# Let a plane P contain two lines

Question:

Let a plane P contain two lines

$\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}), \lambda \in \mathrm{R}$ and

$\overrightarrow{\mathrm{r}}=-\hat{\mathrm{j}}+\mu(\hat{\mathrm{j}}-\hat{\mathrm{k}}), \mu \in \mathrm{R}$

If $\mathrm{Q}(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1,0,1)$ to $P$, then $3(\alpha+\beta+\gamma)$ equals_______________

Solution:

Dr's normal to plane

$=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 1 & 0 \\ 0 & 1 & -1\end{array}\right|=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

Equation of plane

$-1(x-1)+1(y-0)+1(z-0)=0$

$x-y-z-1=0$..........(1)

Now $\frac{\alpha-1}{1}=\frac{\beta-0}{-1}=\frac{\gamma-1}{-1}=-\frac{(1-0-1-1)}{3}$

$\frac{\alpha-1}{1}=\frac{\beta}{-1}=\frac{\gamma-1}{-1}=\frac{1}{3}$

$\alpha=\frac{4}{3}, \beta=-\frac{1}{3}, \gamma=\frac{2}{3}$

$3(\alpha+\beta+\gamma)=3\left(\frac{4}{3}-\frac{1}{3}+\frac{2}{3}\right)=5$