Question:
The Cartesian equation of the plane $\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ is
The Cartesian equation of the plane $\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ is
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To convert a parametric plane equation to Cartesian form quickly, you can directly compute the scalar triple product determinant: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0$. Both methods yield the exact same result.
Updated On: Jun 4, 2026
- $x+y+z=0$
- $5x+2y+3z=3$
- $2x+y+z=0$
- $5x-2y-3z-7=0$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We are given the vector equation of a plane in parametric form and need to convert it into its standard Cartesian equation.
Step 2: Key Formula or Approach:
The equation $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$ represents a plane passing through a point with position vector $\vec{a}$ and parallel to vectors $\vec{b}$ and $\vec{c}$.
The normal vector to the plane is $\vec{n} = \vec{b} \times \vec{c}$.
The scalar equation is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$, or $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Step 3: Detailed Explanation:
From the given equation, identify the vectors:
Position vector $\vec{a} = \hat{i} - \hat{j} + 0\hat{k}$
Parallel vector $\vec{b} = \hat{i} + \hat{j} + \hat{k}$
Parallel vector $\vec{c} = \hat{i} - 2\hat{j} + 3\hat{k}$
Calculate the normal vector $\vec{n}$ using the cross product:
$$\vec{n} = \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix}$$ $$\vec{n} = \hat{i}(3 - (-2)) - \hat{j}(3 - 1) + \hat{k}(-2 - 1)$$ $$\vec{n} = 5\hat{i} - 2\hat{j} - 3\hat{k}$$ Now, apply the plane equation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:
$$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = (\hat{i} - \hat{j}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k})$$ $$5x - 2y - 3z = (1)(5) + (-1)(-2) + (0)(-3)$$ $$5x - 2y - 3z = 5 + 2$$ $$5x - 2y - 3z = 7$$ Rearranging into standard general form:
$$5x - 2y - 3z - 7 = 0$$
Step 4: Final Answer:
The Cartesian equation is $5x-2y-3z-7=0$, matching option (D).
We are given the vector equation of a plane in parametric form and need to convert it into its standard Cartesian equation.
Step 2: Key Formula or Approach:
The equation $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$ represents a plane passing through a point with position vector $\vec{a}$ and parallel to vectors $\vec{b}$ and $\vec{c}$.
The normal vector to the plane is $\vec{n} = \vec{b} \times \vec{c}$.
The scalar equation is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$, or $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Step 3: Detailed Explanation:
From the given equation, identify the vectors:
Position vector $\vec{a} = \hat{i} - \hat{j} + 0\hat{k}$
Parallel vector $\vec{b} = \hat{i} + \hat{j} + \hat{k}$
Parallel vector $\vec{c} = \hat{i} - 2\hat{j} + 3\hat{k}$
Calculate the normal vector $\vec{n}$ using the cross product:
$$\vec{n} = \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix}$$ $$\vec{n} = \hat{i}(3 - (-2)) - \hat{j}(3 - 1) + \hat{k}(-2 - 1)$$ $$\vec{n} = 5\hat{i} - 2\hat{j} - 3\hat{k}$$ Now, apply the plane equation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:
$$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = (\hat{i} - \hat{j}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k})$$ $$5x - 2y - 3z = (1)(5) + (-1)(-2) + (0)(-3)$$ $$5x - 2y - 3z = 5 + 2$$ $$5x - 2y - 3z = 7$$ Rearranging into standard general form:
$$5x - 2y - 3z - 7 = 0$$
Step 4: Final Answer:
The Cartesian equation is $5x-2y-3z-7=0$, matching option (D).
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