Question:medium

The equation of the plane which is parallel to \(x+3y+5z+1=0\) and \(5\) units from the origin is

Show Hint

A plane parallel to \(ax+by+cz+d=0\) has the same \(a,b,c\), and distance from origin is \(\frac{|d|}{\sqrt{a^2+b^2+c^2}}\).
  • \(x+3y+5z+5=0\)
  • \(x+3y+5z+5\sqrt{35}=0\)
  • \(x+3y+5z-5\sqrt{35}=0\)
  • \(x+3y+5z-3\sqrt{35}=0\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
A plane parallel to a given plane $Ax+By+Cz+D=0$ will have the same normal vector, so its equation will be of the form $Ax+By+Cz+k=0$, where $k$ is a constant. We can find $k$ using the given distance from the origin.

Step 2: Key Formula or Approach:

The equation of any plane parallel to $x+3y+5z+1=0$ is $x+3y+5z+k=0$.
The distance of a plane $Ax+By+Cz+D=0$ from the origin $(0,0,0)$ is given by the formula:
\[ \text{Distance} = \frac{|D|}{\sqrt{A^2+B^2+C^2}} \]

Step 3: Detailed Explanation:

The family of parallel planes is given by $x+3y+5z+k=0$.
Here, $A=1, B=3, C=5$, and the constant term is $D=k$.
The distance from the origin is given as 5 units. Using the formula:
\[ 5 = \frac{|k|}{\sqrt{1^2+3^2+5^2}} \] \[ 5 = \frac{|k|}{\sqrt{1+9+25}} \] \[ 5 = \frac{|k|}{\sqrt{35}} \] Now, we solve for $k$:
\[ |k| = 5\sqrt{35} \] This gives two possible values for $k$: $k = 5\sqrt{35}$ and $k = -5\sqrt{35}$.
So, there are two planes that satisfy the conditions:
Plane 1: $x+3y+5z+5\sqrt{35}=0$
Plane 2: $x+3y+5z-5\sqrt{35}=0$
Looking at the options, option (B) is $x+3y+5z+5\sqrt{35}=0$, which is one of the correct solutions. Option (C) is the other valid solution. Since (B) is listed and marked as correct, we choose it.

Step 4: Final Answer:

The equation of the plane is $x+3y+5z+5\sqrt{35}=0$.
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