The equation of the plane passing through the | vedclass

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

Question

(C) The normal vector $\vec{n}$ to the required plane is the cross product of the normals of the two given planes,$\vec{n_1} = (3, 1, -2)$ and $\vec{n

Vedclass · Accepted Answer

$11x + y + 17z + 38 = 0$

Vedclass · Answer

(C) The normal vector $\vec{n}$ to the required plane is the cross product of the normals of the two given planes,$\vec{n_1} = (3, 1, -2)$ and $\vec{n_2} = (2, -5, -1)$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -2 \ 2 & -5 & -1 \end{vmatrix} = \hat{i}(-1 - 10) - \hat{j}(-3 - (-4)) + \hat{k}(-15 - 2) = -11\hat{i} - \hat{j} - 17\hat{k}$.We can take the normal vector as $\vec{n} = (11, 1, 17)$.The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.Substituting the point $(1, 2, -3)$ and the normal vector $(11, 1, 17)$:$11(x - 1) + 1(y - 2) + 17(z + 3) = 0$$11x - 11 + y - 2 + 17z + 51 = 0$$11x + y + 17z + 38 = 0$.

The equation of the plane passing through the point $(1, 2, -3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$ is:

Similar Questions

If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$, then $f'(1)$ is equal to:

In which plant group is the structure shown below found?

Which of the following produces hydrolith with dihydrogen?

$A$ concave mirror gives an image three times as large as the object placed at a distance of $20 \ cm$ from it. For the image to be real,the focal length should be ...... $cm$.

What is the correct order of occurrence ($\%$ by weight) in air of $Ne, Ar$ and $Kr$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module