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प्रश्न
Show that the points (2, –1, 3), (4, 3, 1) and (3, 1, 2) are collinear
उत्तर
Let the given points be A(2, –1, 3), B(4, 3, 1) and C(3, 1, 2)
`vec"OA" = 2hat"i" - hat"j" + 3hat"k"`
`vec"OB" = 4hat"i" + 3hat"j" + hat"k"`
`vec"OC" = 3hat"i" + hat"j" + 2hat"k"`
`vec"AB" = vec"OB" - vec"OA"`
= `(4hat"i" + 3hat"j" + hat"k") - (2hat"i" - hat"j" + 3hat"k")`
= `4hat"i" + 3hat"j" + hat"k" - 2hat"i" + hat"j" - 3hat"k"`
`vec"AB" = 2hat"i" + 4hat"j" - 2hat"k"|`
`|vec"AB"| = |2hat"i" + 4hat"j" - 2hat"k"|`
= `sqrt(2^2 + 4^2 + (-2)^2`
= `sqrt(4 +16 + 4)`
= `sqrt(24)`
AB = `sqrt(6 xx 4)`
= `2sqrt(6)`
`vec"BC" = vec"OC" - vec"OB"`
= `(3hat"i" + hat"j" + 2hat"k") - (4hat"i" + 3hat"j" + hat"k")`
= `3hat"i" + hat"j" + 2hat"k" - 4hat"i" - 3hat"j" - hat"k"`
`vec"BC" = -hat"i" - 2hat"j" + hat"k"`
`|vec"BC"| = |-hat"i" - 2hat"j" + hat"k"|`
= `sqrt((-1)^2 + (-2)^2 + 1^2)`
BC = `sqrt(1 + 4 + 1)`
= `sqrt(6)`
`vec"CA" = vec"OC" - vec"OA"`
= `(3hat"i" + hat"j" + 2hat"k") - (2hat"i" + hat"j" + 3hat"k")`
= `3hat"i" + hat"j" + 2hat"k" - 2hat"i" - hat"j" - 3"k"`
`vec"BC" = -hat"i" - 2hat"j" + hat"k"`
`vec"CA" = |hat"i" + 2hat"j" - hat"k"|`
= `sqrt(1^2 + 2^2 + (-1)^2`
CA = `sqrt(1 + 4 + 1)`
= `sqrt(6)`
AB = `2sqrt(6)`, BC = `sqrt(6)`, CA = `sqrt(6)`
BC + CA =`sqrt(6) + sqrt(6) = 2sqrt(6)`
∴ BC + CA = BA = `2sqrt(6)`
Hence the given points A, B, C are collinear.
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