The coordinates of the foot of the perpendicu | vedclass

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

Question

(D) The equation of the line passing through the origin $(0, 0, 0)$ and perpendicular to the plane $3x + 2y + 6z - 56 = 0$ is given by $\frac{x-0}{3}

Vedclass · Accepted Answer

$\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}\right)$

Vedclass · Answer

(D) The equation of the line passing through the origin $(0, 0, 0)$ and perpendicular to the plane $3x + 2y + 6z - 56 = 0$ is given by $\frac{x-0}{3} = \frac{y-0}{2} = \frac{z-0}{6} = k$.Any point on this line is $(3k, 2k, 6k)$.Since this point lies on the plane,we substitute it into the plane equation:$3(3k) + 2(2k) + 6(6k) = 56$$9k + 4k + 36k = 56$$49k = 56$$k = \frac{56}{49} = \frac{8}{7}$Substituting $k$ back into the coordinates $(3k, 2k, 6k)$:$x = 3 	imes \frac{8}{7} = \frac{24}{7}$$y = 2 	imes \frac{8}{7} = \frac{16}{7}$$z = 6 	imes \frac{8}{7} = \frac{48}{7}$Thus,the foot of the perpendicular is $\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}ight)$.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $3x + 2y + 6z = 56$ are:

Similar Questions

The equation of the plane passing through the point $(1,2,2)$ and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$ is:

The perpendicular distance of the origin from the plane $2x + y - 2z - 18 = 0$ is (in $\text{ units}$)

Let the mirror image of the point $P(1, 3, a)$ with respect to the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - b = 0$ be $Q(-3, 5, 2)$. Then the value of $|a + b|$ is equal to ......

If the foot of the perpendicular drawn from the origin to a plane is $P(-1, -1, 2)$,then the equation of the plane is

Find the vector and Cartesian equations of the plane that passes through the point $(1, 0, -2)$ and the normal to the plane is $\hat{i} + \hat{j} - \hat{k}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module