Question ID: #1003
The line $L_{1}$ is parallel to the vector $\vec{a}=-3\hat{i}+2\hat{j}+4\hat{k}$ and passes through the point $(7, 6, 2)$ and the line $L_{2}$ is parallel to the vector $\vec{b}=2\hat{i}+\hat{j}+3\hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_{1}$ and $L_{2}$ is:
- (1) $\frac{23}{\sqrt{38}}$
- (2) $\frac{21}{\sqrt{57}}$
- (3) $\frac{23}{\sqrt{57}}$
- (4) $\frac{21}{\sqrt{38}}$
Solution:
$$d = \left| \frac{(\vec{a_2} – \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$$
$$L_{1}: \vec{r} = (7\hat{i}+6\hat{j}+2\hat{k}) + \lambda(-3\hat{i}+2\hat{j}+4\hat{k})$$
$$L_{2}: \vec{r} = (5\hat{i}+3\hat{j}+4\hat{k}) + \mu(2\hat{i}+\hat{j}+3\hat{k})$$
$$\vec{a_1} = 7\hat{i}+6\hat{j}+2\hat{k}, \quad \vec{b_1} = -3\hat{i}+2\hat{j}+4\hat{k}$$
$$\vec{a_2} = 5\hat{i}+3\hat{j}+4\hat{k}, \quad \vec{b_2} = 2\hat{i}+\hat{j}+3\hat{k}$$
$$\vec{a_1} – \vec{a_2} = (7-5)\hat{i} + (6-3)\hat{j} + (2-4)\hat{k}$$
$$\vec{a_1} – \vec{a_2} = 2\hat{i} + 3\hat{j} – 2\hat{k}$$
$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix}$$
$$\vec{b_1} \times \vec{b_2} = \hat{i}(6 – 4) – \hat{j}(-9 – 8) + \hat{k}(-3 – 4)$$
$$\vec{b_1} \times \vec{b_2} = 2\hat{i} + 17\hat{j} – 7\hat{k}$$
$$|\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (17)^2 + (-7)^2}$$
$$|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 289 + 49} = \sqrt{342}$$
$$(\vec{a_1} – \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2}) = (2\hat{i} + 3\hat{j} – 2\hat{k}) \cdot (2\hat{i} + 17\hat{j} – 7\hat{k})$$
$$(\vec{a_1} – \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2}) = (2)(2) + (3)(17) + (-2)(-7)$$
$$(\vec{a_1} – \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2}) = 4 + 51 + 14 = 69$$
$$d = \left| \frac{69}{\sqrt{342}} \right|$$
$$d = \frac{69}{\sqrt{9 \times 38}}$$
$$d = \frac{69}{3\sqrt{38}}$$
$$d = \frac{23}{\sqrt{38}}$$
Ans. (1)
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