137.Find a unit normal vector to the plane x + 2y + 3z – 6 = 0.
The equation of plane is x + 2y’ + 3z – 6 = 0 Direction-ratios of a normal to the plane are 1, 2, 3. Dividing each by the direction-cosines of a normal to the given plane are ∴ the normal vector to the given plabne is
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138.Find the normal form of the equation of the plane 3x– 4y + z + 5 = 0.
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139.Find the value of k for which the planes 3 x– 6 y– 2 z = 7 and 2x + y'– k z = 5 are perpendicular to each other.
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140.
The foot of the perpendicular drawn from the origin to the plane is (4, 3, 2). Find the equation of the plane.