If the line joining the points hat{i}+hat{j} | vedclass

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(B) The line passes through $A(1, 1, 0)$ and $B(3, 1, -1)$. The direction vector is $\vec{v} = B - A = 2\hat{i} + 0\hat{j} - 1\hat{k}$. The equation o

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$29 \hat{i}+\hat{j}-14 \hat{k}$

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(B) The line passes through $A(1, 1, 0)$ and $B(3, 1, -1)$. The direction vector is $\vec{v} = B - A = 2\hat{i} + 0\hat{j} - 1\hat{k}$. The equation of the line is $\frac{x-1}{2} = \frac{y-1}{0} = \frac{z}{-1} = r$. Thus,any point $P$ on the line is $(2r+1, 1, -r)$.The plane passes through $(2, 4, 0)$ and is parallel to $\vec{u} = 3\hat{j} + 5\hat{k}$ and $\vec{w} = 3\hat{i} - \hat{k}$. The normal to the plane is $\vec{n} = \vec{u} 	imes \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 3 & 5 \ 3 & 0 & -1 \end{vmatrix} = -3\hat{i} + 15\hat{j} - 9\hat{k}$.The equation of the plane is $-3(x-2) + 15(y-4) - 9(z-0) = 0$,which simplifies to $x - 5y + 3z + 18 = 0$.Substituting $P(2r+1, 1, -r)$ into the plane equation: $(2r+1) - 5(1) + 3(-r) + 18 = 0 \Rightarrow -r + 14 = 0 \Rightarrow r = 14$.The position vector of $P$ is $(2(14)+1)\hat{i} + 1\hat{j} - 14\hat{k} = 29\hat{i} + \hat{j} - 14\hat{k}$.

If the line joining the points $\hat{i}+\hat{j}$ and $3 \hat{i}+\hat{j}-\hat{k}$ meets the plane that passes through the point $2 \hat{i}+4 \hat{j}$ and is parallel to the vectors $3 \hat{j}+5 \hat{k}$ and $3 \hat{i}-\hat{k}$ at point $P$,then the position vector of the point $P$ is

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