The length of the perpendicular from the orig | vedclass

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

Question

(D) The equation of the plane is $3x + 4y + 12z - 52 = 0$. The length of the perpendicular $p$ from a point $(x_1, y_1, z_1)$ to the plane $Ax + By +

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The equation of the plane is $3x + 4y + 12z - 52 = 0$. The length of the perpendicular $p$ from a point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by the formula $p = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Here,the point is the origin $(0, 0, 0)$,so $x_1 = 0, y_1 = 0, z_1 = 0$. Substituting the values,we get $p = \frac{|3(0) + 4(0) + 12(0) - 52|}{\sqrt{3^2 + 4^2 + 12^2}}$. $p = \frac{|-52|}{\sqrt{9 + 16 + 144}} = \frac{52}{\sqrt{169}} = \frac{52}{13} = 4$. Since the value $4$ is not among the options $A, B, C$,the correct option is $D$.

The length of the perpendicular from the origin to the plane $3x + 4y + 12z = 52$ is

Similar Questions

Find the equation of the plane passing through $(a, b, c)$ and parallel to the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$.

The equation of the plane passing through the point $(2, 3, 4)$ and parallel to the plane $x + 2y + 4z = 5$ is:

If $M$ denotes the midpoint of the line segment joining $A(4, 5, -10)$ and $B(-1, 2, 1)$, then the equation of the plane passing through $M$ and perpendicular to $AB$ is:

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

If the given planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are mutually perpendicular,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module