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(A) The line joining $(1, 2, 3)$ and $(2, 3, 4)$ has the direction vector $\vec{p} = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} +

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) The line joining $(1, 2, 3)$ and $(2, 3, 4)$ has the direction vector $\vec{p} = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} + \hat{k}$. The equation of this line is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$.The second line is $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} - \hat{j} + 0\hat{k})$.Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{p} = \hat{i} + \hat{j} + \hat{k}$,and $\vec{q} = 2\hat{i} - \hat{j}$.The cross product $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & 0 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(0 - 2) + \hat{k}(-1 - 2) = \hat{i} + 2\hat{j} - 3\hat{k}$.The magnitude $|\vec{p} 	imes \vec{q}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.The vector $\vec{b} - \vec{a} = (1-1)\hat{i} + (-1-2)\hat{j} + (2-3)\hat{k} = -3\hat{j} - \hat{k}$.The shortest distance $\alpha = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} 	imes \vec{q})}{|\vec{p} 	imes \vec{q}|} ight| = \left| \frac{(0\hat{i} - 3\hat{j} - 1\hat{k}) \cdot (1\hat{i} + 2\hat{j} - 3\hat{k})}{\sqrt{14}} ight| = \left| \frac{0 - 6 + 3}{\sqrt{14}} ight| = \frac{3}{\sqrt{14}}$.Thus,$28\alpha^2 = 28 	imes \left( \frac{3}{\sqrt{14}} ight)^2 = 28 	imes \frac{9}{14} = 2 	imes 9 = 18$.

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$,and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$,then $28 \alpha^2$ is equal to $........$.

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