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प्रश्न
Find the equation of the plane through the intersection of the planes `vecr * (hati + 3hatj) - 6` = 0 and `vecr * (3hati + hatj + 4hatk)` = 0, whose perpendicular distance from origin is unity.
उत्तर
The equation of family of planes passing through the intersection of two given planes
`vecr * (hati + 3hatj) - 6` = 0 and `vecr * (3hati + hatj + 4hatk)` = 0 is given by
`[vecr * (hati + 3hatj) - 6] + lambda[vecr * (3hati - hatj - 4hatk)]` = 0
⇒ `vecr * [(hati + 3hatj) + lambda(3hati - hatj - 4hatk)] - 6` = 0
⇒ `vecr * [(1 + 3lambda)hati + (3 - lambda)hatj - 4lambdahatk] - 6` = 0. ......(1)
Given that the perpendicular distance of the origin from the required plane is unity.
Now, the position vector of origin is `veca = 0hati + 0hatj + 0hatk`
The normal vector to the required plane is
`vecN = (1 + 3lambda)hati + (3 - lambda)hatj - 4lambdahatk` and d = 6.
(Because equation of required plane is `vecr [(1 + 3lambda)hati + (3 - lambda)hatj - 4lambdahatk]` = 6 and the equation of plane in normal form is given by `vecN = (1 + 3lambda)hati + (3 - lambda)hatj - 4lambdahatk` and d = 6)
We know that the perpendicular distance of a point whose position vector is `veca` from the plane
`vecr * vecn` = d is given by `|(veca. vecN - d)/|vecN||`
Therefore, the perpendicular distance of the origin from the required plane is `|(veca. vecN - d)/|vecN||` = 1.
⇒ `|((0hati + 0hatj + 0hatk)*((1 + 3lambda)hati + (3 - lambda)hatj - 4lambdahatk) - 6)/|(1 + 3lambda)hati + (3 - lambda)hatj + -4lambdahatk||` = 1 .....`(because hati*hati = hatj*hatj = hatk*hatk = 1 "and" hati*hatj = hatj*hatk*hati = 0)`
⇒ `|(-6)/sqrt((1 + 3lambda)^2 + (3 - lambda)^2 + (-4lambda)^2)|` = 1
⇒ `sqrt((1 + 3lambda)^2 + (3 - lambda)^2 + (-4k=lambda)^2)` = 6
⇒ `(1 + 3lambda)^2 + (3 - lambda)^2 + (-4lambda)^2` = 36
⇒ `1 + 9lambda^2 + 6lambda + 9 + lambda^2 - 6lambda + 16lambda^2` = 36
⇒ `26lambda^2` = 26
⇒ `lambda^2` = 1
⇒ `lamda = +- 1`.
Therefore, the equation of required plane is
`vecr * [(1 + 3(1)hati + (3 - 1)hatj - 4(1)hatk] - 6` = 0
or
`vecr * [(1 + 3(-1))hati + (3 - (-1))hatj - 4(-1)hatk] - 6` = 0.
Hence, the equation of required plane is
`vecr * [4hati + 2hatj - 4hatk]` = 6 or `vecr * [-2hati + 4hatj + 4hatk]` = 6.
And the Cartesian equation of required plane is 4x + 2y − 4z = or – 2x + 4y + 4z = 6.
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