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Vectors

The plane P1:4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane P2:5x+3y+10z-25=0.

If the plane in its new position is denoted by P3, and then,

What is the distance of the plane from the origin?

Please give a detailed stepwise approach.

Sushant Mehta

15 Years agoGrade

1 Answer

Arun Kumar

12 Years ago

P3 will be parallel to

$(4,7,4)*((4,7,4)*(5,3,10))= (-81,324,-486 )$

a point of intersection of two planes

$4x+7y=-4z-81\\ 5x+3y=-10z+25\\ x=\begin{pmatrix} -4z-81 &7 \\ -10z+25& 3 \end{pmatrix}/\begin{pmatrix} 4 &7 \\ 5 &3 \end{pmatrix} \\y=\begin{pmatrix} 4 &-4z-81 \\ 5& -10z+25 \end{pmatrix}/\begin{pmatrix} 4 &7 \\ 5 &3 \end{pmatrix}$

(x,y,z)

Now we just

$\vec a=(x,y,z)\\ distance\,from\,origin=\vec r.\vec n/|\vec n|=\vec a.\vec n/|\vec n|$

where n is perpendicular vector to plane we calculated a first

and a is (x,y,z)

Arun Kumar

IIT Delhi

Askiitians Faculty

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