vec{r} cdot(hat{i}-hat{j}+hat{k})=5 and vec{r | vedclass

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

Question

(D) The equations of the given planes are $x-y+z=5$ and $2x+y-z=3$. The equation of any plane passing through the line of intersection of these two pl

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(D) The equations of the given planes are $x-y+z=5$ and $2x+y-z=3$. The equation of any plane passing through the line of intersection of these two planes is given by $(x-y+z-5) + \lambda(2x+y-z-3) = 0$. Since this plane passes through the point $(0, 1, 2)$, we substitute $x=0, y=1, z=2$ into the equation: $(0-1+2-5) + \lambda(2(0)+1-2-3) = 0$. $-4 + \lambda(-4) = 0 \Rightarrow -4\lambda = 4 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ into the equation of the plane: $(x-y+z-5) - 1(2x+y-z-3) = 0$. $x-y+z-5-2x-y+z+3 = 0$. $-x-2y+2z-2 = 0 \Rightarrow -x-2y+2z = 2$. Comparing this with $\vec{r} \cdot(a\hat{i}+b\hat{j}+c\hat{k}) = m$, we get $a=-1, b=-2, c=2$. Therefore, $\frac{bc}{a^2} = \frac{(-2)(2)}{(-1)^2} = \frac{-4}{1} = -4$.

Similar Questions

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle QPR$ is a right angle,then $12a^2$ is equal to

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

The coordinates of the point where the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ meets the plane $x + y - z = 3$ are

Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals...

Vedclass Test Series

Exam Paper Generator

Online Exam Module