If the line joining the points overline{i} + | vedclass

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Question

(A) Let the points be $A(1, 2, 0)$ and $B(0, 1, -2)$. The equation of the line passing through $A$ and $B$ is $\overline{r} = (1 - t)(\overline{i} + 2

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Let the points be $A(1, 2, 0)$ and $B(0, 1, -2)$. The equation of the line passing through $A$ and $B$ is $\overline{r} = (1 - t)(\overline{i} + 2\overline{j}) + t(\overline{j} - 2\overline{k}) = (1 - t)\overline{i} + (2 - t)\overline{j} - 2t\overline{k}$.Let the points on the plane be $P(2, -1, 0)$,$Q(0, 2, 3)$,and $R(-2, 0, 1)$. The normal vector $\overline{n}$ to the plane is given by $\vec{PQ} 	imes \vec{PR} = (-2\overline{i} + 3\overline{j} + 3\overline{k}) 	imes (-4\overline{i} + \overline{j} + \overline{k}) = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k} \ -2 & 3 & 3 \ -4 & 1 & 1 \end{vmatrix} = 0\overline{i} - 10\overline{j} + 10\overline{k} = -10\overline{j} + 10\overline{k}$.We can take the normal vector as $\overline{n} = \overline{j} - \overline{k}$.The equation of the plane is $(\overline{r} - (2\overline{i} - \overline{j})) \cdot (\overline{j} - \overline{k}) = 0$,which simplifies to $y - z = -1$.Substituting the coordinates of the line $(1-t, 2-t, -2t)$ into the plane equation: $(2-t) - (-2t) = -1 \implies 2 + t = -1 \implies t = -3$.The intersection point $\overline{r}$ is $(1 - (-3))\overline{i} + (2 - (-3))\overline{j} - 2(-3)\overline{k} = 4\overline{i} + 5\overline{j} + 6\overline{k}$.Finally,$\overline{r} \cdot (\overline{i} + \overline{j} + \overline{k}) = (4\overline{i} + 5\overline{j} + 6\overline{k}) \cdot (\overline{i} + \overline{j} + \overline{k}) = 4 + 5 + 6 = 15$.

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