The equation of the plane passing through (-2 | vedclass

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

Question

(A) Let the equation of the plane be $A(x + 2) + B(y - 2) + C(z - 2) = 0$. Since it passes through $(2, -2, -2)$,we have $A(2 + 2) + B(-2 - 2) + C(-2

Vedclass · Accepted Answer

$5x + 3y + 2z = 0$

Vedclass · Answer

(A) Let the equation of the plane be $A(x + 2) + B(y - 2) + C(z - 2) = 0$. Since it passes through $(2, -2, -2)$,we have $A(2 + 2) + B(-2 - 2) + C(-2 - 2) = 0$,which simplifies to $4A - 4B - 4C = 0$,or $A - B - C = 0$ ... $(i)$. The plane is perpendicular to $9x - 13y - 3z = 0$,so the normal vector of the required plane $(A, B, C)$ is perpendicular to the normal vector of the given plane $(9, -13, -3)$. Thus,$9A - 13B - 3C = 0$ ... $(ii)$. Solving $(i)$ and $(ii)$ using cross product: $\frac{A}{(-1)(-3) - (-1)(-13)} = \frac{B}{(-1)(9) - (1)(-3)} = \frac{C}{(1)(-13) - (-1)(9)}$ $\frac{A}{3 - 13} = \frac{B}{-9 + 3} = \frac{C}{-13 + 9}$ $\frac{A}{-10} = \frac{B}{-6} = \frac{C}{-4}$ Simplifying the ratios,we get $A:B:C = 5:3:2$. Substituting these into the plane equation: $5(x + 2) + 3(y - 2) + 2(z - 2) = 0$. $5x + 10 + 3y - 6 + 2z - 4 = 0$. $5x + 3y + 2z = 0$.

The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

Similar Questions

The length of the foot of the perpendicular from the point $P\left(1, \frac{3}{2}, 2\right)$ to the plane $2x - 2y + 4z + 17 = 0$ is

If from a point $P(a, b, c)$ perpendiculars $PA$ and $PB$ are drawn to the $yz$-plane and $zx$-plane respectively,then the equation of the plane $OAB$ (where $O$ is the origin) is:

$A$ point on the plane determined by the points $A(1,1,-1)$, $B(2,-1,0)$, and $C(-1,0,2)$ among the following is:

$A$ plane meets the coordinate axes at $P, Q,$ and $R$ such that the position vector of the centroid of $\Delta PQR$ is $2i - 5j + 8k$. Then the equation of the plane is:

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 ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module