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Question

(D) The equation of a plane passing through a point $(x_0, y_0, z_0)$ and perpendicular to a normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$

Vedclass · Accepted Answer

$\frac{-4}{5}$

Vedclass · Answer

(D) The equation of a plane passing through a point $(x_0, y_0, z_0)$ and perpendicular to a normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is given by $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$. Given the point $A(-2, 1, 3)$ and the normal vector $\vec{n} = 3\hat{i} + 1\hat{j} + 5\hat{k}$,we have $A=3, B=1, C=5$. Substituting these values into the equation: $3(x - (-2)) + 1(y - 1) + 5(z - 3) = 0$ $3(x + 2) + (y - 1) + 5(z - 3) = 0$ $3x + 6 + y - 1 + 5z - 15 = 0$ $3x + y + 5z - 10 = 0$. Comparing this with $ax + by + cz + d = 0$,we get $a = 3, b = 1, c = 5, d = -10$. Now,calculating the required value: $\frac{a + b}{c + d} = \frac{3 + 1}{5 - 10} = \frac{4}{-5} = -\frac{4}{5}$.

If the equation of the plane passing through the point $A(-2, 1, 3)$ and perpendicular to the vector $3 \hat{i} + \hat{j} + 5 \hat{k}$ is $ax + by + cz + d = 0$,then $\frac{a + b}{c + d} = $

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