Question

determine whether the given planes are parallel or perpendicular,and in case they are neither, find the angles between them. (a)7x+5y+6z+30=0 and 3x-y-10z+4=0 

(b)2x+y+3z-2=0 and x-2y+5=0 

(c)2x-2y+4z+5=0 and 3x-3y+6z-1=0 

(d)2x-y+3z-1=0 and 2x-y+3z+3=0 

(e)4x+8y+z-8=0 and y+z-4=0

Answer 1

The direction ratios of normal to the plane,
L₁:a₁x+b₁y+c₁z=0, are a₁,b₁,c₁ and
L₂:a₁x+b₂y+c2z=0 are a₂,b₂,c₂.

 

\(L1||L2, if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
\(L1|L2, if a_1a_2+b_1b_2+c_1c_2=0\)

The angle between L1 and L2is given by,
\(Qcos^-1|\frac{a_1a_2+b_1b_2+c_1c_2}{√a_1^2+b_1^2+c_1^2.√a_2^2+b_2^2+c_2^2}|\)

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(a)The equations of the planes are 7x+5y+6z+30=0 and 3x-y-10z+4=0

Here,
a₁=7, b₁=5, c₁=6
a₂=3, b₂=-1, c₂=-10

a₁a₂+b₁b₂+c₁c₂
=7×3+5×(-1)+6×(-10)
=-44≠0

Therefore, the given planes are not perpendicular.

=\(cos^{-1}\frac{2}{5}\)

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(b)The equations of the planes are 2x+y+3z-2=0 and x-2y+5=0

Here,
a₁=2, b₁=1, c₁=3 and
a₂=1, b₂=-2, c₂=0

∴a₁a₂+b₁b₂+c₁c₂
=2×1+1×(-2)+3×0
=0

Thus, the given planes are perpendicular to each other.

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(c)The equations of the given planes are 2x-2y+4z+5=0 and 3x-3y+6z-1=0

Here, a₁=2, b₁=-2, c₁=4 and
a₂=3, b₂=-3, c₂=6

a₁a₂+b₁b₂+c₁c₂
=2×3+(-2)(-3)+4×6
=6+6+24
=36≠0.

Thus, the given planes are not perpendicular to each other.
\(∴\frac{a_1}{a_2}=\frac{b^1}{b^2}=\frac{c^1}{c^2}\)

Thus, the given planes are parallel to each other.

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(d)The equations of the planes are 2x-y+3z-1=0 and 2x-y+3z+3=0

Here,
a₁=2, b₁=-1, c₁=3 and
a₂=2, b₂=-1, c₂=3
\(∴\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Thus, the given lines are parallel to each other.

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(e)The equations of the given planes are 4x+8y+z=0 and y+z-4=0

Here,
a₁=4, b₁=8, c₁=1 and
a₂=0, b2=1, c2=1

a₁a2+b1b2+c₁c₂
=4×0+8×1+1
=9≠0

Therefore, the given lines are not perpendicular to each other.
\(∴\frac{a_1}{a_2}≠\frac{b_1}{b_2}≠\frac{c_1}{c_2}\)

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by 45⁰