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प्रश्न
Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
उत्तर
We know that, the equation of a plane passing through three non-collinear points
(x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is
`|(x - x_1, y - y_1, z - z_1),(x_2 - x_1, y_2 - y_1, z_2 - z_1),(x_3 - x_1, y_3 - y_1, z_3 - z_1)|` = 0
⇒ `|(x - 2, y - 1, z - 0),(3 - 2, -2 - 1, -2 - 0),(3 - 2, 1 - 1, 7 - 0)|` = 0
⇒ `|(x - 2, y - 1, z),(1, -3, -2),(1, 0, 7)|` = 0
⇒ (x − 2)(−21 + 0) − (y − 1)(7 + 2) + z(3) = 0
⇒ −21x + 42 − 9y + 9 + 3z = 0
⇒ −21x − 9y + 3z = −51
∴ 7x + 3y − z = 17
So, the required equation of plane is 7x + 3y − z = 17.
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