If a plane passes through the points $(2,0,0)(0,3,0)$ and $(0,0,4)$ the equation of plane is ................ .
We know that, equation of a the plane that cut the coordinate axes at $(a, 0,0)(0, b, 0)$ and $(0,0, c)$ is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.
Hence, the equation of plane passes through the points $(2,0,0),(0,3,0)$ and $(0,0,4)$ is $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$.
The direction cosines of the vector $(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$ are .............. .
Direction cosines of $(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$ are $\frac{2}{\sqrt{4+4+1}}, \frac{2}{\sqrt{4+4+1}}, \frac{-1}{\sqrt{4+4+1}}$ i.e., $\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}$.
The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is ............ .
We have, $\overrightarrow{\mathbf{a}}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
So, the vector equation will be
$$\begin{array}{l} & \vec{r} & =(5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \Rightarrow & & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) & =\lambda(3(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ \Rightarrow & & (x-5) \hat{\mathbf{i}}+(y+4) \hat{\mathbf{j}}+(z-6) \hat{\mathbf{k}} & =\lambda(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{array}$$
The vector equation of the line through the points $(3,4,-7)$ and $(1,-1$, 6) is ................. .
We know that, vector equation of a line passes through two points is represented by $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\lambda(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}})$
Here, $\overrightarrow{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-7 \hat{\mathbf{k}}$
and $$\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+6 \hat{\mathbf{k}}$$
$$\Rightarrow \quad(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}})=-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+13 \hat{\mathbf{k}}$$
So, the required equation is
$$\begin{array}{ll} & x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-7 \hat{\mathbf{k}}+\lambda(-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+13 \hat{\mathbf{k}}) \\ \Rightarrow & (x-3) \hat{\mathbf{i}}+(y-4) \hat{\mathbf{j}}+(z+7) \hat{\mathbf{k}}=\lambda(-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+13 \hat{\mathbf{k}}) \end{array}$$
The cartesian equation of the plane $\overrightarrow{\mathbf{r}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2$ is ............. .
$$\begin{array}{lr} \text { We have, } & \overrightarrow{\mathbf{r}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 \\ \Rightarrow & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 \\ \Rightarrow & x+y-z=2 \end{array}$$
which is the required form