Question

Let $P_1:\overrightarrow{r}.(2\hat{i}+\hat{j}-3\hat{k})=4$ be a plane. Let $P_2$ be another plane which passes through the points $(2,–3,2),(2,–2,–3)$ and $(1,–4,2).$ If the direction ratios of the line of intersection of $P_1$ and $P_2$ be $16,$ $\alpha ,\beta ,$ then the value of $\alpha +\beta$ is equal to

• A. 28
• B. 28.00
• C. 28.0

Answer 1

Answer: (A), (B), (C)

Direction ratio of normal to $P_1\equiv <2,1,-3>$
and that of $P_2\equiv \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& -5\\ -1& -2& 5\end{array}\right|=-5\hat{i}-\hat{j}(-5)+\hat{k}\left(1\right)$

$i.e.<–5,5,1>$
d.r.’s of line of intersection are along vector
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -3\\ -5& 5& 1\end{array}\right|=\hat{i}\left(16\right)-\hat{j}(-13)+\hat{k}\left(15\right)$

$i.e.<16,13,15>$
$\therefore \alpha +\beta =13+15=28$