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(B) The plane containing the two lines must have a normal vector $\vec{n}$ perpendicular to the direction vectors of the lines $\vec{v_1} = (2, 4, 3)$

Vedclass · Accepted Answer

$3$

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(B) The plane containing the two lines must have a normal vector $\vec{n}$ perpendicular to the direction vectors of the lines $\vec{v_1} = (2, 4, 3)$ and $\vec{v_2} = (2, 6, \lambda)$.Thus,$\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 3 \ 2 & 6 & \lambda \end{vmatrix} = (4\lambda - 18)\hat{i} - (2\lambda - 6)\hat{j} + (12 - 8)\hat{k} = (4\lambda - 18, 6 - 2\lambda, 4)$.For the lines to be coplanar,the vector connecting points $(-1, 3, -1)$ and $(-3, -2, 1)$,which is $\vec{a} = (-2, -5, 2)$,must be perpendicular to $\vec{n}$.So,$\vec{a} \cdot \vec{n} = -2(4\lambda - 18) - 5(6 - 2\lambda) + 2(4) = -8\lambda + 36 - 30 + 10\lambda + 8 = 2\lambda + 14 = 0$,which gives $\lambda = -7$.Substituting $\lambda = -7$,$\vec{n} = (4(-7) - 18, 6 - 2(-7), 4) = (-46, 20, 4)$.Dividing by $-2$,we get the normal vector $(23, -10, -2)$,which is parallel to the given plane $23x - 10y - 2z + 48 = 0$.The plane containing the lines passes through $(-1, 3, -1)$,so its equation is $23(x+1) - 10(y-3) - 2(z+1) = 0 \Rightarrow 23x - 10y - 2z + 51 = 0$.The distance between $23x - 10y - 2z + 48 = 0$ and $23x - 10y - 2z + 51 = 0$ is $d = \frac{|51 - 48|}{\sqrt{23^2 + (-10)^2 + (-2)^2}} = \frac{3}{\sqrt{529 + 100 + 4}} = \frac{3}{\sqrt{633}}$.Comparing with $\frac{k}{\sqrt{633}}$,we get $k = 3$.

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to

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