Question

Let a plane $P$contain two lines $\begin{array}{l}\overrightarrow{r}=\hat{i}+\lambda (\hat{i}+\hat{j})\end{array}$, $\lambda \in R$ and $\begin{array}{l}\overrightarrow{r}=-\hat{j}+\mu (\hat{j}-\hat{k})\end{array}$, $\mu \in R$. If $Q(\alpha ,\beta ,\gamma )$ is the foot of the perpendicular drawn from the point $M(1,0,1)$ to $P$, then $3(\alpha +\beta +\gamma )$ equals:

Answer 1

Step 1: Finding the unit vector perpendicular to the plane

Given that the plane containing the vectors

$\begin{array}{l}\overrightarrow{r}=\hat{i}+\lambda (\hat{i}+\hat{j})\end{array}$

And $\begin{array}{l}\overrightarrow{r}=-\hat{j}+\mu (\hat{j}-\hat{k})\end{array}$

Considering the unit vector be perpendicular to the plane be $\hat{n}$

$\Rightarrow \begin{array}{l}\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& -1\end{array}\right|\end{array}$

$\Rightarrow \overrightarrow{n}=-\hat{i}+\hat{j}+\hat{k}$

Therefore, the equation of plane containing the point $M(1,0,1)$ and normal $\hat{n}$

$$\begin{gathered} \Rightarrow 1(x-1)+1(y-0)+1(z-0)=0 \\ \Rightarrow x-y-z-1=0...\left(i\right) \end{gathered}$$

Step 2: Finding the value of $3(\alpha +\beta +\gamma )$

Foot of the perpendicular drawn from the point $M(1,0,1)$ to the plane $\left(i\right)$

$$\begin{gathered} \Rightarrow \frac{(x-1)}{1}=\frac{(y-0)}{-1}=\frac{(z-1)}{-1}=\frac{-(1-0-1-1)}{3} \\ \Rightarrow (x-1)=\frac{1}{3},\frac{y}{-1}=\frac{1}{3},\frac{(z-1)}{-1}=\frac{1}{3} \\ \Rightarrow x=\frac{4}{3},y=-\frac{1}{3},z=\frac{2}{3} \end{gathered}$$

As given $Q(\alpha ,\beta ,\gamma )$lies on foot of perpendicular

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

$$\begin{gathered} \Rightarrow \alpha +\beta +\gamma =\frac{4}{3}-\frac{1}{3}+\frac{2}{3} \\ \Rightarrow \alpha +\beta +\gamma =\frac{5}{3} \\ \Rightarrow 3(\alpha +\beta +\gamma )=5 \end{gathered}$$

Hence, the $3(\alpha +\beta +\gamma )$ is equal to $5$