The vector equation of the plane passing thro | vedclass

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

Question

(B) The equation of a plane passing through the line of intersection of the planes $r \cdot a = \lambda$ and $r \cdot b = \mu$ is given by the family

Vedclass · Accepted Answer

$r \cdot (\lambda b - \mu a) = 0$

Vedclass · Answer

(B) The equation of a plane passing through the line of intersection of the planes $r \cdot a = \lambda$ and $r \cdot b = \mu$ is given by the family of planes equation:$(r \cdot a - \lambda) + k(r \cdot b - \mu) = 0$Rearranging the terms,we get:$r \cdot (a + kb) = \lambda + k\mu$ .....$(i)$Since this plane passes through the origin,the position vector $r = 0$ must satisfy the equation:$0 \cdot (a + kb) = \lambda + k\mu$$0 = \lambda + k\mu$$k = -\frac{\lambda}{\mu}$Substituting the value of $k$ into equation $(i)$:$r \cdot (a - \frac{\lambda}{\mu} b) = \lambda + (-\frac{\lambda}{\mu})\mu$$r \cdot (\frac{\mu a - \lambda b}{\mu}) = 0$Multiplying by $\mu$,we get:$r \cdot (\mu a - \lambda b) = 0$Which is equivalent to:$r \cdot (\lambda b - \mu a) = 0$

The vector equation of the plane passing through the origin and the line of intersection of the planes $r \cdot a = \lambda$ and $r \cdot b = \mu$ is

Similar Questions

If $a, b, c$ are the intercepts made on $X, Y, Z$-axes respectively by the plane passing through the points $(1, 0, -2), (3, -1, 2)$ and $(0, -3, 4)$, then $3a + 4b + 7c =$

Find the equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z - 1 = 0$.

The equation of the plane passing through the points $(2,3,1)$ and $(4,-5,3)$ and parallel to the $y$-axis is:

The equation $xy = 0$ in three-dimensional space represents

The plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$ meets the coordinate axes at points $A, B,$ and $C$. The centroid of the triangle $ABC$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module