Find the equation of the plane which contains | vedclass

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

Question

(A) The equations of the given planes are:$\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$  $(1)$$\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$  $(2)$

Vedclass · Accepted Answer

$\vec{r} \cdot(33 \hat{i}+45 \hat{j}+50 \hat{k})-41=0$

Vedclass · Answer

(A) The equations of the given planes are:$\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$  $(1)$$\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$  $(2)$The equation of the plane passing through the line of intersection of planes $(1)$ and $(2)$ is given by:$[\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4] + \lambda[\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5] = 0$$\vec{r} \cdot[(1+2\lambda) \hat{i} + (2+\lambda) \hat{j} + (3-\lambda) \hat{k}] + (5\lambda-4) = 0$  $(3)$Since this plane is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$,the dot product of their normal vectors is zero:$5(1+2\lambda) + 3(2+\lambda) - 6(3-\lambda) = 0$$5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$$19\lambda - 7 = 0 \Rightarrow \lambda = \frac{7}{19}$Substituting $\lambda = \frac{7}{19}$ into equation $(3)$:$\vec{r} \cdot[(1 + \frac{14}{19}) \hat{i} + (2 + \frac{7}{19}) \hat{j} + (3 - \frac{7}{19}) \hat{k}] + (5(\frac{7}{19}) - 4) = 0$$\vec{r} \cdot[\frac{33}{19} \hat{i} + \frac{45}{19} \hat{j} + \frac{50}{19} \hat{k}] + (\frac{35-76}{19}) = 0$$\vec{r} \cdot(33 \hat{i} + 45 \hat{j} + 50 \hat{k}) - 41 = 0$

Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$.

Similar Questions

The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane

If $\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{k}+\hat{i}, \hat{i}-\hat{j}, \hat{j}-\hat{k}$ are the position vectors of the points $A, B, C, D, E$ respectively, then the point of intersection of the line $AB$ and the plane passing through $C, D, E$ is

Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^{2}$ is equal to ..... .

If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module