Question

Let $P1:\overrightarrow{r}.(2\^i+\^j−3\^k )=4$ be a plane. Let $P_2$ be another plane which passes through 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 $16, α,β,$ then the value of $α + β$ is equal to _____ .

Answer 1

Direction ratio of normal to $P_1≡< 2, 1, – 3 >$
and $P2≡\begin{vmatrix} \hat i & \hat j & \hat k \\\[0.3em] 0 & 1 & -5 \\\[0.3em] -1 & -2 & 5 \end{vmatrix}$
$P_2=−5\hat i−\hat j(−5)+\hat k(1)$
i.e.$< –5, 5, 1 >$
d.r’s of line of intersection are along vector
\begin{vmatrix} \hat i & \hat j & \hat k \\\[0.3em] 2 & 1 & -3 \\\[0.3em] -5 & 5 & 1 \end{vmatrix}$$=\hat i(16)−\hat j(−13)+\hat k(15)
i.e.$< 16, 13, 15 >$
Therefore, $α + β = 13 + 15 = 28$

So, the answer is $28$.