Question
Find the angle between the plane:
2x - 3y + 4z = 1 and -x + y = 4
2x - 3y + 4z = 1 and -x + y = 4
✓
Answer
We know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x - 3y + 4z = 1 and -x + y + 0z = 4 is given by
$\cos\theta=\frac{(2)(-1)+(-3)(1)+(4)(0)}{\sqrt{2^2+(-3)^2+4^2}\sqrt{(-1)^2+1^2+0^2}}$
$=\frac{-2-3+0}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x - 3y + 4z = 1 and -x + y + 0z = 4 is given by
$\cos\theta=\frac{(2)(-1)+(-3)(1)+(4)(0)}{\sqrt{2^2+(-3)^2+4^2}\sqrt{(-1)^2+1^2+0^2}}$
$=\frac{-2-3+0}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$
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