First slide

Planes in 3D

Question

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

Easy

A

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

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

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

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

Solution

The equation of the plane passing through the line of intersection of two planes    $x + y + z = 1$ and $x - 2 y + 2 = 0$   is $(x + y + z - 1) + λ (x - 2 y + 2) = 0$
          It is passing through the point $(1, 0, 2)$
          Hence, $(2) + λ (3) = 0 = λ = \frac{- 2}{3}$
          Therefore, the equation of the required plane is $x + 7 y + 3 z - 7 = 0$ , this can be written as $\vec{r} \cdot (i + 7 j + 3 k) = 7$

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