Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 2: EXERCISE

The ratio in which the plane $\overrightarrow{r}·(i-2j+3k)=17$ divides the line joining the points $-2i+4j+7k$ and $3i-5j+8k$ is

The value of $m$ so that the line $\overrightarrow{r}=i+2k+\lambda (2i-mj-3k)$ is parallel to the plane $\overrightarrow{r}·\left(mi+3j+k\right)=4$ is

The image of the point $(1,3,4)$ in the plane $x+2y-z+3=0$ is

The image of the point $(0,0,0)$ in the plane $3x+4y-6z+1=0$ is

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

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

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

Find the distance of the point $(3, 2, 1)$, from the point where the line $x-3=\frac{y-2}{2}=\frac{z-1}{2}$, intersects the plane $4x-3y+2z+5=0$.