Question:
The Cartesian equation of the plane $\vec{r} = (2\hat{i} - 3\hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$ is \dots}
The Cartesian equation of the plane $\vec{r} = (2\hat{i} - 3\hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$ is \dots}
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You can verify your final Cartesian equation quickly by plugging the original anchor point $(2, -3, 0)$ into it. $5(2) - 3(-3) - 0 = 10 + 9 = 19$. If it doesn't match, you've made a cross-product error!
Updated On: Jun 19, 2026
- $5x - 4y + z = 22$
- $5x - 3y + z = 19$
- $5x - 3y - z = 19$
- $5x - 4y - z = 22$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We are given the vector parametric equation of a plane: $\vec{r} = \vec{a} + \lambda\vec{d_1} + \mu\vec{d_2}$.
We must convert this into the standard Cartesian form $Ax + By + Cz = D$.
Step 2: Key Formula or Approach:
1. Identify the anchor point $\vec{a}$ and the two direction vectors $\vec{d_1}$ and $\vec{d_2}$ that span the plane.
2. The normal vector $\vec{n}$ to the plane is found by taking the cross product of the two direction vectors: $\vec{n} = \vec{d_1} \times \vec{d_2}$.
3. Use the scalar product equation of a plane: $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Step 3: Detailed Explanation:
From the given equation:
Anchor point $\vec{a} = 2\hat{i} - 3\hat{j} + 0\hat{k}$.
Directions: $\vec{d_1} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{d_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.
Calculate the normal vector $\vec{n}$:
$$\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 3 & 1 \end{vmatrix}$$
Expand the determinant:
$$\vec{n} = \hat{i}(2(1) - (-1)(3)) - \hat{j}(1(1) - (-1)(2)) + \hat{k}(1(3) - 2(2))$$
$$\vec{n} = \hat{i}(2 + 3) - \hat{j}(1 + 2) + \hat{k}(3 - 4)$$
$$\vec{n} = 5\hat{i} - 3\hat{j} - \hat{k}$$
Now, apply $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:
LHS: $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (5\hat{i} - 3\hat{j} - \hat{k}) = 5x - 3y - z$
RHS: $(2\hat{i} - 3\hat{j} + 0\hat{k}) \cdot (5\hat{i} - 3\hat{j} - \hat{k}) = (2)(5) + (-3)(-3) + (0)(-1) = 10 + 9 = 19$
Equating LHS and RHS:
$$5x - 3y - z = 19$$
Step 4: Final Answer:
The Cartesian equation is $5x - 3y - z = 19$, matching option (c).
We are given the vector parametric equation of a plane: $\vec{r} = \vec{a} + \lambda\vec{d_1} + \mu\vec{d_2}$.
We must convert this into the standard Cartesian form $Ax + By + Cz = D$.
Step 2: Key Formula or Approach:
1. Identify the anchor point $\vec{a}$ and the two direction vectors $\vec{d_1}$ and $\vec{d_2}$ that span the plane.
2. The normal vector $\vec{n}$ to the plane is found by taking the cross product of the two direction vectors: $\vec{n} = \vec{d_1} \times \vec{d_2}$.
3. Use the scalar product equation of a plane: $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Step 3: Detailed Explanation:
From the given equation:
Anchor point $\vec{a} = 2\hat{i} - 3\hat{j} + 0\hat{k}$.
Directions: $\vec{d_1} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{d_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.
Calculate the normal vector $\vec{n}$:
$$\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 3 & 1 \end{vmatrix}$$
Expand the determinant:
$$\vec{n} = \hat{i}(2(1) - (-1)(3)) - \hat{j}(1(1) - (-1)(2)) + \hat{k}(1(3) - 2(2))$$
$$\vec{n} = \hat{i}(2 + 3) - \hat{j}(1 + 2) + \hat{k}(3 - 4)$$
$$\vec{n} = 5\hat{i} - 3\hat{j} - \hat{k}$$
Now, apply $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:
LHS: $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (5\hat{i} - 3\hat{j} - \hat{k}) = 5x - 3y - z$
RHS: $(2\hat{i} - 3\hat{j} + 0\hat{k}) \cdot (5\hat{i} - 3\hat{j} - \hat{k}) = (2)(5) + (-3)(-3) + (0)(-1) = 10 + 9 = 19$
Equating LHS and RHS:
$$5x - 3y - z = 19$$
Step 4: Final Answer:
The Cartesian equation is $5x - 3y - z = 19$, matching option (c).
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