Question

Equation of the plane passing through the point $(2, 0, 5)$ and parallel to the vectors $\hat{i} - \hat{j} + \hat{k}$ and $3\hat{i} + 2\hat{j} - \hat{k}$ is

• A. $x - 4y - z + 3 = 0$
• B. $x + 4y + 5z - 27 = 0$
• C. $x - 4y - 5z + 23 = 0$
• D. $x - 4y + z - 7 = 0$

Hint:

Save time by plugging the given point $(2, 0, 5)$ directly into the options to eliminate incorrect choices!
Let's check option (C): $2 - 4(0) - 5(5) + 23 = 2 - 25 + 23 = 0$. Since it satisfies the equation, it is a highly viable candidate. Checking option (A): $2 - 0 - 5 + 3 = 0$, which also works. To break the tie, verify which option's coefficients are orthogonal to $\hat{i}-\hat{j}+\hat{k}$ via a fast mental dot product: $(1)(1) + (-4)(-1) + (-5)(1) = 1 + 4 - 5 = 0$. This confirms option (C) instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to determine the Cartesian equation of a flat plane that passes through a given point $(x_1, y_1, z_1) = (2, 0, 5)$ and runs completely parallel to two non-parallel directional vectors.

Step 2: Key Formula or Approach:
The normal vector $\bar{n}$ of a plane is perpendicular to any vector lying along or parallel to that plane. Since the plane is parallel to vectors $\bar{u} = \hat{i} - \hat{j} + \hat{k}$ and $\bar{v} = 3\hat{i} + 2\hat{j} - \hat{k}$, the normal vector can be derived via their cross product:
$$\bar{n} = \bar{u} \times \bar{v}$$ Once the normal vector components $\bar{n} = A\hat{i} + B\hat{j} + C\hat{k}$ are established, the structural equation of the plane is given by:
$$A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$$

Step 3: Detailed Explanation:
Let's find the cross product using the determinant method:
$$\bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & 2 & -1 \end{vmatrix}$$ Expanding along the first row:
$$\bar{n} = \hat{i}\big((-1)(-1) - (1)(2)\big) - \hat{j}\big((1)(-1) - (1)(3)\big) + \hat{k}\big((1)(2) - (-1)(3)\big)$$ $$\bar{n} = \hat{i}(1 - 2) - \hat{j}(-1 - 3) + \hat{k}(2 + 3)$$ $$\bar{n} = -\hat{i} + 4\hat{j} + 5\hat{k}$$ We can multiply the entire normal vector by $-1$ to simplify our leading coefficient: $\bar{n}' = \hat{i} - 4\hat{j} - 5\hat{k}$. Thus, $A = 1, B = -4, C = -5$.
Now, substitute these coefficients and the point coordinates $(2,0,5)$ into the plane formula:
$$1(x - 2) - 4(y - 0) - 5(z - 5) = 0$$ $$x - 2 - 4y - 5z + 25 = 0$$ Combine the constant scalar terms:
$$x - 4y - 5z + 23 = 0$$

Step 4: Final Answer:
The equation of the plane is $x - 4y - 5z + 23 = 0$, matching option (C).