Question

Show that 2x + 3y + 7 = 0 represents a plane perpendicular to xy-plane.
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Answer 1

Answer:

We have,

2x+5y+7=0

Let, the plane=2,5,0

Equation of xy-plane z=0

So, ratio of normal to the plane are 0,0,1

∴a

1

×a

2

+a

1

×a

2

+a

1

×a

2

=2×0+5×0+0×7

=0+0+0

=0

Hence, this is answer.

Step-by-step explanation:

hopes it help uu

Answer 2

Given:

plane P = 2x + 3y + 7 = 0

To Find:

the equation represents a plane perpendicular to XY plane

Solution:

Ratios of directions of P = ($a_{1},b_{1},c_{1}$)

                                        = (2,3,0)

Equations of XY plane: z = 0

Direction ratios of XY ⇒($a_{2},b_{2},c_{2}$)= (0,0,1)

Angle between the two plane,

cosθ = $\frac{|a_{1}a_{2} + b_{1}b_{2} +c_{1}c_{2} }{\sqrt{a_{1} ^{2}+b_{1} ^{2}+c_{1} ^{2} }. \sqrt{a_{1} ^{2}+b_{1} ^{2}+c_{1} ^{2} } }$

Now, $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}$ = 2.0 + 3.0 + 0.1

                                     = 0

cosθ = 0

⇒ θ = 90°

Therefore, 2x+ 3y+7 = 0 is perpendicular to XY plane.