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

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

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

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

(C) $\vec{r} \cdot(\hat{i}-7 \hat{j}+3 \hat{k})=\frac{7}{3}$

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Three Dimensional Geometry

The vector equation of the plane passing through the intersection of the planes

r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1

and

r \cdot (\hat{i} - 2 \hat{j}) = - 2

, and the point

(1, 0, 2)

is:

Answer: (D)

A plane through the line of intersection has the form

(x + y + z - 1) + λ (x - 2 y + 2) = 0

. Putting

(1, 0, 2)

into it gives

(1 + 0 + 2 - 1) + λ (1 - 0 + 2) = 0

, so

2 + 3 λ = 0

and

λ = - \frac{2}{3}

. The plane is then

(x + y + z - 1) - \frac{2}{3} (x - 2 y + 2) = 0

, and multiplying by

3

gives

3 x + 3 y + 3 z - 3 - 2 x + 4 y - 4 = 0

, that is

x + 7 y + 3 z = 7

. In vector form the plane is

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

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option