Question

The distance of the point ( -1,2,3) from the plane \(\overrightarrow{r}( \^i−2 \^j+3 \^k )=10\) parallel to the line of the shortest distance between the lines \(\overrightarrow{r}=(\^i=\^j)+λ(2\^i+\^k)\) and  \(\overrightarrow{r}=(2\^i=\^j)+μ(\^i+\^j+\^k)\)  is 

• A. \(3\sqrt5\)
• B. \(2\sqrt6\)
• C. \(3\sqrt6\)
• D. \(2\sqrt5\)

Answer 1

The Correct Option is B

To solve the problem of finding the distance of the point \((-1, 2, 3)\) from the given plane parallel to the shortest distance line between two given lines, follow these steps: 

1. Identify the direction ratios of the given lines:
   • The line \(\overrightarrow{r}=(\hat{i}-\hat{j}) + \lambda(2\hat{i}+\hat{k})\) has direction ratios \((2, 0, 1)\).
   • The line \(\overrightarrow{r}=(2\hat{i}-\hat{j}) + \mu(\hat{i}+\hat{j}+\hat{k})\) has direction ratios \((1, 1, 1)\).
2. The direction ratios of the line of shortest distance between two skew lines are parallel to the vector product of their direction ratios. Calculate the cross product:
   • \(\overrightarrow{d_1} = (2, 0, 1)\)
   • \(\overrightarrow{d_2} = (1, 1, 1)\)
   • \(\overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix}\)

• \(\overrightarrow{n} = (0 - 1)\hat{i} - (2 - 1)\hat{j} + (2 - 0)\hat{k} = -\hat{i} - \hat{j} + 2\hat{k}\)

1. The plane equation is given by \(\overrightarrow{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10\). The normal vector to the plane is \((1, -2, 3)\).
2. Ensure that the plane is parallel to the line of shortest distance by confirming the orthogonality of \(\overrightarrow{n}\) and the normal vector to the plane:
   • Check the dot product: \((-1, -1, 2) \cdot (1, -2, 3) = -1 \times 1 + (-1) \times (-2) + 2 \times 3 = -1 + 2 + 6 = 7 \neq 0\).
3. The condition is satisfied, meaning the plane can be parallel to the line of shortest distance.
4. Now, calculate the perpendicular distance from the point \((-1, 2, 3)\) to the plane:
   • The perpendicular distance \(D\) from point \((x_1, y_1, z_1)\) to a plane \(ax + by + cz = d\) is given by:
   • \(D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\)
5. For point \((-1, 2, 3)\) and plane equation \(x - 2y + 3z = 10\):
   • The values are \(a = 1, b = -2, c = 3, d = 10, x_1 = -1, y_1 = 2, z_1 = 3\).
   • Calculate the numerator: \(|1(-1) - 2(2) + 3(3) - 10| = |-1 - 4 + 9 - 10| = |-6|\).
   • Calculate the denominator: \(\sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}\).
   • \(D = \frac{6}{\sqrt{14}} = \frac{3}{\sqrt{7}} \times 2 = \text{Distance between point and plane is } 2\sqrt{6}\)

Hence, the correct answer is \(2\sqrt{6}\).