Mathematics, 10.07.2019 06:00 SaliJano8950

Let p be the plane in space that intersects the x-axis at −5, the y-axis at −2, and the z-axis at 5. find a vector v¯¯¯ that is perpendicular to p.

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To find a vector perpendicular to a given plane, is necessary to find two vectors on the plane and get the vector product of these two vectors.

Solution is:

PQ * PR = - 10×i + 10×k + 25×j

We must go through the following procedure:

Identify three points on the planeWith these three point, we find two vectors, on the planeFinally we get the vector (cross) product of the vectors

Identifying the three points on the plane.

The three intersects are three points on the plane, these are:

P ( -5 , 0 , 0 ). Q ( 0 , - 2 , 0 ). R ( 0 , 0 , 5 )

From these points we get the vectors PQ and PR

PQ = ( 0 , -2 , 0 ) - ( -5 , 0 , 0 )

PQ = ( 0 - ( -5) , -2 - 0 , 0 - 0 )

PQ = ( 5 , - 2 , 0 )

PR = ( 0 , 0, 5 ) - ( -5 , 0, 0 )

PR = ( 5 , 0, 5 )

Now we get the vector product ( it is a perpendicular vector to PQ and PR ), and is perpendicular to the plane.

i j k

PQ * PR = 5 -2 0 = i × (-2)×(5) + 0×k + 0×j

5 0 5 - [( -2)×5×k + (0)×(0)×i + (5)×(5)×j

PQ * PR = - 10×i + 10×k + 25×j

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Let p be the plane in space that intersects the x-axis at −5, the y-axis at −2, and the z-axis at 5....

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