Explanation Given: The plane passes through the point ( 0 , 0 , 0 ) and perpendicular to the vector n = − 3 i + 2 k . Formula used: The equation of a plane in standard form is: a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0 Calculation: The equation of a plane can be determined if a point through which it is passing and the normal vector to the plane is known. Since, the provided point is ( 0 , 0 , 0 ) and normal vector is n = − 3 i + 2 k . Now, use x 1 = 0 , y 1 = 0 and z 1 = 0 as the provided point is ( 0 , 0 , 0 ) and the direction numbers a = − 3 , b = 0 and c = 2 as the provided normal vector is n = − 3 i + 2 k in the standard equation of the plane

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Chapter 11.5, Problem 42E

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Chapter 11.5, Problem 41E

Chapter 11.5, Problem 43E

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Chapter 11 Solutions

Calculus, Early Transcendentals

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The equation of the plane passes through the point ( 0 , 0 , 0 ) and perpendicular to the vector n = − 3 i + 2 k .

Solution Summary: The author explains that the equation of a plane can be determined if the point through which it is passing and the normal vector to the plane is known.

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To calculate: The equation of the plane passes through the point (0,0,0) and perpendicular to the vector n=−3i+2k.

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