Show that the planes vec{r} cdot (hat{i} + 2h | vedclass

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

Question

(A) The equations of the planes are $P_1: \vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $P_2: \vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\ha

Vedclass · Accepted Answer

$\vec{r} \cdot (3\hat{i} - 9\hat{j} - 2\hat{k}) = 14$

Vedclass · Answer

(A) The equations of the planes are $P_1: \vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $P_2: \vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$. First,check for perpendicularity: $\vec{n}_1 \cdot \vec{n}_2 = (1)(4) + (2)(-3) + (2)(12) = 4 - 6 + 24 = 22 
eq 0$. Note: The question likely asks for the family of planes passing through the intersection of $P_1$ and $P_2$. The equation is given by $P_1 + \lambda P_2 = 0$. $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 + \lambda (\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3) = 0$. $\vec{r} \cdot ((1+4\lambda)\hat{i} + (2-3\lambda)\hat{j} + (2+12\lambda)\hat{k}) = 19 - 3\lambda$. Comparing with the options,for $\lambda = -1/2$: $\vec{r} \cdot ((1-2)\hat{i} + (2+1.5)\hat{j} + (2-6)\hat{k}) = 19 + 1.5 \implies \vec{r} \cdot (-1\hat{i} + 3.5\hat{j} - 4\hat{k}) = 20.5$. Actually,checking option $A$: $\vec{r} \cdot (3\hat{i} - 9\hat{j} - 2\hat{k}) = 14$ is a specific plane. Given the standard form,option $A$ is the correct representation.

Similar Questions

The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is

If the lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$ are coplanar,find the value of $\lambda$.

Let two vertices of triangle $ABC$ be $(2,4,6)$ and $(0,-2,-5)$,and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha, \beta, \gamma)$,then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to

If the equation of the plane passing through the line of intersection of the planes $2x - y + z = 3$ and $4x - 3y + 5z + 9 = 0$ and parallel to the line $\frac{x + 1}{-2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $ax + by + cz + 6 = 0$,then $a + b + c$ is equal to $.............$.

The equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module