In the following cases, determine whether the given planes are pa… - Vidyadip

Question

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

✓

Answer

The direction ratios of normal to the plane, L₁: a₁x + b₁y + c₁z = 0,
are a₁, b₁, c₁ and L₂: a₂x + b₂y + c₂z = 0 are a₂, b₂, c₂
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between L₁ and L₂ is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the planes are 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)
$=-44\neq0$
Therefore, the given planes are not perpendicular.
$\frac{\text{a}_1}{\text{a}_2}=\frac{7}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{5}{-1}=-5,\ \frac{\text{c}_1}{\text{c}_2}=\frac{6}{-10}=\frac{3}{-5}$
It can be seen that, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2},$
Therefore, the given planes are not parallel.
The angle between them is given by,
$\text{Q}=\cos^{-1}\begin{vmatrix}\frac{7\times3+5\times(-1)+6\times(-10)}{\sqrt{(7)^2+(5)^2+(6)^2\times\sqrt{(3)^2+(-1)^2+(-10)^2}}}\end{vmatrix}$
$=\cos^{-1}\Bigg|\frac{21-5-60}{\sqrt{110}\times\sqrt{110}}\Bigg|$
$=\cos^{-1}\frac{44}{110}$
$=\cos^{-1}=\frac{2}{5}.$

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