In what ratio does the plane ax + by + cz + d | vedclass

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Question

(A) Let the plane $ax + by + cz + d = 0$ divide the line segment joining the points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $k:1$. Usin

Vedclass · Accepted Answer

$-\frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 + d}$

Vedclass · Answer

(A) Let the plane $ax + by + cz + d = 0$ divide the line segment joining the points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $k:1$. Using the section formula,the coordinates of the point of division $P$ are: $P = \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1}, \frac{kz_2 + z_1}{k+1} \right)$. Since this point $P$ lies on the plane $ax + by + cz + d = 0$,it must satisfy the equation: $a\left( \frac{kx_2 + x_1}{k+1} \right) + b\left( \frac{ky_2 + y_1}{k+1} \right) + c\left( \frac{kz_2 + z_1}{k+1} \right) + d = 0$. Multiplying by $(k+1)$,we get: $a(kx_2 + x_1) + b(ky_2 + y_1) + c(kz_2 + z_1) + d(k+1) = 0$. Rearranging the terms to solve for $k$: $k(ax_2 + by_2 + cz_2 + d) + (ax_1 + by_1 + cz_1 + d) = 0$. $k(ax_2 + by_2 + cz_2 + d) = -(ax_1 + by_1 + cz_1 + d)$. $k = -\frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 + d}$. Thus,the plane divides the line segment in the ratio $-\frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 + d} : 1$.

In what ratio does the plane $ax + by + cz + d = 0$ divide the line segment joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$?

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