The line l_1 passes through the point (2, 6, | vedclass

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(C) The line $l_1$ passes through $A(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. The direction vector of the plane is $\vec{n} = 2

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) The line $l_1$ passes through $A(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. The direction vector of the plane is $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$. Thus,the equation of line $l_1$ is $\frac{x - 2}{2} = \frac{y - 6}{1} = \frac{z - 2}{-2}$.The second line is $l_2: \frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$,which passes through $B(-1, -4, 0)$ with direction vector $\vec{v_2} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.The shortest distance $d$ between two skew lines is given by $d = \left| \frac{(\vec{b_2} - \vec{b_1}) \cdot (\vec{v_1} 	imes \vec{v_2})}{|\vec{v_1} 	imes \vec{v_2}|} ight|$.Here,$\vec{b_1} = 2\hat{i} + 6\hat{j} + 2\hat{k}$ and $\vec{b_2} = -\hat{i} - 4\hat{j} + 0\hat{k}$.$\vec{b_2} - \vec{b_1} = (-1 - 2)\hat{i} + (-4 - 6)\hat{j} + (0 - 2)\hat{k} = -3\hat{i} - 10\hat{j} - 2\hat{k}$.$\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 2 & -3 & 2 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(4 - (-4)) + \hat{k}(-6 - 2) = -4\hat{i} - 8\hat{j} - 8\hat{k}$.$|\vec{v_1} 	imes \vec{v_2}| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$.$d = \left| \frac{(-3\hat{i} - 10\hat{j} - 2\hat{k}) \cdot (-4\hat{i} - 8\hat{j} - 8\hat{k})}{12} ight| = \left| \frac{12 + 80 + 16}{12} ight| = \frac{108}{12} = 9$.

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$ is :

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