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

(C) The equation of a plane passing through the intersection of two planes $P_1: \overrightarrow{r} \cdot \overrightarrow{n_1} = d_1$ and $P_2: \overr

Vedclass · Accepted Answer

$\overrightarrow{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$

Vedclass · Answer

(C) The equation of a plane passing through the intersection of two planes $P_1: \overrightarrow{r} \cdot \overrightarrow{n_1} = d_1$ and $P_2: \overrightarrow{r} \cdot \overrightarrow{n_2} = d_2$ is given by $(\overrightarrow{r} \cdot \overrightarrow{n_1} - d_1) + \lambda(\overrightarrow{r} \cdot \overrightarrow{n_2} - d_2) = 0$.Given planes are $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 = 0$ and $\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) + 2 = 0$.The equation of the required plane is $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 + \lambda(\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) + 2) = 0$.Rearranging the terms,we get $\overrightarrow{r} \cdot [\hat{i}(1 + \lambda) + \hat{j}(1 - 2\lambda) + \hat{k}(1)] = 1 - 2\lambda$.Since the plane passes through the point $(1, 0, 2)$,the position vector is $\overrightarrow{r} = \hat{i} + 2\hat{k}$.Substituting this into the equation: $(\hat{i} + 2\hat{k}) \cdot [\hat{i}(1 + \lambda) + \hat{j}(1 - 2\lambda) + \hat{k}(1)] = 1 - 2\lambda$.$(1 + \lambda) + 2(1) = 1 - 2\lambda$.$3 + \lambda = 1 - 2\lambda$.$3\lambda = -2 \implies \lambda = -\frac{2}{3}$.Substituting $\lambda = -\frac{2}{3}$ back into the equation: $\overrightarrow{r} \cdot [\hat{i}(1 - \frac{2}{3}) + \hat{j}(1 + \frac{4}{3}) + \hat{k}(1)] = 1 - 2(-\frac{2}{3})$.$\overrightarrow{r} \cdot [\hat{i}(\frac{1}{3}) + \hat{j}(\frac{7}{3}) + \hat{k}] = 1 + \frac{4}{3} = \frac{7}{3}$.Multiplying by $3$,we get $\overrightarrow{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$.

The vector equation of the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) = -2$,and the point $(1, 0, 2)$ is:

Similar Questions

If $n=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$m=\hat{i}-\hat{j}$,and $l=2 \hat{i}-\hat{j}+\hat{k}$,then the Cartesian equation of the plane passing through the line of intersection of two planes $r \cdot n=1$ and $r \cdot m=-4$ and perpendicular to the plane $r \cdot l=-8$ is

Let $P$ be a plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane $P$,then the value of $|5\alpha|$ is equal to ....... .

Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .

The plane passing through the points $(1, 2, 1), (2, 1, 2)$ and parallel to the line $2x = 3y, z = 1$ also passes through the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module