Question

The distance of the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$ from the origin is ______.

• A. $\frac{7}{\sqrt{38}}$ units
• B. $\frac{1}{\sqrt{38}}$ units
• C. $\frac{5}{\sqrt{38}}$ units
• D. $\frac{2}{\sqrt{38}}$ units

Hint:

Standard distance $d = \frac{|D|}{\sqrt{A^2+B^2+C^2}}$.

Answer 1

The Correct Option is C

Step 1: Find Normal Vector
$\vec{n} = \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(5) - \hat{j}(2) + \hat{k}(-3) = 5\hat{i} - 2\hat{j} - 3\hat{k}$.
Step 2: Equation of Plane
$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
$\vec{a} = \hat{i} - \hat{j} \implies (\hat{i} - \hat{j}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = 5 + 2 = 7$.
Plane: $5x - 2y - 3z - 7 = 0$.
Step 3: Distance from Origin
$d = \frac{|-7|}{\sqrt{5^2 + (-2)^2 + (-3)^2}} = \frac{7}{\sqrt{25+4+9}} = \frac{7}{\sqrt{38}}$.
Wait, let's re-calculate $a \cdot n$: $1(5) - 1(-2) = 5 + 2 = 7$. Correct. Result is (A).
Final Answer: (A)