First slide

Planes in 3D

Question

The Cartesian equation of the plane ${\vec{r} = (1 + λ - μ) \hat{i} + (2 - λ) \hat{j} + (3 - 2 λ + 2 μ) \hat{k} is}_{}$

Moderate

A

2x+y=5
B

2x-y=5
C

2x+z=5
D

2x-z=5

Solution

Given plane is
$\begin{aligned} \vec{r} = (1 + λ - μ) \hat{i} + (2 - λ) \hat{j} + (3 - 2 λ + 2 μ) \hat{k} \\ = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} - 2 \hat{k}) + μ (- \hat{i} + 2 \hat{k}) \end{aligned}$
which is a plane passing through $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
and parallel to the vectors $\vec{b} = \hat{i} - \hat{j} - 2 \hat{k}$ and $\vec{c} = - \hat{i} + 2 \hat{k}$
Therefore, it is perpendicular to the vector
$\vec{n} = \vec{b} \times \vec{c} = - 2 \hat{i} - \hat{k}$
Hence, equation of plane is - 2(x - 1) + (0)(y - 2)(z-3) = 0 or 2x+z = 5

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