Question:
If the equation of the plane passing through the point \(A(-2,1,3)\) and perpendicular to the vector \(3\vec{i}+\vec{j}+5\vec{k}\) is \(ax+by+cz+d=0\), then \(\dfrac{a+b}{c+d}=\)
If the equation of the plane passing through the point \(A(-2,1,3)\) and perpendicular to the vector \(3\vec{i}+\vec{j}+5\vec{k}\) is \(ax+by+cz+d=0\), then \(\dfrac{a+b}{c+d}=\)
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If a plane is perpendicular to a given vector, then that vector acts as the normal vector of the plane.
Updated On: Jun 22, 2026
- \(\dfrac{4}{5}\)
- \(\dfrac{2}{3}\)
- \(1\)
- \(-\dfrac{4}{5}\)
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The Correct Option is D
Solution and Explanation
Step 1: Identify the normal vector of the plane.
The plane is perpendicular to the vector
\[ 3\vec{i}+\vec{j}+5\vec{k} \] Therefore, this vector is normal to the plane.
So, the normal vector of the plane is
\[ \vec{n}=3\vec{i}+\vec{j}+5\vec{k} \]
Step 2: Compare with the general equation of a plane.
The general equation of a plane is
\[ ax+by+cz+d=0 \] Its normal vector is
\[ a\vec{i}+b\vec{j}+c\vec{k} \] Therefore, comparing with \(3\vec{i}+\vec{j}+5\vec{k}\), we get
\[ a=3,\quad b=1,\quad c=5 \]
Step 3: Use the point \(A(-2,1,3)\).
Since the plane passes through \(A(-2,1,3)\), substitute \(x=-2,\;y=1,\;z=3\) in
\[ 3x+y+5z+d=0 \] Thus,
\[ 3(-2)+1+5(3)+d=0 \] \[ -6+1+15+d=0 \] \[ 10+d=0 \] \[ d=-10 \]
Step 4: Calculate \(\dfrac{a+b}{c+d}\).
Now,
\[ a=3,\quad b=1,\quad c=5,\quad d=-10 \] Therefore,
\[ \frac{a+b}{c+d} = \frac{3+1}{5+(-10)} \] \[ = \frac{4}{-5} \] \[ = -\frac{4}{5} \]
Step 5: Final conclusion.
Hence,
\[ \boxed{-\frac{4}{5}} \]
The plane is perpendicular to the vector
\[ 3\vec{i}+\vec{j}+5\vec{k} \] Therefore, this vector is normal to the plane.
So, the normal vector of the plane is
\[ \vec{n}=3\vec{i}+\vec{j}+5\vec{k} \]
Step 2: Compare with the general equation of a plane.
The general equation of a plane is
\[ ax+by+cz+d=0 \] Its normal vector is
\[ a\vec{i}+b\vec{j}+c\vec{k} \] Therefore, comparing with \(3\vec{i}+\vec{j}+5\vec{k}\), we get
\[ a=3,\quad b=1,\quad c=5 \]
Step 3: Use the point \(A(-2,1,3)\).
Since the plane passes through \(A(-2,1,3)\), substitute \(x=-2,\;y=1,\;z=3\) in
\[ 3x+y+5z+d=0 \] Thus,
\[ 3(-2)+1+5(3)+d=0 \] \[ -6+1+15+d=0 \] \[ 10+d=0 \] \[ d=-10 \]
Step 4: Calculate \(\dfrac{a+b}{c+d}\).
Now,
\[ a=3,\quad b=1,\quad c=5,\quad d=-10 \] Therefore,
\[ \frac{a+b}{c+d} = \frac{3+1}{5+(-10)} \] \[ = \frac{4}{-5} \] \[ = -\frac{4}{5} \]
Step 5: Final conclusion.
Hence,
\[ \boxed{-\frac{4}{5}} \]
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