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

Step 1: Find the direction of the line of shortest distance.
The two given lines can be written in vector form as \[ \begin{aligned} L_1:\;& \mathbf{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k}),\\[5pt] L_2:\;& \mathbf{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} + \hat{k}). \end{aligned} \] Hence their direction vectors are \[ \mathbf{a}_1 = \langle 2,\,0,\,1\rangle, \quad \mathbf{a}_2 = \langle 1,\,-1,\,1\rangle. \] The line of shortest distance between two skew lines is parallel to the cross product \(\mathbf{a}_1 \times \mathbf{a}_2\). Compute: \[ \mathbf{n} = \mathbf{a}_1 \times \mathbf{a}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(0\cdot1 - 1\cdot(-1)) - \hat{j}(2\cdot1 - 1\cdot1) + \hat{k}(2\cdot(-1) - 0\cdot1). \] \[ \mathbf{n} = \hat{i}(0 + 1) - \hat{j}(2 - 1) + \hat{k}(-2) = \hat{i} - \hat{j} - 2\hat{k}. \] So \(\mathbf{n} = \langle 1,\,-1,\,-2\rangle\).

 Step 2: Parametric form from the point to the plane along \(\mathbf{n}\).
We want the distance from the point \((-1,\,2,\,3)\) to the plane \(\mathbf{r}\cdot(\hat{i}-2\hat{j}+3\hat{k})=10\) in the direction \(\mathbf{n}\). Set up the line \[ \mathbf{R}(t) = \bigl\langle -1,\,2,\,3\bigr\rangle + t\,\langle 1,\,-1,\,-2\rangle. \] We must find \(t\) such that \(\mathbf{R}(t)\) lies on the plane. Let \(\mathbf{n}_{\text{plane}} = \langle 1,\,-2,\,3\rangle\) be the plane’s normal. Then \[ \mathbf{R}(t)\cdot \mathbf{n}_{\text{plane}} = 10. \] Explicitly: \[ \bigl(-1 + t,\; 2 - t,\; 3 - 2t\bigr) \cdot \langle 1,\,-2,\,3\rangle = 10. \] Compute the dot product: \[ (-1 + t)(1) + (2 - t)(-2) + (3 - 2t)(3) = (-1 + t) - 2(2 - t) + 3(3 - 2t). \] \[ = (-1 + t) - (4 - 2t) + (9 - 6t) = (-1 + t - 4 + 2t + 9 - 6t) = 4 - 3t. \] Thus \(\;4 - 3t = 10\implies -3t = 6\implies t=-2.\)

 Step 3: Distance in that direction.
Hence the intersection point with the plane is \(\mathbf{R}(-2) = \langle -1,2,3\rangle + (-2)\langle 1,-1,-2\rangle = \langle -3,4,7\rangle.\) The vector from \(\langle -1,2,3\rangle\) to \(\langle -3,4,7\rangle\) is \[ \langle -3 - (-1),\;4 - 2,\;7 - 3\rangle = \langle -2,\,2,\,4\rangle = -2\,\langle 1,\,-1,\,-2\rangle, \] which is indeed \(-2\,\mathbf{n}\). Its magnitude is \[ \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\,\sqrt{6}. \] Therefore, the required distance along the direction \(\mathbf{n}\) from the point to the plane is \(\boxed{2\sqrt{6}}.\)