The locus represented by $x y+y z=0$ is
If the plane $2 x-3 y+6 z-11=0$ makes an angle $\sin ^{-1} \alpha$ with $X$-axis, then the value of $\alpha$ is
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
$$\begin{array}{l} & \vec{r} & =(5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \Rightarrow & & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) & =\lambda(3(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ \Rightarrow & & (x-5) \hat{\mathbf{i}}+(y+4) \hat{\mathbf{j}}+(z-6) \hat{\mathbf{k}} & =\lambda(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{array}$$