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 _____ .
Correct Answer: 28
Solution and Explanation
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 \(28\).
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