

If the line $\overrightarrow{r}=(\hat{i}-2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}+2\hat{k})$ is parallel to the plane $\overrightarrow{r}·(3\hat{i}-2\hat{j}+m\hat{k})=10$, then value of $m$ is
 

Find the angle between the line $\overrightarrow{r}=(\hat{i}+2\hat{ȷ}-k)+\lambda (\hat{i}-\hat{j}+\hat{k})$ and the plane $\overrightarrow{r}\left(2\stackrel{⏜}{i}-\stackrel{⏜}{j}+\stackrel{⏜}{k}\right)=5$.

Find the angle between the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z+2}{4}$ and the plane $2x+y-3z+4=0$

Find the angle between the planes $\overrightarrow{r}.\left(\hat{i}-2\hat{j}+3\hat{k}\right)=5$ and the line $\overrightarrow{r}=\left(\hat{i}+\hat{j}-\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)$

The angle between the line $\overrightarrow{r}=\left(i+\hat{\mathit{j}}-\hat{\mathit{k}}\right)+\lambda \left(3\hat{i}+\hat{\mathit{j}}\right)$ and the plane $\overrightarrow{r}·\left(\hat{\mathit{ı}}+2\hat{\mathit{ȷ}}+3\hat{\mathit{k}}\right)=8$ is

The angle between the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}$ and the plane $\overrightarrow{r}·\left(6\hat{\mathrm{ı}}-2\hat{\mathrm{ȷ}}-3\hat{\mathrm{k}}\right)=5$ is

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

Show that the line of intersection of the planes $\overrightarrow{r}·(\hat{i}+3\hat{j}-2\hat{k})=0$ and $\overrightarrow{r}\left(2\hat{i}+4\hat{j}-3\hat{k}\right)=0$ is equally inclined to $\hat{i}$ & $\hat{j}$. Also find the angle it makes with $\hat{k}$.

Find the value of $k$. If the  angle between the line $\overrightarrow{r}=(\hat{i}+2\hat{ȷ}-\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and the plane $\overrightarrow{r}\left(2\hat{i}-\hat{j}+\hat{k}\right)=5$ is ${\sin }^{-1}\left(\frac{2\sqrt{2}}{k}\right)$.

Find the angle at which the normal vector to the plane $4x+8y+z=5$ is inclined to the coordinate axes.  

If the line $\overrightarrow{r}=\hat{i}+\mathrm{\lambda }\left(2\hat{i}-m\hat{j}-3\hat{k}\right)$ is parallel to the plane $\overrightarrow{r}·\left(m\hat{i}+3\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)=4$, then find the value of $m$.

If the line $\overrightarrow{r}=\left(\hat{i}-2\hat{j}+\hat{k}\right)+\mathrm{\lambda }\left(2\hat{i}+\hat{j}+2\hat{k}\right)$ is parallel to the plane $\overrightarrow{r}·\left(3\hat{i}-2\hat{j}+m\hat{k}\right)=3$, then find the value of $m$.

Find the angle between the line $\overrightarrow{r}=\left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\mathrm{\lambda }\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)$ and plane $\stackrel{\rightharpoonup }{\mathrm{r}}·\left(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)=4$

Find the angle between the line $\overrightarrow{r}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\mathrm{\lambda }\left(\hat{i}-\hat{j}+\hat{k}\right)$ and plane $\overrightarrow{r}·\left(2\hat{i}-\hat{j}+\hat{k}\right)=4$

 Find the angle between the line $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-2}{2}$ and plane $3x+4y+z+5=0$.

Find the angle between the line $\frac{x+1}{3}=\frac{y-1}{2}=\frac{z-2}{4}$ and plane $2x+y-3z+4=0$