The Cartesian equation of the plane {{r}}=(2 {{i}}-3 {{j}})+λ({{i}}+2 {{j}}-{{k}})+(2 {{i}}+3 {{j}}+{{k}}) is

Correct Answer is : 5x−3y−z=195 x-3 y-z=195x−3y−z=19

Correct Answer is : 5x−3y−z=195 x-3 y-z=195x−3y−z=19

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MHTCET 25 Apr 2025 Shift 2 Paper

Question : 124 of 150

Marks: +1, -0

\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})

Solution:

Concept:

A plane given in parametric form can be converted to Cartesian form using a normal vector obtained by the cross product of the two direction vectors.

Explanation:

The plane passes through the point

(2, -3, 0)

The direction vectors are

\vec{p} = \hat{i} + 2\hat{j} - \hat{k}

and

\vec{q} = 2\hat{i} + 3\hat{j} + \hat{k}

The normal vector is:

\vec{n} = \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 3 & 1 \end{vmatrix}

\vec{n} = \hat{i}(2+3) - \hat{j}(1+2) + \hat{k}(3-4) = 5\hat{i} - 3\hat{j} - \hat{k}

The plane equation is

\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}

, where

\vec{a} = 2\hat{i} - 3\hat{j}

\vec{r} \cdot (5\hat{i} - 3\hat{j} - \hat{k}) = (2\hat{i} - 3\hat{j}) \cdot (5\hat{i} - 3\hat{j} - \hat{k})

\Rightarrow \vec{r} \cdot (5\hat{i} - 3\hat{j} - \hat{k}) = 10 + 9 = 19

Thus, the Cartesian equation is

5x - 3y - z = 19

Answer:

Option C:

5x - 3y - z = 19

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