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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. Find a vector v¯¯¯ that is perpendicular to p.

Respuesta :

frika

Let P be the plane that intersects

  • x-axis at point (-5,0,0);
  • y-axis at point (0,-2,0);
  • z-axis at point (0,0,5).

Write the equation of the plane P:

[tex]\left|\begin{array}{ccc}x-(-5)&y-0&z-0\\0-(-5)&-2-0&0-0\\0-(-5)&0-0&5-0\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x+5&y&z\\5&-2&0\\5&0&5\end{array}\right|=0.[/tex]

Then

[tex]-10(x+5)+10z-25y=0,\\ \\2(x+5)-2z+5y=0,\\ \\2x+5y-2z=-10.[/tex]

The coefficients at variables x, y and z are the coordinates of perpendicular vector to the plane. Thus

[tex]\vec{v}=(2,5,-2)\perp P.[/tex]

Answer: [tex]\vec{v}=(2,5,-2).[/tex]

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 plane
  • With these three point, we find two vectors, on the plane
  • Finally 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

Related Link:https://brainly.com/app/ask?q=vector+perpendicular+to+a+plane

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