Answer to Question #210882 in Analytic Geometry for Ree

Question #210882

Question 7


(7.1) (Find a point-normal form of the equation of the plane passing

through P = (1,2,−3) and having n =< 2,−1,2 > as a normal.


(7.2) Determine in each case whether the given planes are parallel or

perpendicular: (a) x +y +3z +10=0andx +2y −z =1,


(b)3x −2y +z −6=0 and 4x +2y −4z =0, (c)3x +y +z −1=0 and−x +2y

+z+3=0,


(d)x −3y +z+1=0 and 3x −4y +z−1=0.


1
Expert's answer
2021-07-16T07:44:57-0400


(7.1)

Point-normal form of the plane is



2(x1)(y2)+2(z+3)=0.2(x − 1) -(y − 2) + 2(z +3) = 0.

We can also write this as



2xy+2z=62x-y+2z=-6



(7.2)

(a). x+y+3z+10=0 and x+2y-z=1



n1=1,1,3,n2=1,2,1\vec n_1=\langle1,1,3\rangle, \vec n_2=\langle1,2,-1\rangle1121\dfrac{1}{1}\not=\dfrac{2}{1}

The vectors n1\vec n_1 and n2\vec n_2 are not collinear.




n1n2=1(1)+1(2)+3(1)=0\vec n_1\cdot\vec n _2=1(1)+1(2)+3(-1)=0

The vectors n1\vec n_1 and n2\vec n_2 are orthogonal.

Hence, the given planes are perpendicular.


(b). 3x-2y+z-6=0 and 4x+2y-4z=0



n1=3,2,1,n2=4,2,4\vec n_1=\langle3,-2,1\rangle, \vec n_2=\langle4,2,-4\rangle4322\dfrac{4}{3}\not=\dfrac{2}{-2}

The vectors n1\vec n_1 and n2\vec n_2 are not collinear.




n1n2=3(4)+(2)(2)+1(4)=40\vec n_1\cdot\vec n _2=3(4)+(-2)(2)+1(-4)=4\not=0

The vectors n1\vec n_1 and n2\vec n_2 are not orthogonal.

Hence, the given planes are neither parallel nor perpendicular.


(c). 3x+y+z-1=0 and -x+2y+z+3=0



n1=3,1,1,n2=1,2,1\vec n_1=\langle3,1,1\rangle, \vec n_2=\langle-1,2,1\rangle1321\dfrac{-1}{3}\not=\dfrac{2}{1}

The vectors n1\vec n_1 and n2\vec n_2 are not collinear.




n1n2=3(1)+1(2)+1(1)=0\vec n_1\cdot\vec n _2=3(-1)+1(2)+1(1)=0

The vectors n1\vec n_1 and n2\vec n_2 are orthogonal.

Hence, the given planes are perpendicular.


(d). x-3y+z+1=0 and 3x-4y+z-1=0



n1=1,3,1,n2=3,4,1\vec n_1=\langle1,-3,1\rangle, \vec n_2=\langle3,-4,1\rangle3143\dfrac{3}{1}\not=\dfrac{-4}{-3}

The vectors n1\vec n_1 and n2\vec n_2 are not collinear.




n1n2=1(3)+(3)(4)+1(1)=160\vec n_1\cdot\vec n _2=1(3)+(-3)(-4)+1(1)=16\not=0

The vectors n1\vec n_1 and n2\vec n_2 are not orthogonal.

Hence, the given planes are neither parallel nor perpendicular.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment