CUET Preparation Today
Distance between two parallel plane \(2x+y+2z=8\) and \(4x+2y+4z+5=0\) is |
\(\frac{5}{2}\) \(\frac{7}{2}\) \(\frac{9}{2}\) \(\frac{3}{2}\) |
\(\frac{7}{2}\) |
\(2x+y+2z=8\) \(2x+y+2z=-\frac{5}{2}\) $D=\frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$ $=\frac{\left|8+\frac{5}{2}\right|}{\sqrt{9}}=\frac{21}{6}=\frac{7}{2}$ |
