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 equation of a plane passing through the intersection of two planes $P_1: \overline{r} \cdot \overline{n}_1 = d_1$ and $P_2: \overline{r} \cdot

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(C) The equation of a plane passing through the intersection of two planes $P_1: \overline{r} \cdot \overline{n}_1 = d_1$ and $P_2: \overline{r} \cdot \overline{n}_2 = d_2$ is given by $(\overline{r} \cdot \overline{n}_1 - d_1) + \lambda(\overline{r} \cdot \overline{n}_2 - d_2) = 0$. Given planes are $P_1: \overline{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k}) - 1 = 0$ and $P_2: \overline{r} \cdot(\hat{i}-\hat{j}) + 4 = 0$. The equation of the required plane is $\overline{r} \cdot((2+\lambda) \hat{i} + (-3-\lambda) \hat{j} + 4 \hat{k}) - 1 + 4\lambda = 0$. This plane is perpendicular to $\overline{r} \cdot(2 \hat{i}-\hat{j}+\hat{k}) + 8 = 0$. Therefore,the dot product of their normal vectors is zero: $(2+\lambda)(2) + (-3-\lambda)(-1) + (4)(1) = 0$. $4 + 2\lambda + 3 + \lambda + 4 = 0 \implies 3\lambda + 11 = 0 \implies \lambda = -\frac{11}{3}$. Substituting $\lambda$ into the plane equation: $\overline{r} \cdot((2-\frac{11}{3}) \hat{i} + (-3+\frac{11}{3}) \hat{j} + 4 \hat{k}) - 1 + 4(-\frac{11}{3}) = 0$. $\overline{r} \cdot(-\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + 4 \hat{k}) - 1 - \frac{44}{3} = 0$. Multiply by $3$: $\overline{r} \cdot(-5 \hat{i} + 2 \hat{j} + 12 \hat{k}) - 3 - 44 = 0$. $\overline{r} \cdot(-5 \hat{i} + 2 \hat{j} + 12 \hat{k}) = 47$. Comparing with $\overline{r} \cdot(-5 \hat{i} + 2 \hat{j} + 12 \hat{k}) = \mu$,we get $\mu = 47$.

Similar Questions

The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$. Then the value of $\alpha \beta$ is:

If the three planes $x = 5$,$2x - 5ay + 3z - 2 = 0$,and $3bx + y - 3z = 0$ pass through a common line,then the value of $(a, b)$ is:

If $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+4 \bar{k})=5$ and $\bar{r} \cdot(\bar{i}+\bar{j}-\bar{k})=7$ are two planes and $(16, -9, 0)$ is a point common to both the planes, then the vector equation of the line of intersection of the planes is $\bar{r}=$

Find the equation of the plane passing through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$.

The Cartesian equation of a line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module