Let L be a line passing through the points 2 | vedclass

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(B) The line $L$ passes through $A_1(2, 3, 8)$ and $A_2(1, 6, 4)$. The direction vector of $L$ is $\vec{v} = (1-2)\hat{i} + (6-3)\hat{j} + (4-8)\hat{k

Vedclass · Accepted Answer

$-\hat{i}+12 \hat{j}-4 \hat{k}$

Vedclass · Answer

(B) The line $L$ passes through $A_1(2, 3, 8)$ and $A_2(1, 6, 4)$. The direction vector of $L$ is $\vec{v} = (1-2)\hat{i} + (6-3)\hat{j} + (4-8)\hat{k} = -\hat{i} + 3\hat{j} - 4\hat{k}$.The equation of line $L$ is $\vec{r} = (2\hat{i} + 3\hat{j} + 8\hat{k}) + t(-\hat{i} + 3\hat{j} - 4\hat{k}) = (2-t)\hat{i} + (3+3t)\hat{j} + (8-4t)\hat{k}$.The plane $P$ passes through $\vec{a} = -5\hat{i} + 19\hat{j} - 14\hat{k}$ and is parallel to $\vec{u} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}$.The normal to the plane is $\vec{n} = \vec{u} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & -2 & 3 \end{vmatrix} = \hat{i}(-3+2) - \hat{j}(3-1) + \hat{k}(-2+1) = -\hat{i} - 2\hat{j} - \hat{k}$.The equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$, which is $(x+5)(-1) + (y-19)(-2) + (z+14)(-1) = 0$, simplifying to $-x-5-2y+38-z-14 = 0$, or $x+2y+z = 19$.Substituting the coordinates of $L$ into the plane equation: $(2-t) + 2(3+3t) + (8-4t) = 19$.$2-t+6+6t+8-4t = 19 \implies t+16 = 19 \implies t = 3$.Substituting $t=3$ into the line equation: $\vec{r} = (2-3)\hat{i} + (3+3(3))\hat{j} + (8-4(3))\hat{k} = -\hat{i} + 12\hat{j} - 4\hat{k}$.

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