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 \(\dfrac{a+b}{c+d}=\)

• A. \(\dfrac{4}{5}\)
• B. \(\dfrac{2}{3}\)
• C. \(1\)
• D. \(-\dfrac{4}{5}\)

Hint:

If a plane is perpendicular to a given vector, then that vector acts as the normal vector of the plane.

Answer 1

The Correct Option is D

Step 1: Spot the normal vector.
The plane is perpendicular to $3\vec i+\vec j+5\vec k$, so this is the normal $\vec n=(3,1,5)$.
Step 2: Write the point-normal form.
A plane through $A(-2,1,3)$ with normal $(3,1,5)$ is \[ 3(x+2)+1(y-1)+5(z-3)=0. \] Step 3: Expand the equation.
Expanding, $3x+6+y-1+5z-15=0$, that is $3x+y+5z-10=0$.
Step 4: Read off the coefficients.
Comparing with $ax+by+cz+d=0$ gives $a=3$, $b=1$, $c=5$, $d=-10$.
Step 5: Form the required ratio.
Then \[ \frac{a+b}{c+d}=\frac{3+1}{5+(-10)}=\frac{4}{-5}. \] Step 6: Final answer.
Hence $\dfrac{a+b}{c+d}=-\dfrac{4}{5}$.
\[ \boxed{-\dfrac{4}{5}} \]