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 ______.
$\frac{5}{\sqrt{38}}$ units
$\frac{7}{\sqrt{38}}$ units
To find the distance of the given plane from the origin, we first need to understand the equation provided in the vector form:
\(\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})\)
This equation is in parametric form, which represents a line in space rather than a plane. To convert this into the normal form of a plane equation, we recognize that the vectors \((\hat{i} + \hat{j} + \hat{k})\)and \((\hat{i} - 2\hat{j} + 3\hat{k})\)are direction vectors. Hence, the normal vector \(\vec{n}\)of the plane can be obtained by taking the cross product of these two vectors.
Compute the cross product \(\vec{a} = (\hat{i} + \hat{j} + \hat{k})\)and \(\vec{b} = (\hat{i} - 2\hat{j} + 3\hat{k})\):
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(1 \cdot 3 - 1 \cdot (-2)) - \hat{j}(1 \cdot 3 - 1 \cdot 1) + \hat{k}(1 \cdot (-2) - 1 \cdot 1)\)
\(= \hat{i}(3 + 2) - \hat{j}(3 - 1) + \hat{k}(-2 - 1)\)
\(= 5\hat{i} - 2\hat{j} - 3\hat{k}\)
The normal vector of the plane, thus, is \(5\hat{i} - 2\hat{j} - 3\hat{k}\).
To find the distance of the plane from the origin, use the point \((1, -1, 0)\)which lies on the plane (given in vector form as constant vector) in the point-normal form of the plane equation:
\(\vec{n} \cdot \vec{r_0} = d\), where \(\vec{r_0}\)is a point on the plane.
\(5 \cdot 1 + (-2) \cdot (-1) + (-3) \cdot 0 = d\)
\(5 + 2 + 0 = 7\)
Thus, the equation of the plane in normal form is \(5x - 2y - 3z = 7\).
The perpendicular distance from the origin \((0, 0, 0)\)to the plane \(Ax + By + Cz + D = 0\)is given by:
\(\frac{|A \cdot 0 + B \cdot 0 + C \cdot 0 + D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|7|}{\sqrt{5^2 + (-2)^2 + (-3)^2}}\)
\(= \frac{7}{\sqrt{25 + 4 + 9}} = \frac{7}{\sqrt{38}}\)
Therefore, the distance of the plane from the origin is \(\frac{7}{\sqrt{38}}\)units.