Question

Find the acute angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5.

• A. $60°$
• B. $30°$
• C. $\theta ={\cos }^{-1}\left(\frac{5\sqrt{3}}{21}\right)$
• D. $\theta ={\cos }^{-1}\left(\frac{-5\sqrt{3}}{21}\right)$

Answer 1

The correct option is C $\theta ={\cos }^{-1}\left(\frac{5\sqrt{3}}{21}\right)$

Comparing the given equations of the planes with the equations
$A_{1}x+B_{1}y+C_{1}z+D_1=0\mathrm{and}{A}_{2}x+B_{2}y+C_{2}z+D_2=0\mathrm{We}\mathrm{get}{A}_1=3,B_1=-6,C_1=2A_2=2,B_2=2,C_2=-2$
$\mathrm{The}\mathrm{angle}\mathrm{between}\mathrm{these}\mathrm{planes}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{cos\theta }=\left(\frac{A_1\times A_2+B_1\times B_2+C_1\times C_2}{\left(\sqrt{{A_1}^2+{B_1}^2+{C_1}^2}\right)\times \left(\sqrt{{A_2}^2+{B_2}^2+{C_2}^2}\right)}\right|$

$\mathrm{cos\theta }=\left(\frac{3\times 2+\left(-6\right)\times 2+2\times \left(-2\right)}{\left(\sqrt{3^2+{\left(-6\right)}^2+2^2}\right)\times \left(\sqrt{2^2+2^2+{\left(-2\right)}^2}\right)}\right|\mathrm{cos\theta }=\frac{5\sqrt{3}}{21}\theta ={\cos }^{-1}\left(\frac{5\sqrt{3}}{21}\right)$