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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Concepts Used:
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The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.
Read More: Distance Between Two Points