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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Use point-normal form and substitute coordinates directly to get the plane equation.
Updated On: May 19, 2025
- $4/5$
- $2/3$
- $1$
- $-4/5$
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The Correct Option is D
Solution and Explanation
Normal vector to the plane is $\vec{n} = \langle 3, 1, 5 \rangle$
So plane equation: $3(x + 2) + 1(y - 1) + 5(z - 3) = 0$
$3x + 6 + y - 1 + 5z - 15 = 0 \Rightarrow 3x + y + 5z - 10 = 0$
Thus, $a = 3$, $b = 1$, $c = 5$, $d = -10$
Then: $\dfrac{a + b}{c + d} = \dfrac{3 + 1}{5 - 10} = \dfrac{4}{-5} = -\dfrac{4}{5}$
So plane equation: $3(x + 2) + 1(y - 1) + 5(z - 3) = 0$
$3x + 6 + y - 1 + 5z - 15 = 0 \Rightarrow 3x + y + 5z - 10 = 0$
Thus, $a = 3$, $b = 1$, $c = 5$, $d = -10$
Then: $\dfrac{a + b}{c + d} = \dfrac{3 + 1}{5 - 10} = \dfrac{4}{-5} = -\dfrac{4}{5}$
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