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प्रश्न
Let the position vectors of two points P and Q be `3hat"i" - hat"j" + 2hat"k"` and `hat"i" + 2hat"j" - 4hat"k"`, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, –1, 2) and (–2, 1, –2) respectively. Let lines PR and QS intersect at T. If the vector `vec("TA")` is perpendicular to both `vec("PR")` and `vec("QS")` and the length of vector `vec("TA")` is `sqrt(5)` units, then the modulus of a position vector of A is ______.
विकल्प
`sqrt(5)`
`sqrt(171)`
`sqrt(227)`
`sqrt(482)`
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उत्तर
Let the position vectors of two points P and Q be `3hat"i" - hat"j" + 2hat"k"` and `hat"i" + 2hat"j" - 4hat"k"`, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, –1, 2) and (–2, 1, –2) respectively. Let lines PR and QS intersect at T. If the vector `vec("TA")` is perpendicular to both `vec("PR")` and `vec("QS")` and the length of vector `vec("TA")` is `sqrt(5)` units, then the modulus of a position vector of A is `underlinebb(sqrt(171))`.
Explanation:
Given: `vec"p" = 3hat"i" - hat"j" + 2hat"k"; "P"(3, -1, 2)`
and `vec("Q") = hat"i" + 2hat"j" - 4hat"k"; "Q"(1, 2, -4)`
`vec("PR") || 4hat"i" - hat"j" + 2hat"k"` and `vec("QS") || -2hat"i" + hat"j" - 2hat"k"`
Direction ratios of normal to the plane containing P, T and Q will proportional to
`[(hat"i", hat"j", hat"k"),(4, -1, 2),(-2, 1, -2)]`
⇒ `0hat"i" + 4hat"j" + 2hat"k"`
∴ `"l"/0 = "m"/4 = "m"/2`
For the point, T:
`vec("PT") = (x - 3)/4 = ("y" + 1)/(-1) = ("z" - 2)/2` = λ
and `vec("QT") = (x - 1)/(-2) = ("y" - 2)/1 = ("z" + 4)/(-2)` = μ
T: (4λ + 3 – λ – 1, 2λ + 2) ≃ (–2μ + 1, μ + 2, –2μ – 4)
After comparing, we get
4λ + 3 = –2μ + 1 ⇒ 2λ + μ = –1
–λ – 1 = μ + 2 ⇒ λ + μ = –3
2λ + 2 = –2μ – 4 ⇒ λ + μ = –3
⇒ λ = 2 and μ = –3
So, point T(11, –3, 6)
`vec("OT") = 11hat"i" - 3hat"j" + 6hat"k"`
Now, `vec("TA") = vec("OA") - vec("OT")`
`vec("OT") = vec("OT") + vec("TA")`
`vec("OA") = (11hat"i" - 3hat"j" + 6hat"k") + ((4hat"j" + 2hat"k"))/sqrt(4^2 + 2^2) xx sqrt(5)`
`vec("OA") = (11hat"i" - 3hat"j" + 6hat"k") + ((4hat"j" + 2hat"k"))/(2sqrt(5)) xx sqrt(5)`
`vec("OA") = (11hat"i" - 3hat"j" + 6hat"k") + ((2hat"j" + hat"k")/sqrt(5))(sqrt(5))`
`vec("OA") = (11hat"i" - 3hat"j" + 6hat"k") + (2hat"j" + hat"k")`
`vec("OA") = 11hat"i" - hat"j" + 7hat"k"`
⇒ `vec("OA") = sqrt(171)`
