Question:
The distance between the planes \(2x+y+2z=8\) and \(4x+2y+4z+5=0\) is
The distance between the planes \(2x+y+2z=8\) and \(4x+2y+4z+5=0\) is
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Before finding distance between parallel planes, make the coefficients of \(x,y,z\) identical.
Updated On: Jul 3, 2026
- \(\dfrac{7}{2}\)
- \(\dfrac{5}{2}\)
- \(\dfrac{3}{2}\)
- \(\dfrac{1}{2}\)
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The Correct Option is A
Solution and Explanation
Concept:
The distance between two parallel planes \[ ax+by+cz+d_1=0 \] and \[ ax+by+cz+d_2=0 \] is \[ \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}} \]
Step 1: Write both planes in the same normal form. First plane: \[ 2x+y+2z=8 \] \[ 2x+y+2z-8=0 \] Second plane: \[ 4x+2y+4z+5=0 \] Divide by \(2\): \[ 2x+y+2z+\frac{5}{2}=0 \]
Step 2: Identify constants. \[ d_1=-8,\qquad d_2=\frac{5}{2} \] and \[ a=2,\quad b=1,\quad c=2 \]
Step 3: Apply distance formula. \[ \text{Distance} = \frac{\left|-8-\frac{5}{2}\right|}{\sqrt{2^2+1^2+2^2}} \] \[ = \frac{\left|-\frac{16}{2}-\frac{5}{2}\right|}{\sqrt{4+1+4}} \] \[ = \frac{\frac{21}{2}}{3} \] \[ = \frac{7}{2} \]
Step 4: Final answer. \[ \boxed{\frac{7}{2}} \]
The distance between two parallel planes \[ ax+by+cz+d_1=0 \] and \[ ax+by+cz+d_2=0 \] is \[ \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}} \]
Step 1: Write both planes in the same normal form. First plane: \[ 2x+y+2z=8 \] \[ 2x+y+2z-8=0 \] Second plane: \[ 4x+2y+4z+5=0 \] Divide by \(2\): \[ 2x+y+2z+\frac{5}{2}=0 \]
Step 2: Identify constants. \[ d_1=-8,\qquad d_2=\frac{5}{2} \] and \[ a=2,\quad b=1,\quad c=2 \]
Step 3: Apply distance formula. \[ \text{Distance} = \frac{\left|-8-\frac{5}{2}\right|}{\sqrt{2^2+1^2+2^2}} \] \[ = \frac{\left|-\frac{16}{2}-\frac{5}{2}\right|}{\sqrt{4+1+4}} \] \[ = \frac{\frac{21}{2}}{3} \] \[ = \frac{7}{2} \]
Step 4: Final answer. \[ \boxed{\frac{7}{2}} \]
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