Question:

Let P1:r.(2iˆ+jˆ3kˆ)=4P1:\overrightarrow{r}.(2\^i+\^j−3\^k )=4 be a plane. Let P2P_2 be another plane which passes through points  (2,3,2)(2, - 3, 2)(2,2,3)(2, - 2, -3) and (1,4,2)(1, -4, 2). If the direction ratios of the line of intersection of P1P_1 and P2P_2 be 16, α,β,16, α,β, then the value of α+ βα + β is equal to _____ .

Updated On: Mar 2, 2024
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Correct Answer: 28

Solution and Explanation

Direction ratio of normal to P1<2,1,3>P_1≡< 2, 1, – 3 >
and P2i^j^k^015125P2≡\begin{vmatrix} \hat i & \hat j & \hat k \\[0.3em] 0 & 1 & -5 \\[0.3em] -1 & -2 & 5 \end{vmatrix}
P2=5i^j^(5)+k^(1)P_2=−5\hat i−\hat j(−5)+\hat k(1)
i.e.<5,5,1>< –5, 5, 1 >
d.r’s of line of intersection are along vector
i^j^k^213551\begin{vmatrix} \hat i & \hat j & \hat k \\[0.3em] 2 & 1 & -3 \\[0.3em] -5 & 5 & 1 \end{vmatrix}=i^(16)j^(13)+k^(15)=\hat i(16)−\hat j(−13)+\hat k(15)
i.e.<16,13,15>< 16, 13, 15 >
Therefore, α+β=13+15=28α + β = 13 + 15 = 28

So, the answer is 2828.

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