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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $(x_2, y_2, z_2)$?

Similar Questions

The lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar. Find the equation of the plane in which they lie.

$X$ intercept of the plane containing the line of intersection of the planes $x-2y+z+2=0$ and $3x-y-z+1=0$ and also passing through $(1,1,1)$ is

$A$ point $P$ lies on a line passing through $Q(1, -2, 3)$ and is parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$. If $P$ lies on the plane $2x + 3y - 4z + 22 = 0$, then the length of the segment $PQ$ is:

Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module