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: Understand the plane.
The plane passes through $(2,0,5)$ and is parallel to two vectors $\bar{u} = \hat{i} - \hat{j} + \hat{k}$ and $\bar{v} = 3\hat{i} + 2\hat{j} - \hat{k}$.
Step 2: Get the normal direction.
The plane's normal is at right angles to both vectors that lie in it. So the normal is their cross product $\bar{n} = \bar{u}\times\bar{v}$.
Step 3: Compute the cross product.
\[ \bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & 2 & -1 \end{vmatrix} \] \[ = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(3)) + \hat{k}((1)(2) - (-1)(3)) \] \[ = \hat{i}(1-2) - \hat{j}(-1-3) + \hat{k}(2+3) = -\hat{i} + 4\hat{j} + 5\hat{k} \]
Step 4: Tidy the normal.
Multiply by $-1$: $\bar{n} = \hat{i} - 4\hat{j} - 5\hat{k}$, so $A = 1$, $B = -4$, $C = -5$.
Step 5: Write the plane equation.
Using point $(2,0,5)$: \[ 1(x-2) - 4(y-0) - 5(z-5) = 0 \]
Step 6: Simplify.
\[ x - 2 - 4y - 5z + 25 = 0 \;\Rightarrow\; x - 4y - 5z + 23 = 0 \] \[ \boxed{x - 4y - 5z + 23 = 0} \]