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

(A) The direction ratios of the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ are $(2 - (-3)), (3 - 1), (4 - 2)$,which are $5, 2, 2$.Since the

Vedclass · Accepted Answer

$\bar{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 1$

Vedclass · Answer

(A) The direction ratios of the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ are $(2 - (-3)), (3 - 1), (4 - 2)$,which are $5, 2, 2$.Since the plane is perpendicular to this line,the normal vector to the plane is $\vec{n} = 5\hat{i} + 2\hat{j} + 2\hat{k}$.The equation of a plane passing through a point $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ with normal $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Substituting the values,we get $(\vec{r} - (-\hat{i} + 2\hat{j} + \hat{k})) \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 0$.This simplifies to $\vec{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = (-\hat{i} + 2\hat{j} + \hat{k}) \cdot (5\hat{i} + 2\hat{j} + 2\hat{k})$.Calculating the dot product on the right side: $(-1)(5) + (2)(2) + (1)(2) = -5 + 4 + 2 = 1$.Thus,the equation of the plane is $\vec{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 1$.

The equation of the plane passing through the point $(-1, 2, 1)$ and perpendicular to the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ is $.........$

Similar Questions

If the cartesian equation of the plane passing through the point $\hat{i}+2 \hat{j}+\hat{k}$ and parallel to the vectors $2 \hat{i}+3 \hat{j}+\hat{k}$ and $-\hat{i}+2 \hat{j}-3 \hat{k}$ is $a x+b y+c z=1$, then $18(a+b+c)$ is equal to

Consider the three planes $P_{1}: 3x + 15y + 21z = 9$; $P_{2}: x - 3y - z = 5$; and $P_{3}: 2x + 10y + 14z = 5$. Then,which one of the following is true?

The equation of the plane in Cartesian form, which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector drawn from the origin being $2 \hat{i}-3 \hat{j}+4 \hat{k}$, is

The coordinates of the foot of the perpendicular drawn from the origin to a plane is $(2, 4, -3)$. The equation of the plane is

$A$ plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module