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,
$Qcos^-1|\frac{a_1a_2+b_1b_2+c_1c_2}{√a_1^2+b_1^2+c_1^2.√a_2^2+b_2^2+c_2^2}|$

---

(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.

=$cos^{-1}\frac{2}{5}$

---

(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.

---

(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.

---

(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.

---

(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